# Lectures in Engineering Mathematics by slappypappy115

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```									Engineering Mathematics / The Johns Hopkins University

Lectures in Engineering
Mathematics
George Nakos

Engineering Mathematics
The Johns Hopkins University

Fall 2004

1
Engineering Mathematics / The Johns Hopkins University

Part 1: Linear Algebra

1. Matrices: Addition and Scalar Multiplication

2. Matrix Multiplication

3. Linear Systems and Gauss Elimination

4. The Rank of a Matrix; Linear Independence

2
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6.1 Matrices: Deﬁnitions

Basic Deﬁnitions

A matrix is a rectangular arrangement of numbers called en-
tries. A matrix has rows that are numbered top to bottom and
columns that are numbered left to right. The (i, j ) entry is the
entry at the ith row and jth column.

A matrix has size m × n (pronounced ‘m by n’), if it has m rows
and n columns. If m = n, then the matrix is called square. In
this case, n is the size of the square matrix.

Lecture 1 / @ Copyright: George Nakos                      3
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6.1 Matrices: Examples

The following are matrices of respective sizes 4 × 2, 2 × 3, 3 × 3,
5 × 1, and 1 × 2.
         
                                                               7.1
1 −2                                               
a11 a12 a13      3.2 
      
   −3  5          7 √ −1
21                                     
 −1.5  ,
,                        ,  a21 a22 a23  ,                       a b
                                    
0  6          9  5 4
                                                              
a31 a23 a33      4.9 

2 −8                                                     
6.9
The (3, 2) entry of the ﬁrst matrix is 6. The third matrix is square
of size 3.

Lecture 1 / @ Copyright: George Nakos                                      4
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6.1 Matrices: General Form

A general matrix A of size m × n with (i, j ) entry aij is denoted
by
                                   
a11 a12 a13         · · · a1n
       a21 a22 a23         · · · a2n   
A=
                                       
        .
.
.   .
.
.   .
.
.          ...    .
.
.


am1 am2 am3         · · · amn
This is abbreviated by
A = aij
where i and j are indices such that 1 ≤ i ≤ m and 1 ≤ j ≤ n.

Lecture 1 / @ Copyright: George Nakos                              5
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6.1 Vectors
If n = 1, then A is called a column matrix, or a m-vector, or
a vector. If m = 1, then A is called a row matrix, or a n-row
vector, or a row vector. The entries of vectors are usually
called components.

The following are vectors. The ﬁrst is a 2-vector, the second is
a 4-vector, and the third is a n-vector.
                       
4               v1
7                     −3             v2   
,                     ,
                       
−3

    2   

    .
.
.


−1               vn
Here are some row vectors.
√
[ −3] ,  1.2   3 ,                    a b c ,              −2 3 0 4 1

Lecture 1 / @ Copyright: George Nakos                                 6
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6.1 Zero Matrices; Equal Matrices

A zero matrix, denoted by 0, is a matrix with zero entries. Here
are some examples.
                                    
0 0                              0
0 0
0 = [ 0] ,   0=               ,    0 =  0 0 ,          0=   0 0 ,   0= 0 
0 0
                                  
0 0                              0

We say that two matrices A and B are equal and we write A = B,
if A and B have the same size and their corresponding entries
are equal. So, if
1 2                c d
A=               ,    B=
a b                3 4
then A = B, only if a = 3, b = 4, c = 1, and d = 2.
Lecture 1 / @ Copyright: George Nakos                                  7
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We can add two matrices of the same size by adding the cor-
responding entries. The resulting matrix is the sum of the two
matrices.

Example We have
1 −3 0              0 4  5              1 1 5
+                     =
2 −4 7             −1 4 −2              1 0 5

In general, if A = aij and B = bij , for 1 ≤ i ≤ m and 1 ≤ j ≤ n,
then
A + B = aij + bij

Lecture 1 / @ Copyright: George Nakos                            8
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6.1 Scalar Multiplication
We also multiply any real number c, times a matrix A, by multi-
plying all entries of A by c.

Example We have
                            
1  0        2  0
2  −3  4  =  −6  8 
                 
5 −1       10 −2

In general, if A = aij , then

cA = caij
This operation is called scalar multiplication. The multiplier c
is often called a scalar, because it scales A.
Lecture 1 / @ Copyright: George Nakos                        9
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6.1 Matrices: Opposite, Diﬀerence
The matrix (−1) A is called the opposite of A and it is denoted
by −A. For example,
0 4  5            0 −4 −5
−                 =
−1 4 −2            1 −4  2

The matrix A + (−1) B is denoted by A − B and it is called the
diﬀerence between A and B. This is the subtraction operation.
A − B = A + (−1) B

Example We have
1 −2       1 −1        0 −1
                         
 7  4   6    3     1   1 
 5 −5  −  7  0  =  −2 −5 
8  0      −3  7       11 −7

Lecture 1 / @ Copyright: George Nakos                    10
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6.1 Properties of Operations
Theorem
1. (A + B) + C = A + (B + C)                             (Associative Law)

2. A + B = B + A                                        (Commutative Law)

3. A + 0 = 0 + A = A

4. A + (−A) = (−A) + A = 0

5. c(A + B) = cA + cB                                    (Distributive Law)

6. (a + b)C = aC + bC                                    (Distributive Law)

7. (ab)C = a(bC) = b(aC)

8. 1A = A

9. 0A = 0

Lecture 1 / @ Copyright: George Nakos                                11
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6.1 Matrix Transpose

Let A be any m × n matrix. The transpose of A, denoted by AT ,
is the n × m matrix obtained from A by switching all columns of
A to rows and maintaining the same order.

Example We have
       
     T                                                     a               T
1 2

1
1 3 5                           T         b   
3  =    1
 3 4  =                     ,     a b c d           =         ,                     −8
                                                                  
2 4 6                                      c            −8        3
5 6
         
d

Lecture 1 / @ Copyright: George Nakos                                          12
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6.1 Properties of Transposition

Theorem

Let A and B be m × n matrices and let c be any scalar. Then

1. (A + B )T = AT + B T

2. (cA)T = cAT

T
3.   AT       =A

Lecture 1 / @ Copyright: George Nakos                    13
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6.1 Matrices: Symmetric, skew-Symmetric
A matrix A such that AT = A is called symmetric. Examples
               
                            a b c   d
0 −1 3
5 −7                                       b e   f   g   
,     −1  4 9 ,
                                       
−7  6                                       c f   h   i
               
3  9 6

d g i   j
Note the mirror symmetry of a symmetric matrix with respect
to the upper-left to lower-right diagonal line.

A matrix A such that AT = −A is called skew-symmetric. Ex-
amples
                       
                             0 −b −c −d
0 −1  3
0 7                                         b 0 −f −g 
,     1   0 −9  ,
                                       
−7 0                                          c f  0 −i 
              
−3  9  0

d g   i 0
Lecture 1 / @ Copyright: George Nakos                                            14
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6.1 Special Square Matrices

Let A be a square matrix of size n. The entries aii, 1 ≤ i ≤ n
form the main diagonal. A is called upper triangular, if all
entries below the main diagonal are zero, i.e., if aij = 0 for j < i.
Matrix A is called lower triangular, if the entries above the main
diagonal are all zero, so aij = 0 for i < j. If the main diagonal is
also zero, we talk about strictly upper triangular and strictly
lower triangular matrices.

A, D, E are upper triangular. B, C, D, E are lower triangular. C is
strictly lower triangular.
                             
a 0 0          0 0 0
a b                                               1  0          7 0
A=            , B =  b c 0 , C =  1 0 0 , D =               , E=
0 c                                               0 −2          0 7
d e f          1 1 0

Lecture 1 / @ Copyright: George Nakos                             15
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6.1 More Special Square Matrices

If the nondiagonal entries of a square matrix are zero, then the
matrix is called diagonal. If all entries of a diagonal matrix are
equal, then we have a scalar matrix. Matrices D and E are
diagonal. Matrix E is a scalar matrix.

A scalar matrix of size n with common diagonal entry 1 is called
an identity matrix and it is denoted by In, or by I.
                 
                         1   0   ··· 0
1 0 0
1 0                                                 0   1   ··· 0 
I = I2 =              ,        I3 =  0 1 0  , . . . ,   In = 
                                          
0 1                                                  .
.
.   .
.
.       . 
... . 
.
0 0 1

0   0   ··· 1

Lecture 1 / @ Copyright: George Nakos                                      16
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6.2 Matrix Multiplication
Let A be a m×k matrix and B be a k×n matrix. The productAB
is the m × n matrix C = [cij ] = AB, with entries cij are given by

                                 
a11 a12 · · · a1k                                                          

          .
.
.   .
.
.     .
.
.    .
.
.

             b11 · · ·       b1j   · · · b1n 
                                                  b21 · · ·       b2j   · · · b2n 
A=         ai1 ai2 · · · aik               B=
                                                                                  
               .
.
.    .
.
.          .
.
.      .
.
.    . 
. 
. 
.    .     .     .
                                          
         .
.    .
.     .
.     .
.               
                                                   bk1 · · ·       bkj   · · · bkn
am1 am2 · · · amk

k
cij = ai1 b1j + ai2 b2j + · · · + aik bkj =                  air brj
r=1

Lecture 1 / @ Copyright: George Nakos                                             17
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6.2 Matrix Multiplication

Examples

                 
3 2  4
2 0 1        −2 4                   6 7 6
5 =
2 1 2                                4 14 9
0 3 −2

    
1
 −2 
4 −1 −2 1          3 =5
5

4 −1  −2
                                          
1                                       1
 −2                          −8  2   4 −2 
4 −1 −2 1          =
 3                            12 −3  −6  3 
5                            20 −5 −10  5

Lecture 1 / @ Copyright: George Nakos                             18
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6.2 Properties of Matrix Multiplication
Theorem

1. (AB)C = A(BC)                                              (Associative law)

2. A(B + C) = AB + AC                                    (Left Distributive law)

3. (B + C)A = BA + CA                                   (Right Distributive law)

4. a(BC) = (aB)C = B(aC)

5. Im A = AIn = A                                       (Multiplicative identity)

6. 0A = 0 and A0 = 0

7. (AB)T = B T AT

Lecture 1 / @ Copyright: George Nakos                                       19
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6.2 Caution with Matrix Multiplication
AB may not equal BA. In fact, if AB is deﬁned, then BA may
not be deﬁned. If BA is deﬁned, then it may not have the same
size as AB. If it does have the same size, it may still not equal
AB.

1. We say that matrix multiplication is noncommutative.

2. If two matrices A and B satisfy AB = BA, then we say that
they commute.

0 0                    1 0
Example A =                    and B =                   commute.
1 1                    2 3
Lecture 1 / @ Copyright: George Nakos                               20
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6.2 Powers of Square Matrix
Let A be a square matrix. The product AA is also denoted by
A2. Likewise, AAA = A3 and AA · · · A = An for n factors of A. In
addition, we write A1 = A and if A is nonzero, we write A0 = I.
An = AA · · · A ,         A1 = A,         A0 = I
n factors

Example
1 −1                    3 −4                   11 −15
A1 =                , A2 =                , A3 =                  , ···
−2  3                   −8 11                  −30  41
1 2                 1 2                  1 2
B1 =            ,   B2 =            ,    B3 =            , ···
0 0                 0 0                  0 0
0 1                 0 0                  0 0
C1 =            ,   C2 =            ,   C3 =             , ···
0 0                 0 0                  0 0

Lecture 1 / @ Copyright: George Nakos                                            21
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6.2 Motivation for Matrix Multiplication
Let x = (x1, x2) , y = (y1, y2) , and z = (z1, z2) be coordinate
frames. Suppose we go from frame y to frame z by using the
linear transformation
z1 = a11y1 + a12y2
z2 = a21y1 + a22y2
and from frame x to frame y by the linear transformation
y1 = b11x1 + b12x2
y2 = b21x1 + b22x2
If we want to go from frame x to frame z, we substitute
z1 = a11 (b11x1 + b12x2) + a12 (b21x1 + b22x2)
z2 = a21 (b11x1 + b12x2) + a22 (b21x1 + b22x2)

Lecture 1 / @ Copyright: George Nakos                        22
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6.2 Motivation for Matrix Multiplication

and rearrange to get

z1 = (a11b11 + a12b21) x1 + (a11b12 + a12b22) x2
z2 = (a21b11 + a22b21) x1 + (a21b12 + a22b22) x2

Now if A and B are coeﬃcient matrices of the ﬁrst two trans-
formations and C is the coeﬃcient matrix of the last one, then
we see that
C = AB

Lecture 1 / @ Copyright: George Nakos                         23
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6.2 Applications of Matrix Multiplication
Example

Each of three appliances outlets receive and sell daily from three
factories TVs and VCRs according to the following table.
TV      VCR
Factory 1       40       50
Factory 2       70       80
Factory 3       60       65
Each outlet charges the following dollar amounts per appliance.
Outlet 1      Outlet 2 Outlet 3
TV         215           258      319
VCR         305           282      264

Lecture 1 / @ Copyright: George Nakos                         24
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6.2 Applications of Matrix Multiplication

Example (cont.) If A and B are the matrices of these tables,
compute and interpret the product AB.

                                                                  
40 50                                     23 850 24 420 25 960
215 258 319
AB =  70 80                                  =  39 450 40 620 43 450 
                                                                
305 282 264
60 65                                     32 725 33 810 36 300
The (1, 1) entry 40 · 215 + 50 · 305 = 23, 850 is the ﬁrst outlet’s
revenue from selling all the appliances coming from the ﬁrst
factory. The remaining entries are interpreted similarly.

Lecture 1 / @ Copyright: George Nakos                                  25
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6.3 Linear Systems and Gauss Elimination
Deﬁnition A linear system of m equations in n unknowns
x1, . . . , xn, is a set of m linear equations
a11x1 + a12x2 + · · · + a1nxn = b1
a21x1 + a22x2 + · · · + a2nxn = b2
.
.            (1)
.
am1x1 + am2x2 + · · · + amnxn = bm

The unknowns are also called variables, or indeterminants.
The numbers aij are the coeﬃcients and the numbers bi are
the constant terms. If all constant terms are zero, then the
system is called homogeneous. The homogeneous system that
has the same coeﬃcients as system (1) is said to be associated
with (1). If m = n, then the system is called square.
Lecture 1 / @ Copyright: George Nakos                    26
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6.3 Example of Linear System

Example The system
x1 + 2x2       = −3
2x1 + 3x2 − 2x3 = −10                          (2)
−x1       + 6x3 =   9
is linear square with coeﬃcients 1, 2, 0, 2, 3, −2, −1, 0, 6, constant
terms −3, −10, 9, and associated homogeneous system
x1 + 2x2       = 0
2x1 + 3x2 − 2x3 = 0
−x1       + 6x3 = 0

Lecture 1 / @ Copyright: George Nakos                           27
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6.3 More Examples of Linear Systems

The following three systems are linear.                  The ﬁrst is from an
ancient Chinese text.∗

3x + 2y + z = 39                   x1 + x2 = 5                y1 + y2 + y3 = −2
2x + 3y + z = 34                  x1 − 2x2 = 6               y1 − 2y2 + 7y3 = 6
x + 2y + 3z = 26                 −3x1 + x2 = 1

∗A third century BC book titled Nine Chapters of Mathematical Art. See
Carl Boyer’s A History of Mathematics (New York: Wiley).

Lecture 1 / @ Copyright: George Nakos                                   28
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6.3 Augmented and Coeﬃcient Matrix
The matrix that consists of the coeﬃcients and constant terms,
is called the augmented matrix of the system. The augmented
matrix of system (2) is
                           
1 2  0 −3
2 3 −2 −10 
               

−1 0  6   9
The matrix with entries the coeﬃcients is the coeﬃcient matrix
of the system. The vector of all constant terms is the vector of
constants. The coeﬃcient matrix and the vector of constants
of system (2) are
                                       
1 2 0                                −3
 2 3 −2               and           −10 
                                        
−1 0 6                                 9

Lecture 1 / @ Copyright: George Nakos                              29
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6.3       Solution of Linear System

A sequence r1, r2, . . . , rn of scalars is a (particular) solution of
system (1), if all the equations are satisﬁed when we substitute
x1 = r1, . . . , xn = rn. The set of all possible solutions is the
solution set. Any generic element of the solution set is called
the general solution.

If a system has solutions, it is called consistent, otherwise it is
called inconsistent.

Two linear systems with the same solution sets are called equiv-
alent. A solution that consists of zeros only is called a trivial
solution.
Lecture 1 / @ Copyright: George Nakos                           30
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6.3       Elementary Row Operations
A basic solution method of a linear system is to eliminate unknowns so that
an equivalent “triangular” system is obtained. This is done by performing
elementary equation operations: (a) adding to an equation a multiple of
another, (b) multiplying an equation by a nonzero scalar, (c) switching two
equations.

For economy these operations are performed on the augmented matrix.

The elementary row operations of any matrix are:

Elimination: add a constant multiple of one row to another Ri + cRj → Ri

Scaling: multiply a row by a nonzero constant cRi → Ri

Interchange: interchange two rows Ri ↔ Rj

Lecture 1 / @ Copyright: George Nakos                                31
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6.3 Example of Gauss Elimination
Example We solve the system by elimination.

x1 + 2x2        = −3
2x1 + 3x2 − 2x3 = −10
−x1       + 6x3   =9

                                                                         
1 2  0 −3                                                1   2   0 −3
R2 − 2R1 → R2
2 3 −2 −10                                           0 −1 −2 −4 
                                                                   
R3 + R1 → R3

−1 0   6    9                                              0 6 2      6
                                                                     
1   2  0 −3                                          1  2  0 −3
 0 −1 −2 −4                R3 + 2R2 → R3            0 −1 −2 −4
                                                                     

0   0  2 −2                                          0  0  2 −2

Lecture 1 / @ Copyright: George Nakos                                      32
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6.3 Example of Gauss Elimination
The system is in triangular form. Start at the bottom and work upwards to
eliminate unknowns above the leading variables (ﬁrst variables with nonzero
coeﬃcients) of each equation (back-substitution).

                                                                   
1   2     0 −3                                   1       2 0 −3
 0 −1 −2 −4              R2 + R3 → R2           0 −1 0 −6             R1 + 2R2 →
                                                          
0  0   2 −2                                        0      0 2 −2
                                                                  
1  0 0 −15                                               1 0 0 −15
(−1) R2 → R2
 0 −1 0   −6                                            0 1 0   6 
                                                                  
(1/2)R3 → R3
0  0 2 −2                                                0 0 1 −1

x1 = −15, x2 = 6, x3 = −1
Lecture 1 / @ Copyright: George Nakos                                         33
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6.3       Inﬁnitely Many Solutions

Example Find the intersection of the three planes.
x + 2y − z = 4
2x + 5y + 2z = 9
x + 4y + 7z = 6

By
Solution:  elimination the augmented matrix of the system

1 0 −9 2
reduces to  0 1  4 1  . We get x−9z = 2, y +4z = 1. Hence,
          
0 0  0 0

x = 9r + 2
y = −4r + 1         r∈R
z=r

Lecture 1 / @ Copyright: George Nakos                    34
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6.3       No Solutions

Example [No Solutions] Find the intersection of the three planes
in the (p, q, k)-coordinate system.
q − 2k = −5
2p − q + k = −2
4p − q      = −4

Solution: The augmented matrix of the system reduces to
                
2 −1     1 −2
 0    1 −2 −5  . The last row corresponds to the false ex-
                
0    0   0   5
pression 0 = 5. Hence, the system is inconsistent. Therefore,
the planes do not have a common intersection.

Lecture 1 / @ Copyright: George Nakos                     35
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6.3 Matrix Form of Linear System

System (1) can be written as equality of two vectors. By using
the matrix-vector product we have
                                                
a11 a12 · · · a1n            x1             b1
   a21 a22 · · · a2n          x2           b2   
=
                                                

    .
.
.   .
.
.  ...    .
.
.

    .
.
.            .
.
.


am1 am2       amn            xn             bm
This is also abbreviated as

Ax = b                                 (3)
where A is the coeﬃcient matrix, x is the vector of the unknowns,
and b is the vector of constants.

Lecture 1 / @ Copyright: George Nakos                                     36
Engineering Mathematics / The Johns Hopkins University

6.3 Matrix Form of Linear System

Example Write the linear system in matrix-vector product form.

7x1 + 4x2 + 5x3 =            1
2x1 − 3x2 + 9x3 = −8

Solution: We have
      
x1
7  4 5                          1
 x2  =
    
2 −3 9                         −8
x3

Lecture 1 / @ Copyright: George Nakos                         37
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6.3 Echelon Forms
A zero row of a matrix is a row that consists entirely of zeros. The ﬁrst
nonzero entry of a nonzero row is called a leading entry. If a leading entry
happens to be 1, we call it a leading 1. Similarly, we can talk about zero
columns.

Deﬁnitions Consider the following conditions on a matrix A.

1. All zero rows are at the bottom of the matrix.

2. The leading entry of each nonzero row after the ﬁrst occurs to the right
of the leading entry of the previous row.

3. The leading entry in any nonzero row is 1.

4. All entries in the column above and below a leading 1 are zero.

If A satisﬁes the ﬁrst two conditions, we call it row echelon form. If it
satisﬁes all four conditions, we call it reduced row echelon form. We often
omit the word “row” and just say echelon form, or reduced echelon form.
Lecture 1 / @ Copyright: George Nakos                                 38
Engineering Mathematics / The Johns Hopkins University

6.3 Echelon Forms
Example
Consider the matrices.
                                                              
1 0 0 0                  1 0 0 −6                          1 0 1
A= 0 0 1 0          ,     B= 0 1 0  0 ,                 C =  0 0 1 ,
0 0 0 0                  0 0 1 −1                          0 0 1
                                                                    
1 1 0 0        2                                      1 7 0    9 0
0 0
D= 0 0 1 0            3 ,    E=             ,       F   =  0 0 1 −8 0  ,
1 0
0 0 0 1        4                                      0 0 0    0 1
                                                 
1     0 −1 0                1            0 0    0
G= 0        1  0 0 ,         H= 0            0 1    0 
0     0  1 0                0            0 0 −2
Matrices A, B, D, F, G, H are in echelon form. Out of these, A, B, D, F are in
reduced echelon form. Matrices G and H are not in reduced echelon form.
For G condition 4 fails. For H condition 3 fails. Matrices C and E are not in
echelon form. For C condition 2 fails. For E condition 1 fails.
Lecture 1 / @ Copyright: George Nakos                                         39
Engineering Mathematics / The Johns Hopkins University

6.3 Gauss Elimination
Algorithm [Gauss Elimination] To reduce any matrix to reduced row echelon
form apply the following steps.

1. Find the leftmost nonzero column.

2. If the ﬁrst row has a zero in the column of step 1, interchange it with
one that has a nonzero entry in the same column.

3. Obtain zeros below the leading entry by adding suitable multiples of the
top row to the rows below that.

4. Cover the top row and repeat the same process starting with step 1
applied to the leftover submatrix. Repeat this process with the rest of
the rows, until the matrix is in echelon form.

5. Starting with the last nonzero row work upward: For each row obtain a
leading 1 and introduce zeros above it, by adding suitable multiples to
the corresponding rows.

Lecture 1 / @ Copyright: George Nakos                                40
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6.3 Gauss Elimination

Example Apply Gauss elimination to ﬁnd a reduced echelon form
of the matrix.
                         
0    3 −6 −4 −3
 −1     3 −10 −4 −4 
                         
 4 −9       34    0   1 
                         

2 −6     20    8   8

Solution:
                       

−1  3 −10 −4 −4 
    0  3 −6 −4 −3 
R1 ↔ R2                            

    4 −9  34  0  1 

2 −6  20  8  8

Lecture 1 / @ Copyright: George Nakos                        41
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6.3 Gauss Elimination

The pivot now is −1, at pivot position (1, 1) .
                        
−1   3 −10 −4 −4
R3 + 4R1 → R3              
    0   3 −6 −4 −3      

R4 + 2R1 → R4                   0   3 −6 −16 −15
                        
                        
0   0   0  0   0

                                                                         
−1 3 −10 −4 −4                                       −1   3 −10 −4 −4
0 3 −6 −4 −3                                          0 3 −6 −4 −3
                                                                         
                                                                         
                                      Step 1                             

   0 3 −6 −16 −15               
       − −→
−− −       
      0 3 −6 −16 −15    

0 0  0   0   0                                        0 0   0  0   0

Lecture 1 / @ Copyright: George Nakos                                42
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6.3 Gauss Elimination

The next pivot is 3, at position (2, 2) .
−1   3 −10 −4 −4                                             −1   3 −10    −4 −4
                                                                                     

0   3  −6 −4 −3                                            0   3 −6     −4 −3    
 R3 − R2 → R3 
                                                                                     
0   3 −6 −16 −15                                             0   0   0   -12 −12
                                                                                       
                                                                                     
0   0   0  0   0                                             0   0   0     0   0
STEP 5: Starting with the last nonzero row work upward: For
each row obtain a leading 1 and introduce zeros above it, by
adding suitable multiples to the corresponding rows.
                              
−1   3 −10 −4 −4
           0   3 −6 −4 −3                   R2 + 4R3 → R2
(−1/12)R3 → R3 
                           
0   0   0  1  1                  R1 + 4R3 → R1


0   0   0  0  0

Lecture 1 / @ Copyright: George Nakos                                         43
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6.3 Gauss Elimination

                         
−1 3 −10 0 0
                         
−1   3 −10 0 0
3 −6 0 1                                                    1 
                           
    0                                                 0 1    −2 0 3  R − 3R →
(1/3)R2 → R2 
                         

0   0   0 1 1
                                             1     2
0 0     0 1 1 
                                         

0   0   0 0 0                                       0 0        0 0 0
                                                                         
−1 0 −4 0 −1                                         1 0    4 0 1
1                           0 1 −2 0 1 
                                                                

     0 1 −2 0            3      (−1)R1 → R1                   3 
0 0  0 1            1                           0 0  0 1 1 
                                                                

0 0  0 0            0                               0 0    0 0 0

Lecture 1 / @ Copyright: George Nakos                                           44
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6.4 Linear Combinations of Vectors

Deﬁnition Let v1, v2, . . . , vk be given n-vectors and let c1, c2, . . . , ck
be any scalars. The n-vector v

v = c1v1 + c2v2 + · · · + ck vk
is called a linear combination of v1, . . . , vk . The scalars c1, . . . , ck
are called the coeﬃcients of the linear combination. If not all ci
are zero, we have a nontrivial linear combination. If all ci are
zero, we have the trivial linear combination. The trivial linear
combination represents the zero vector.

The concept of linear combination is simple: we scale a few
vectors and then we add them.
Lecture 1 / @ Copyright: George Nakos                                  45
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6.4 Linear Combinations of Vectors
Example Check that the following are linear combinations of the
vectors v1, v2, and v3.
−v1 + 3v2 + 4v3,             v1 + 1.5v2 − 9v3,   v1 − v3

Solution: We have
−v1 + 3v2 + 4v3 = (−1) v1 + 3v2 + 4v3
v1 + 1.5v2 − 9v3 = 1v1 + (1.5) v2 + (−9) v3
v1 − v3 = 1v1 + 0v2 + (−1) v3

Note that the diﬀerence v1 − v2 between two vectors v1 and v2
is the linear combination 1v1 + (−1) v2 with coeﬃcients 1 and
−1.
Lecture 1 / @ Copyright: George Nakos                                 46
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6.4 Application of Linear Combinations
A sports company owns two factories each making aluminum and titanium
mountain bikes. The ﬁrst factory makes 150 aluminum and 15 titanium bikes
a day. For the second factory the numbers are 220 and 20, respectively. If
150              220
v1 =         and v2 =         , compute and discuss the meaning of:
15               20

(a) v1 + v2

(b) v2 − v1

(c) 10v1

(d) av1 + bv2, for a, b > 0.

Lecture 1 / @ Copyright: George Nakos                               47
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6.4 Application of Linear Combinations
Solution:
370
(a) v1 + v2 =            represents the total number of aluminum
35
(370) and titanium (35) bikes produced by the two factories in
one day.
70
(b) v2 − v1 =         represents how many more bikes the second
5
factory makes over the ﬁrst one in one day.
1500
(c) 10v1 =             represents how many bikes the ﬁrst factory
150
makes in 10 days.
150a + 220b
(d) av1 + bv2 =                     represents the total number of
15a + 20b
bikes produced if the ﬁrst factory operates for a days and the
second for b days.
Lecture 1 / @ Copyright: George Nakos                       48
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6.4 Linear Dependence
Deﬁnition The sequence of m-vectors v1 , . . . , vk is linearly dependent (or
the vectors are linearly dependent), if there are scalars c1 , . . . , ck not all zero
such that
c1 v1 + · · · + ck vk = 0                            (4)
So, there is a nontrivial linear combination of the vis representing the zero
vector. Equation (4) with not all ci zero is called a linear dependence
relation of the vis.
Example The vectors
                
1      1        4
 −1   2   14 
 3  ,  0  ,  −6 
4      2        4
are linearly dependent, because if we let c1 = 2, c2 = −6, and c3 = 1, then
                                  
1             1            4        0
 −1           2        14   0 
2       + (−6)  0  + 1  −6  =  0 
3
4             2            4        0

Lecture 1 / @ Copyright: George Nakos                                          49
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6.4 Linear Dependence
                 

   0       1        3 
Example Let S =  −2  ,  2  ,  14  .
                 
                       
   3       7        9 

(a) Show that S is linearly dependent.

(b) Find a linear dependence relation.

Solution:      (a) We seek c1, c2, c3 not all zero such that
                                       
0          1          3        0
c1  −2  + c2  2  + c3  14  =  0 
                             
3          7          9        0

Lecture 1 / @ Copyright: George Nakos                                50
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6.4 Linear Dependence
Equivalently, we seek nontrivial solutions to the homogeneous linear system
                         
0 1     3      c1       0
 −2 2 14   c2  =  0 
3 7     9      c3       0
We solve this system to get c1 = 4r, c2 = −3r, c3 = r. There are nontrivial
solutions, hence the set is linearly dependent.

(b) To get a particular   linear dependence relation we assign a nonzero value
to the parameter r. For   example, if r = 1, then we have
                                    
0               1           3       0
4  −2      + (−3)  2  + 1  14  =  0 
3               7           9       0
This is one of inﬁnitely many linear dependence relations.

Lecture 1 / @ Copyright: George Nakos                                  51
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6.4 Linear Independence

Deﬁnition The set of m-vectors {v1, . . . , vk } is called linearly
independent, if it is not linearly dependent. This is the same
as saying that there is no linear dependence relation among
v1, . . . , vk . So, all nontrivial linear combinations of the vis yield
nonzero vectors. Equivalently, we have

if   c1v1 + · · · + ck vk = 0,      then c1 = 0, . . . , ck = 0
In other words, the homogeneous system [v1 · · · vk ] c = 0 has
only the trivial solution. We often say that v1, . . . , vk are linearly
independent.

Lecture 1 / @ Copyright: George Nakos                                      52
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6.4 Linear Independence
1         5
Example Show that                       ,           is linearly independent in
−2         3
R2 .

Solution: Let c1 and c2 be scalars such that c1e1 + c2e2 = 0. In
other words,
1                5        0
c1          + c2           =
−2                3        0

1 5 0
We solve the system with augmented matrix                  to get
−2 3 0
c1 = 0 and c2 = 0. Therefore, the set is linearly independent.

Lecture 1 / @ Copyright: George Nakos                                   53
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6.4 Linear Independence
Example Show that S is linearly independent.
           
 2
        8     −4 

 3   −6   3 
                
S =  ,     ,
           
 2   5   1 


                 


4         0 
−6

Solution: We only need to count the number of pivots of the
coeﬃcient matrix.
                                   
2  8 −4                2  8 −4
   3 −6  3              0 −3   5   
∼
                                   
2  5  1                0  0 −21
                                     
                                   
4  0 −6                0  0   0
This number is 3, the same as the number of columns, so the
set is linearly independent.
Lecture 1 / @ Copyright: George Nakos                          54
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6.4 Rank of Matrix
The rank of a m × n matrix A is the maximum number of linearly
independent rows of A.

1. The rank equals the maximum number of linearly indepen-
dent columns of A.

2. The rank is the same as the number of the pivots of A.

3. A and AT have the same rank.

To compute the rank of A it we reduce A to echelon form
and count the number of nonzero rows or the number of pivot
columns.
Lecture 1 / @ Copyright: George Nakos                    55
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6.4 Example of Rank
                    
1 2     2 −1
 1 3  1 −2 
           
           
The rank of A = 
 1 1  3  0  is 2.

 0 1 −1 −1 
           
1 2     2 −1

because, A reduces to the row echelon form matrix

                      
1 2       2 −1
 0 1 −1 −1 
           
           
B= 0 0
      0  0 

 0 0  0  0 
           
0 0       0   0

Lecture 1 / @ Copyright: George Nakos                        56
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Lecture 2 in Engineering
Mathematics
George Nakos

Engineering Mathematics
The Johns Hopkins University

Fall 2004

Lecture 1 / @ Copyright: George Nakos                    57
Engineering Mathematics / The Johns Hopkins University

Part 1: Linear Algebra

1. The Rank of a Matrix; Linear Independence

2. Vector Space, Basis, Dimension

3. Determinants; Cramer’s Rule

4. Matrix Inversion

5. Inner Product Spaces

6. Linear Transformations

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6.4 Vector Space: Deﬁnition
Deﬁnition Let V be a set equipped with two operations named
addition and scalar multiplication. Addition is a map that
associates any two elements u and v of V with a third one,
called the sum of u and v and denoted by u + v.
V × V → V,                (u, v) → u + v
Scalar multiplication is a map that associates any real scalar c
and any element u of V with another element of V, called the
scalar multiple of u by c and denoted by cu.
R × V → V,                (c, u) → cu
Such a set V is called a (real) vector space, if the two operations
satisfy the following properties, known as axioms for a vector
space.
Lecture 2 / @ Copyright: George Nakos                          59
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6.4 Vector Space: Deﬁnition

(A1) u + v belongs to V for all u, v ∈ V.

(A2) u + v = v + u for all u, v ∈ V.                      (Commutative Law)

(A3) (u + v) + w = u + (v + w) for all u, v, w ∈ V.        (Associative Law)

(A4) There exists a unique element 0 ∈ V, called the zero of V, such that for
all u in V
u+0=0+u=u

(A5) For each u ∈ V there exists a unique element −u ∈ V, called the negative
or opposite of u, such that
u + (−u) = (−u) + u = 0

Lecture 2 / @ Copyright: George Nakos                                60
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6.4 Vector Space: Examples
Scalar Multiplication

(M1) c u belongs to V for all u ∈ V and all c ∈ R.

(M2) c(u + v) = cu + cv for all u, v ∈ V and all c ∈ R.   (Distributive Law)

(M3) (c + d)u = cu + du for all u ∈ V and all c, d ∈ R.   (Distributive Law)

(M4) c(du) = (cd)u for all u ∈ V and all c, d ∈ R.

(M5) 1u = u for all u ∈ V.

The elements of a vector space are called vectors. Axioms (A1) and (M1)
are also expressed by saying that V is closed under addition and is closed
under scalar multiplication. Note that a vector space is a nonempty set,
because it has a zero by (A4).

Lecture 2 / @ Copyright: George Nakos                                61
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6.4 Vector Space: Examples

1. The set Rn of all n-vectors with real components.
Operations:    The usual vector addition and scalar multiplication. Zero:
The zero n-vector 0. Axioms:         For the axioms see the Properties of
Matrix Operations Theorem.

2. The set Mmn of all m × n matrices with real entries.
Operations:    The usual matrix addition and scalar multiplication. Zero:
The m × n zero matrix 0. Axioms: For the axioms see the Properties of
Matrix Operations Theorem.

3. The set P of all polynomials with real coeﬃcients.

Lecture 2 / @ Copyright: George Nakos                                   62
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6.4 Vector Space: Examples
Operations:

1. Addition: The sum of two polynomials is formed by adding the coeﬃ-
cients of the same powers of x of the polynomials. Explicitly, if
p1 = a0 + a1 x + · · · + anxn,    p2 = b0 + b1 x + · · · + bm xm ,    n≥m
we write p2 as p2 = b0 + b1 x + · · · + bnxn, by adding zeros if necessary, and
form the sum
p1 + p2 = (a0 + b0 ) + (a1 + b1 ) x + · · · + (an + bn) xn

2. Scalar multiplication: This is multiplication of a polynomial through by
a constant.
cp1 = (ca0 ) + (ca1 ) x + · · · + (can) xn

3. Zero: The zero polynomial, 0, is the polynomial with zeros as coeﬃ-
cients.

4. Axioms: The veriﬁcation of the axioms is left as exercise.
Lecture 2 / @ Copyright: George Nakos                                          63
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6.4 Vector Space: Examples
The set F (R) of all real valued functions deﬁned on R.
Operations: Let f and g be two real valued functions with domain R and let
c be any scalar.

1. Addition: We deﬁne the sum f + g of f and g as the function whose
values are given by
(f + g)(x) = f (x) + g(x)         for all x ∈ R

2. Scalar multiplication: The scalar product cf is deﬁned by
(c f )(x) = c f (x)   for all x ∈ R

3. Zero: The zero function 0 is the function whose values are all zero.
0(x) = 0     for all x ∈ R

4. Axioms: The veriﬁcation of the axioms is left as exercise.
Lecture 2 / @ Copyright: George Nakos                                    64
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6.4 Vector Space: Examples
Is R2 with the usual addition and the following scalar multiplica-
tion, denoted by , a vector space?
a1         ca1
c               =
a2          a2

Solution: It is not a vector space, because
a1            (c + d) a1               ca1 + da1
(c + d)                 =                        =
a2                a2                       a2
and
a1              a1               ca1        da1           ca1 + da1
c          +d                =               +              =
a2              a2                a2        a2               2a2
a1                  a1              a1
So, (c + d)              =c                  +d                 and axiom (M3) fails.
a2                  a2              a2
Lecture 2 / @ Copyright: George Nakos                                             65
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6.4 Subspaces
Deﬁnition A subset W of a vector space V is called a subspace of
V, if W itself is a vector space under the same addition and scalar
multiplication as V . In particular, a subspace always contains the
zero element.

Theorem Let W be a nonempty subset W of a vector space
V. Then W is a subspace of V if and only if it is closed under
addition (axiom (A1)) and scalar multiplication (axiom (M1)),
that is, if and only if

1. If u and v are in W, then u + v is in W.

2. If c is any scalar and u is in W, then c u is in W .

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6.4 Examples of Subspaces

1. The set W = {cv, c ∈ R} of all scalar multiples of the ﬁxed
vector v of a vector space V is a subspace of V.

2. {0} and V are subspaces of V. These are the trivial sub-
spaces of V. {0} is called the zero subspace.

3. The set Pn that consists of all polynomials of degree ≤ n and
the zero polynomial is a subspace of P.

4. The set C(R) of all continuous real valued functions deﬁned
on R is a subspace of F (R).

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6.4 Linear Combinations and Span
If v1, . . . , vn are vectors from a vector space V and c1, . . . , cn are
scalars, then the expression
c1 v1 + · · · + cn vn
is well deﬁned and is called a linear combination of v1, . . . , vn.
If not all ci are zero, we have a nontrivial linear combination.
If all ci are zero, we have the trivial linear combination. The
trivial linear combination represents the zero vector.

The set of all linear combinations of v1, . . . , vk is called the span
of these vectors and it is denoted by
Span {v1, v2, . . . , vk }
If V = Span{v1, . . . , vk }, we say that v1, . . . , vk span V and that
{v1, . . . , vk } is a spanning set of V.
Lecture 2 / @ Copyright: George Nakos                               68
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6.4 Linear Combinations; Span
Example Let V be a vector space and let v1 , v2 be in V. The following vectors
are in Span{v1 , v2 }.
0,     v1,    v2 ,   v1 + v2 ,     −2v1 ,   3v1 − 2v2
Example Let V be a vector space and v be in V. Span{v} is the set of all
scalar multiples of v.
Span{v} = {cv , c ∈ R}
Example Let p = −1 + x − 2x2 in P3 . Show that p ∈ Span {p1 , p2 , p3 } , where
p1 = x − x2 + x3 ,     p2 = 1 + x + 2x3 ,    p3 = 1 + x
Solution: Let c1 , c2 , c3 be scalars such that
−1 + x − 2x2 = c1 x − x2 + x3 + c2 1 + x + 2x3 + c3 (1 + x)
Then
−1 + x − 2x2 = (c2 + c3 ) + (c1 + c2 + c3 ) x − c1 x2 + (c1 + 2c2 ) x3
Equating coeﬃcients of the same powers of x yields the linear system
c2 + c3 = −1, c1 + c2 + c3 = 1, −c1 = −2, c1 + 2c2 = 0
with solution c1 = 2, c2 = −1, c3 = 0. Therefore, p is in the span of p1 , p2 , p3 .
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6.4 Linear Independence
Deﬁnition Let v1 , . . . , vn be vectors of a vector space V. Then {v1 , . . . , vn} is
linearly dependent, if there are scalars c1 , . . . , cn not all zero such that
c1 v1 + · · · + cn vn = 0                             (5)
So, there are nontrivial linear combinations that represent the zero vector.
Equation (5) with not all ci zero is called a linear dependence relation of
the vis.
Example Show that the set {2 − x + x2 , 2x + x2 , 4 − 4x + x2 } is linearly
dependent in P3 .
Solution: This true, because
2(2 − x + x2 ) + (−1) (2x + x2 ) + (−1) 4 − 4x + x2 = 0

Deﬁnition The set of vectors {v1 , . . . , vn} from a vector space V is called lin-
early independent, if it is not linearly dependent. This is the same as saying
that there is no linear dependence relation among v1 , . . . , vk . Equivalently,
c1 v1 + · · · + ck vk = 0 ⇒ c1 = 0, . . . , ck = 0
So, every nontrivial linear combination is nonzero.
Lecture 2 / @ Copyright: George Nakos                                           70
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6.4 Linear Independence Example                                Show that set {E11 , E12 , E21 , E22 }
is linearly independent in M22 .
Solution: Let
1 0            0 1            0 0             0 0             0 0
c1             + c2           + c3           + c4              =
0 0            0 0            1 0             0 1             0 0
c1 c2         0 0
⇒             =
c3 c4         0 0
Hence, c1 = c2 = c3 = c4 = 0. So, the set is linearly independent.
Example Are 1 + x, −1 + x, 4 − x2 , 2 + x3 linearly independent in P3 ?
Solution: If a linear combination in these polynomials is the zero polynomial,
then
c1 (1 + x) + c2 (−1 + x) + c3 4 − x2 + c4 2 + x3 = 0 ⇒
(c1 − c2 + 4c3 + 2c4 ) + (c1 + c2 ) x + (−c3 ) x2 + c4 x3 = 0
Equating coeﬃcients yields,
c1 − c2 + 4c3 + 2c4 = 0,     c1 + c2 = 0,      −c3 = 0,       c4 = 0
We solve this linear system to get c1 = c2 = c3 = c4 = 0. So, the vectors are
linearly independent in P3 .
Lecture 2 / @ Copyright: George Nakos                                               71
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6.4 Basis
Deﬁnition A subset {v1, . . . , vn} of a nonzero vector space V is
a basis of V, if

1. it is linearly independent, and

2. it spans V.

The empty set is, by deﬁnition, the only basis of the zero vector
space {0}.

Here are some examples of bases.
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6.4 Examples of Bases

1. The standard basis vectors e1, e2, . . ., en in Rn form a basis of
Rn .

2. {1, x, x2, . . . , xn} is a basis of Pn, called the standard basis
of Pn.

3. {E11, E12, E13, . . . , Emn} is a basis of Mmn, called the stan-
dard basis of Mmn.

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6.4 Examples of Bases
Example Show that B = {1 + x, −1 + x, x2} is a basis of P2.

(a) To show that B spans P2, we want every polynomial p =
a + bx + cx2 to be a linear combination in B. So, we look for
scalars c1, c2, c3 such that
c1(1 + x) + c2(−1 + x) + c3x2 = a + bx + cx2 ⇒
(c1 − c2) + (c1 + c2)x + c3x2 = a + bx + cx2
which leads to the system c1 − c2 = a, c1 + c2 = b, c3 = c. We
have
1a + 1b
     
1 0 0 2
                    
1 −1 0 a                2   
 1  1 0 b ∼ 0 1 0 − 1a + 1b 
           
2    2 
0   0 1 c                0 0 1    c
Lecture 2 / @ Copyright: George Nakos                         74
so the system is consistent for all choices of a, b, c. Thus, B
spans P2.

(b) To show that B is linearly independent, let c1, c2, c3 be such
that
c1(1 + x) + c2(−1 + x) + c3x2 = 0 ⇒
(c1 − c2) + (c1 + c2)x + c3x2 = 0
Hence, we have the system    c1 − c2 = 0, c1 + c2 = 0, c3 = 0.
Now
                              
1 −1 0 0          1 0 0 0
 1   1 0 0     ∼ 0 1 0 0 
                         
0   0 1 0         0 0 1 0
So the system has only the trivial solution c1 = c2 = c3 = 0.
Thus, B is linearly independent.
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6.4 Characterization of Basis

One of the main characterizations of a basis is described in the
following theorem.

Theorem A subset B = {v1, . . . , vn} of a vector space V is a
basis of V if and only if for each vector v in V there are unique
scalars c1, . . . , cn such that

v = c1v1 + · · · + cnvn

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6.4 Dimension

Deﬁnitions If a vector space V has a basis with n elements,
then V is called ﬁnite dimensional and we say that n is the
dimension of V. We write

dim(V ) = n

the dimension is a well deﬁned number and does not depend
on the choice of basis. The dimension of the zero space {0} is
deﬁned to be zero. A vector space that has no ﬁnite spanning
set it is called inﬁnite dimensional.

By counting the number of elements of the standard bases we
see that dim(Rn) = n, dim(Pn) = n + 1, dim(Mmn) = m · n.
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6.6 Determinants
a11 a12
Let A =                      . The determinant, det (A) , of A is the
a21 a22
number
det(A) = a11a22 − a12a21

Let A be
                      
a11 a12 a13
A =  a21 a22 a23 
             
a31 a32 a33
The determinant of A in terms of 2 × 2 determinants is the
number
a22 a23                a21 a23         a21 a22
det(A) = a11                − a12                  + a13
a32 a33                a31 a33         a31 a32

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6.6 Determinants

In the same manner we can deﬁne determinants of 4×4 matrices.
a11    a12   a13   a14
a22 a23 a24       a21 a23 a24
a21    a22   a23   a24
= a11 a32 a33 a34 − a12 a31 a33 a34 +
a31    a32   a33   a34
a42 a43 a44       a41 a43 a44
a41    a42   a43   a44
a21 a22 a24       a21 a22 a23
+ a13 a31 a32 a34 − a14 a31 a32 a33
a41 a42 a44       a41 a42 a43

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6.6 Determinants

Example Find det (C ) , if
                      
1   2   0  1
       −1   1   2  0      
C=
                          
−2   1   0 −2

                          
1   0   2 −1

Solution: det (C ) equals
1 2  0     −1 2  0     −1 1  0     −1 1 2
1 1 0 −2 − 2 −2 0 −2 + 0 −2 1 −2 − 1 −2 1 0
0 2 −1      1 2 −1      1 0 −1      1 0 2
= 1 · 6 − 2 · (−12) + 0 · (−3) − 1 · 0 = 30

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6.6 Cofactor Expansion
We have introduced what is known as the cofactor expansion of a deter-
minant about its ﬁrst row. Each entry of the ﬁrst row is multiplied by the
corresponding minor and each such product is multiplied by ±1 depending on
the position of the entry. The signed products were added together. Actually,
instead of the ﬁrst row can use any row or column. Here is how: Let
                    
a11 a12 · · · a1n
 a21 a22 · · · a2n 
A= .
 .      .
.  ...   . 
. 
.    .        .
an1 an2 · · · ann

First we assign the sign (−1)i+j to the entry aij of A. This is a checkerboard
pattern of ±’s.
                 
+ − + ···
 − + − ··· 
                 
 + − + ··· 
.
.
.   .
.
.   . ...
.
.
Then we pick a row or column and multiply each entry aij of it by the corre-
sponding signed minor (−1)i+j Mij . Lastly, we add all these products.
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6.6 Determinants and Row Reduction
The signed minor (−1)i+j Mij is called the (i, j) cofactor, of A and is denoted
by Cij .
Cij = (−1)i+j Mij

1. Cofactor Expansion about the ith row The determinant of A can be
expanded about the ith row in terms of the cofactors as follows.
det A = ai1 Ci1 + ai2 Ci2 + · · · + ainCin

2. Cofactor Expansion about the jth column The determinant of A can
be expanded about the jth column in terms of the cofactors as follows.
det A = a1j C1j + a2j C2j + · · · + anj Cnj

This method of computing determinants by using cofactors is called the co-
factor, or Laplace expansion and it is attributed to Vandermonde and to
Laplace.

Lecture 2 / @ Copyright: George Nakos                                   81
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6.6 Properties of Determinants
1. A and its transpose have the same determinant, det(A) = det(AT ). For example,
a1   a2   a3            a1    b1    c1
b1   b2   b3    =       a2    b2    c2
c1   c2   c3            a3    b3    c3

2. Let B be obtained from A by multiplying one of its rows (or columns) by a nonzero
constant. Then det(B) = k det(A). For example,
a1     a2    a3              a1   a2   a3             a1    a2    ka3            a1    a2    a3
kb1    kb2   kb3     =k      b1   b2   b3        ,    b1    b2    kb3      =k    b1    b2    b3
c1     c2    c3              c1   c2   c3             c1    c2    kc3            c1    c2    c3

3. Let B be obtained from A by interchanging any two rows (or columns). Then det(B) =
− det(A). For example,
a1   a2    a3              b1   b2   b3            a1    a2    a3             a3    a2    a1
b1   b2    b3    =−        a1   a2   a3    ,       b1    b2    b3     =−      b3    b2    b1
c1   c2    c3              c1   c2   c3            c1    c2    c3             c3    c2    c1

4. Let B be obtained from A by adding a multiple of one row (or column) to another.
Then det(B) = det(A). For example,
a1          a2            a3                a1   a2   a3
ka1 + b1    ka2 + b2      ka3 + b3       =     b1   b2   b3
c1          c2            c3                c1   c2   c3

Lecture 2 / @ Copyright: George Nakos                                                                        82
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6.6 Properties of Determinants
Note that

1. Elimination Ri + cRj → Ri, does not change the determinant.

2. Scaling, cRi → Ri, scales the determinant by c.

3. Interchange, Ri ↔ Rj , changes the sign of the determinant.

The properties of determinants can be used to compute a de-
terminant as follows. We convert it to triangular form by Gauss
elimination and then multiply the diagonal entries of the trian-
gular form.
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6.6 Determinants by Row Reduction
1  2 3 −1  8                  1  2 3 −1   8
0  0 4  2 −1                  0  0 4  2  −1
0 −5 5  3  7            =     0 −5 5  3   7          −R1 + R5 → R5
0  0 0  1  6                  0  0 0  1   6
1  2 3 −2 −9                  0  0 0 −1 −17
1  2 3 −1   8
0 −5 5  3   7
= −        0  0 4  2  −1       R2 ↔ R3
0  0 0  1   6
0  0 0 −1 −17
1  2 3 −1   8
0 −5 5  3   7
= −        0  0 4  2  −1     R4 + R5 → R5
0  0 0  1   6
0  0 0  0 −11
= 1 · (−5) · 4 · 1 · (−11)
= −220

Lecture 2 / @ Copyright: George Nakos                                 84
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6.6 Cramer’s Rule
Let Ax = b be a square      system with
                                                 
a11       . . . a1n                  x1          b1
A= .   .
.        ...    . , x = 
.
.                    . ,
.
.      b= . 
.
.
an1       . . . ann                  xn          bn
Let Ai denote the matrix obtained from A by replacing the ith column with
b.
                                      
a11 · · · a1,i−1 b1 a1,i+1 · · · a1n
Ai =  ..
.    .
.
.     .
.
.     .
.
.    .
.
.     .
.
.    . 
.
.
an1 · · · an,i−1 bn an,i+1 · · · ann

Cramer’s Rule gives an explicit formula for the solution of a consistent square
system.
Cramer’s Rule If det(A) = 0, then the system Ax = b has a unique solution
x = (x1, . . . , xn) given by
det(A1 )            det(A2 )                     det(An)
x1 =            ,   x2 =            , .     .. , xn =
det(A)              det(A)                       det(A)

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6.6 Cramer’s Rule
Example Use Cramer’s Rule to solve the system.
x 1 + x 2 − x3 = 2
x 1 − x 2 + x3 = 3
−x1 + x2 + x3 = 4

Solution: We compute the determinant of the coeﬃcient matrix
A and the determinants of
                                                          
2  1 −1             1 2 −1             1 1 2
A1 =  3 −1  1  , A2 =  1 3   1  , A3 =  1 −1 3 
                                          
4  1  1            −1 4  1            −1 1 4
to get det(A) = −4, det(A1) = −10, det(A2) = −12, det(A3) =
−14. Hence,
det(A1)  5       det(A2)           det(A3)   7
x1 =         = , x2 =         = 3, x3 =         =
det(A)   2       det(A)            det(A)    2
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6.7 Matrix Inverse

Deﬁnition An n×n matrix A is invertible, if there exists a matrix
B such that
AB = I        and      BA = I
In such case B is called an inverse of A. If no such B exists
for A, then we say that A is noninvertible. Another name for
invertible is nonsingular and another name for noninvertible is
singular.

Note that the deﬁnition forces B to be square of size n (why?).

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6.7 Matrix Inverse

Theorem An invertible matrix has only one inverse.

Proof: Suppose that the invertible matrix A has two inverses B
and C. Then

B = BIn = B(AC) = (BA)C = InC = C
Therefore, B = C.

The unique inverse of an invertible matrix A is denoted by A−1.
So
AA−1 = I          and       A−1A = I

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6.7 Matrix Inverse
Next we see how to compute the inverse of an invertible matrix A. The idea
is simple: If A−1 has unknown columns xi, then AA−1 = I takes the form
[Ax1 · · · Axn] = [e1 · · · en]
This matrix equation splits into n linear systems
Ax1 = e1 , . . . , Axn = en
which we solve to ﬁnd each column xi of A−1 . These systems have the same
coeﬃcient matrix A. Solving each system separately would amount into
n − 1 unnecessary row reductions of A. It is smarter to solve the systems
simultaneously, by simply row reducing the matrix
[A : I]
If we get a matrix of the form [I : B] , then the ith column of B would be xi.
Thus, B = A−1 . So, in order to compute A−1 , we just row reduce [A : I] .

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6.7 Matrix Inverse
             
1 0 −1
Example Compute A−1 , if A =  3 4 −2  .
3 5 −2

Solution: We row reduce [A : I].

1   0    −1   1    0   0            1       0       −1     1    0    0        1   0   −1        1     0    0
3   4    −2   0    1   0    ∼       0       4        1    −3    1    0    ∼   0   4    1       −3     1    0    ∼
3   5    −2   0    0   1            0       5        1    −3    0    1        0   0   −1        3
−5    1
4        4     4
1    0    −1    1   0    0               1       0    0    −2     5       −4       1   0    0    −2     5       −4
0    4     1   −3   1    0       ∼       0       4    0     0    −4        4   ∼   0   1    0     0    −1        1
0    0     1   −3   5   −4               0       0    1    −3     5       −4       0   0    1    −3     5       −4
Therefore,
−2     5   −4
−1
A           =         0    −1    1
−3     5   −4

Lecture 2 / @ Copyright: George Nakos                                                                          90
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6.7 Matrix Inverse
Theorem Let A and B be invertible n × n matrices and let c be a nonzero
scalar. Then

1. AB is invertible and
(AB)−1 = B −1 A−1

2. A−1 is invertible and
(A−1 )−1 = A

3. cA is invertible and
1 −1
(cA)−1 =      A
c

4. AT is invertible and
T
(AT )−1 = A−1

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6.7 Cancellation Laws
Recall that AB = AC does not imply that B = C. However, if A is invertible,
then the implication is true.

Theorem Let A, B, and C be n × n matrices and A is invertible. Then the
cancellation laws hold:

AB = AC ⇒ B = C,          BA = CA ⇒ B = C

Proof: Let AB = AC. Since A−1 exists, we can multiply on the left by A−1
to get

A−1 (AB) = A−1 (AC) ⇒ (A−1 A)B = (A−1 A)C ⇒ IB = IC ⇒ B = C

The second implication is proved similarly.

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6.7 Determinants and Inversion
Cauchy’s Theorem The determinant of a product of two n×n
matrices is the product of the determinants of the factors.
det(AB) = det(A) det(B)

Cauchy’s Theorem has the following implication.

Theorem

A square matrix is invertible if and only if its determinant is
nonzero.

Furthermore, If A is invertible, then
1
det(A−1) =
det(A)
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6.7 Invertibility and Linear Systems

Theorem      Let A be an invertible matrix, so det(A) = 0. Then

1. Ax = b has a unique solution given by
x = A−1b

2. Ax = 0 has only the trivial solution.

Theorem      Let A be a n × n matrix. Then the following are equivalent.

1. det(A) = 0

2. Ax = 0 has nontrivial solutions

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Deﬁnition Let A be an n × n matrix. The matrix whose (i, j)
entry is the cofactor Cij of A is the matrix of cofactors of A.
Its transpose is the adjoint of A and it is denoted by Adj(A).
                       
C11 C21 · · · Cn1
       C12 C22 · · · Cn2   
                           
        .
.
.   .
.
.  ...    .
.
.


C1n C2n · · · Cnn

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Example Find the adjoint of A, where
                  
−1 2  2
A=         4 3 −2 
                
−5 0  3

Solution: The cofactors of A are
C11 = 9,    C12 = −2,       C13 = 15
C21 = −6,    C22 = 7,       C23 = 10
C31 = −10,       C32 = 6,    C33 = −11
Hence,
                          
C11 C21 C31        9  −6 −10
Adj(A) = [Cij ]T =  C12 C22 C32  =  −2   7   6 
C13 C23 C33       15 −10 −11

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6.7 Adjoint and Inverse
Theorem

1. Let A be an n × n matrix. Then
A Adj(A) = det(A)In = Adj(A) A

2. Let A be an invertible matrix. Then
−1 =     1
det(A)

Example For the above A, we have det(A) = 17. Hence,
        9          6     10   

−6 −10                     − 17   − 17
1  9
17
−1     1                                          2          7      6
A =          Adj(A) =      −2     7    6  =  − 17
                           
det(A)          17                                     17     17   
15 −10 −11             15
− 10     11
− 17
17     17

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6.8 The Dot Product
The dot product u · v of two n-vectors u = (u1, ..., un) and
v = (v1, ..., vn) is the matrix-vector product
u · v = uT v
The matrix in this case is the row vector obtained by transposing
u. In terms of components the dot product is the number
     
v1
u · v = [ u1 · · · u n ]  .  = u1 v 1 + · · · + u n v n
 . 
.                                    (6)
vn
If the dot product of two vectors is zero, we call these vectors
orthogonal.

Note that in equation (6) for convenience we identiﬁed a 1 × 1
matrix [a] with its single entry a.
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6.8 The Dot Product
Example Let u = (−3, 2, 1) , v = (4, −1, 5) , and w = (−2, 1, −8) .

(a) Find u · v.

(b) Are u and w are orthogonal?

Solution:

(a) We have
      
4
u·v =   −3 2 1        −1  = (−3) 4 + 2 (−1) + (1) (5) = −9
5

(b) Vectors u and w are orthogonal, because
u · w = (−3, 2, 1) · (−2, 1, −8) = 0

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6.8 The Length of n-Vectors
Deﬁnition The norm, or length, or magnitude of an n-vector u = (u1 , . . . , un)
is
√         2
1
2 2
u = u · u = u1 + · · · + u n
The (Euclidean) distance between two n-vectors u and v is
u−v
A n-vector is a unit vector, if its norm is 1.
1               1
Example Let v = (1, 2, −3, 1) and u = 2 , − 1 , 1 , − 2 . (a) Find the length of
2 2
v. (b) Find the distance between v and u. (c) Is u a unit vector?
Solution: We have

2
1       √
(a)   v =    12   +   22   + (−3) +    12      2
=       15

1 5
√
(b)   v−u =            , , −7, 3
2 2   2 2
=       21

1
(c)   u =     2
1
, −2, 1, −1
2   2
= 1. So, u is a unit vector.

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6.8 Dot Product and Angle
The dot product for plane and space vectors is related to the
length and angle between the vectors by the following formula

u·v = u           v cos θ                   (7)
This can be seen by using the law of cosines on the triangle OP Q
with OP = u and OQ = v.
1
u    v cos θ =        u 2 + v 2 − PQ 2
2                                    
1 3 2      3
2−
3
=       ui +     vi      ( v i − u i )2 
2 i=1      i=1     i=1
3
=         ui v i = u · v
i=1

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6.8 Main Properties of Dot Product
Let u, v, w be n-vectors and c be a scalar. Then

1. u · v = v · u                                                            (Symmetry)

2. u · (v + w) = u · v + u · w                                              (Additivity)

3. c (u · v) = (cu) · v = u · (cv)                                       (Homogeneity)

4. u · u ≥ 0. Also, u · u = 0 if and only if u = 0.               (Positive Deﬁniteness)

5. (Pythagorean Theorem) u and v are orthogonal if and only if
2          2         2
u+v         = u        + v

6. (Cauchy-Bunyakovsky-Schwarz Inequality)
|u · v | ≤ u       v                            (8)

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6.8 Inner Product
Deﬁnition An inner product on a (real) vector space V is a function that
to each pair of vectors u and v of V associates a real number, denoted by
u, v .
,   : V × V → R,   (u, v) → u, v
This function satisﬁes the following properties, or axioms.

For any vectors u, v, w of V and any scalar c, we have

1.   u, v = v, u                                               (Symmetry)

2.   u + w , v = u, v + w , v                                   (Additivity)

3.   cu, v = c u, v                                          (Homogeneity)

4.   u, u ≥ 0. Furthermore, u, u = 0 if and only if u = 0.      (Positivity)

A real vector space with an inner product is called an inner product space.

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6.8 Properties of Inner Product
Theorem Let u, v, and w be any vectors in an inner product
space and let c be any scalar. Then

1. u, v + w = u, v + u, w

2. u, cv = c u, v

3. u − w, v = u, v − w, v

4. u, v − w = u, v − u, w

5. 0, v = v, 0 = 0

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6.8 Examples of Inner Product Spaces
1. The dot product of Rn is an inner product.

2. (Weighted Dot Product) Let w1 , . . . , wn be any positive numbers and let
u = (u1, . . . , un) and v = (v1, . . . , vn) be any n-vectors. The following
deﬁnes an inner product in Rn.
u, v = w 1 u 1 v 1 + · · · + w n u n v n        (9)

3. Let A and B be 2 × 2 matrices with real entries.
a1 a2                         b1 b2
A=                  ,         B=
a3 a4                         b3 b4
The following function deﬁnes an inner product in M22 .
A, B = a1 b1 + a2 b2 + a3 b3 + a4 b4

4. Let f (x) and g(x) be in C[a, b], the vector space of the continuous real-
valued functions deﬁned on [a, b]. Then the following deﬁnes an inner
product on C[a, b].
b
f, g =              f (x)g(x) dx
a

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6.8 Length and Orthogonality
Let V be an inner product space. Two vectors u and v are called orthogonal
if their inner product is zero.
u and v are orthogonal if u, v = 0
The norm (or length, or magnitude) of v is the nonnegative number v ,
deﬁned by
v =      v, v                             (10)
We also deﬁne the distance, d(u, v), between two vectors u and v by
d(u, v) = u − v                         (11)

Note that
d(0, v) = d(v, 0) = v
A vector with norm 1 is called a unit vector. The set S of all unit vectors of
V is called the unit circle or the unit sphere.
S = {v , v ∈ V and v = 1}                     (12)

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6.8 Properties of Norm

The norm in an inner product space V satisﬁes the following
basic properties.

For all vectors u and v of V and all scalars c, we have

1.   cu = |c| u

2.   u+v ≤ u + v                       (the Triangle Inequality)

3.   u ≥ 0 and u = 0 if and only if u = 0

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6.8 Examples of Length and Orthogonality
Example (a) Are the functions sin (x) and sin (2x) orthogonal in C [−π, π]
π
under the integral inner product f, g = −π f (x) g (x) dx?
(b) What is the norm of sin (2x) with respect to this inner product?

(a) We have

π
1 π
sin (x) , sin (2x)       =          sin (x) sin (2x) dx =      (cos x − cos 3x) dx
−π                       2 −π
π
1           1
=          sin x − sin 3x        =0
2           3          −π
so the functions are orthogonal.
(b) The norm is
π                  1/2           π                        1/2
2                     1                                      √
sin (2x) =               sin (2x) dx          =            (1 − cos (4x)) dx         =       π
−π                               2   −π

1                                           1
Note: (a) sin (a) sin (b) =       (cos (a − b) − cos (a + b)) (b) sin2 (a) = (1 − cos 2a)
2                                           2
Lecture 2 / @ Copyright: George Nakos                                                          108
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6.8 Matrix Transformations
A matrix transformation T : Rn → Rm, is a transformation for which there
is an m × n matrix A such that
T (x) = Ax
for all x in Rn.
3   −7     8
Example Consider the matrix transformation T : R3 → R2 with A =     2    1    −4
                           
x1                       x
3 −7     8  1 
T  x2  =                   x2
2   1 −4
x3                       x3
So
    
x1
3x1 − 7x2 + 8x3
T  x2  =
2x1 + x2 − 4x3
x3
For example, the image of the vector (−2, 3, 5) under this transformation is
                              
−2                        −2
3 −7      8                 13
T 3 =                        3 =
2    1 −4                  −21
5                         5
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6.8 Linear Transformations
Deﬁnition A linear transformation or linear map from a vector
space V to a vector space W is a transformation T : V → W such
that for all vectors u and v of V and any scalar c, we have

1. T (u + v) = T (u) + T (v)

2. T (cu) = cT (u)

The addition in u + v is addition in V, whereas the addition in
T (u) + T (v) is addition in W. Likewise, scalar multiplications cu
and cT (u) occur in V and W, respectively. In the special case
where V = W, the linear transformation T : V → V is called a
linear operator of V.
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6.8 Examples of Linear Transformations
• Matrix transformations. Because if A is the matrix of the transformation,
then
T (x1 + x2 ) = A (x1 + x2 ) = Ax1 + Ax2 = T (x1 ) + T (x2 )
and
T (c1 x1 ) = A (c1 x1 ) = c1 Ax1 = c1 T (x1 )

• The special matrix transformations with matrices
−1 0            1  0           −1  0             cos θ − sin θ
,               ,                ,
0 1            0 −1            0 −1             sin θ  cos θ

are linear and represent reﬂection about the y-axis and the x-axis, reﬂec-
tion about the origin and rotation by θ radians about the origin.

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6.8 Examples of Linear Transformations
a    b
•   T : M22 → P3 , T    c    d        = d + cx + (b − a) x3 is linear.

a1   b1       a2       b2                a1 + a2     b1 + b2
T    c1   d1   +   c2       d2        =T      c1 + c2     d1 + d2

= (d1 + d2 ) + (c1 + c2 ) x + {(b1 + b2 ) − (a1 + a2 )} x3

=    d1 + c1 x + (b1 − a1 ) x3     + d2 + c2 x + (b2 − a2 ) x3

a1   b1               a2   b2
=T      c1   d1     +T        c2   d2

and
a1   b1                ca1   cb1
T    c        c1   d1      =T        cc1   cd1

= cd1 + cc1 x + (cb1 − ca1 ) x3

= c d1 + c1 x + (b1 − a1 ) x3

a1    b1
= cT      c1    d1

Lecture 2 / @ Copyright: George Nakos                                                                 112
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Lecture 3 in Engineering
Mathematics
George Nakos

Engineering Mathematics
The Johns Hopkins University

Fall 2004

113
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Part 1: Linear Algebra

1. Eigenvalues and Eigenvectors

2. Diagonalization and Similarity

3. Symmetric and Orthogonal Matrices

4. Hermitian and Unitary Matrices

114
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7.1 Eigenvalues

Deﬁnition Let A be an n × n matrix. A nonzero vector v is called
an eigenvector of A, if for some scalar λ

Av = λv               (13)
The scalar λ (which may zero) is called an eigenvalue of A
corresponding to (or associated with) the eigenvector v.

Geometrically, if v is an eigenvector of A, then v and Av are on
the same line through the origin.

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7.1 Eigenvalues
Example Let
2  2                     2                    1
A=               ,      v1 =         ,       v2 =
2 −1                     1                   −2

(a) Show that v1 and v2 are eigenvectors of A.

(b) What are the eigenvalues corresponding to v1 and v2 ?

Solution: We have
2  2        2          6            2
Av1 =                       =        =3            = 3v1
2 −1        1          3            1

2  2         1             −2               1
Av2 =                       =            = −2              = −2v2
2 −1        −2              4              −2
Therefore, v1 is an eigenvector with corresponding eigenvalue λ = 3 and v2
is an eigenvector with corresponding eigenvalue λ = −2.

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7.1 Eigenvalues
Example Find all the eigenvalues and eigenvectors of A geometrically, if

0 1
(a) A =            .
1 0

(b) A is the standard matrix of the rotation by 30◦ in R3 about the z-axis in
the positive direction.

(a) Ax is the reﬂection of x about the line y = x. The only vectors that remain
on the same line after rotation are the vectors along the lines y = x and
y = −x. These without the origin are the only eigenvectors. For v along
y = x we have Av = 1v, so v is an eigenvector with corresponding
eigenvalue 1. For v along y = −x, Av = −1v, so v is an eigenvector with
corresponding eigenvalue −1.

(b) The only vectors that remain on the same line after rotation are all vectors
along the z-axis. These without the origin are the only eigenvectors. The
corresponding eigenvalue is 1.

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7.1 Computation of Eigenvalues
Theorem Let A be a square matrix.

1. A vector v is an eigenvector of A corresponding to eigenvalue λ if and
only if v is a nontrivial solution of the system
(A − λI)v = 0                       (14)

2. A scalar λ is an eigenvalue of A if and only if
det(A − λI) = 0                         (15)

Equation (15) is called the characteristic equation of A. The determinant
det(A − λI) is a polynomial of degree n in λ and is called the characteristic
polynomial of A. The matrix A − λI is called the characteristic matrix of
A. If an eigenvalue λ is a root of the characteristic equation of multiplicity k,
we say that λ has algebraic multiplicity k.

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7.1 Proof of Theorem

1. We have
Av = λv ⇒ Av = λI v
⇒ Av − λI v = 0
⇒ (A − λI)v = 0
Hence, v is an eigenvector if and only if it is a nontrivial
solution of the homogeneous system (A − λI)v = 0.

2. The homogeneous linear system (14) has a nontrivial solution
if and only if the determinant of the coeﬃcient matrix is zero.
Thus, λ is an eigenvalue of A if and only if det(A − λI) = 0.

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7.1 Eigenspace

Fact Let A be a n × n and let λ be an eigenvalue of A. Let Eλ
be the set that consists of all eigenvectors of A corresponding
to λ and the zero n-vector. Then Eλ is a subspace of Rn.

The subspace Eλ of Rn mentioned above consisting of the zero
vector and the eigenvectors of A with eigenvalue λ is called an
eigenspace of A. It is the eigenspace with eigenvalue λ.

The dimension of Eλ is called the geometric multiplicity of λ.

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7.2 Examples of Finding Eigenvalues
In the next examples we compute the eigenvalues, the eigenvectors and ﬁnd
bases for each eigenspace of the given matrix A.
              
1 −1 −1
Example A =  −2       0    4 .
−2    6 −2

Solution: The characteristic equation is

1−λ  −1  −1
−2 0−λ   4              = −λ3 − λ2 + 30λ = −λ (λ − 5) (λ + 6) = 0
−2   6 −2 − λ
Hence, the eigenvalues are
λ1 = 0,     λ2 = 5,   λ3 = −6
Next, we ﬁnd the eigenevectors. For λ1 = 0 we have

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7.1 Examples of Finding Eigenvalues
                                 
1 −1 −1 0       1 0 −2 0
[A − 0I : 0] =  −2  0  4 0  ∼  0 1 −1 0 
−2  6 −2 0       0 0  0 0

The general solution is (2r, r, r) for r ∈ R. Hence,
                             
 2r                           2 
E0 =  r  , r ∈ R = Span              1 
 r                            1 
   
2
and eigenvector v1 =  1  deﬁnes the basis {v1 } of     E0 .
1

For λ2 = 5 we have
                                 
−4 −1 −1 0       1 0 1/2 0
[A − 5I : 0] =  −2 −5  4 0  ∼  0 1 −1 0 
−2  6 −7 0       0 0   0 0

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7.1 Examples of Finding Eigenvalues
The general solution is (−r/2, r, r) for r ∈ R. Hence,
−r/2                          −1/2
E5 =          r     , r∈R     = Span        1
r                             1
Any nonzero vector of E5 is a basis of E5 . We may choose a fraction free one. So, v2 =
−1
2   deﬁnes the basis {v2 } of E5 .
2

For λ = −6 we have
7   −1   −1   0        1    0   −1/20   0
[A − (−6) I : 0] =     −2    6    4   0   ∼    0    1   13/20   0
−2    6    4   0        0    0     0     0

The general solution is (r/20, −13r/20, r) for r ∈ R. Hence,
r/20                                1/20
E−6 =      −13r/20      , r∈R     = Span       −13/20
r                                   1
1
Any nonzero vector of E−6 is a basis of E−6 . For example, v3 =            −13       deﬁnes the basis
20
{v3 } of E−6 .

Lecture 3 / @ Copyright: George Nakos                                                        123
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7.1 Examples of Finding Eigenvalues
           
0 0 1
Example A =  0 1 0 .
0 0 1

Solution: The characteristic equation is
−λ 0   1
det(A − λI) =      0 1−λ  0            = −λ (1 − λ)2 = 0
0  0  1−λ
Hence, the eigenvalues are
λ1 = 0 , λ2 = λ3 = 1
Next, we ﬁnd the eigenevectors.      For λ1 = 0 we have
                            
0      0 1 0         0 1 0 0
[A − 0I : 0] =  0      1 0 0 ∼ 0 0 1 0 
0      0 1 0         0 0 0 0

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7.1 Examples of Finding Eigenvalues
The general solution is (r, 0, 0) for r ∈ R. Hence,
                        
 r                        1 
E0 =  0  , r ∈ R = Span  0 
 0                        0 
    
1
and eigenvector v1 =  0  deﬁnes the basis {v1 } of E0 (Fig. Ex. ).
0
For λ2 = λ3 = 1 with algebraic multiplicity     2, we have
                   
−1        0 1 0
[A − 1I : 0] =  0         0 0 0 
0      0 0 0
The general solution is (r, s, r) for r ∈ R. But (r, s, r) =   r(1, 0, 1) + s(0, 1, 0),
so                                                       
 r                        1             0 
E1 =  s  , r ∈ R = Span  0  ,               1 
 r                        1             0 

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7.1 Examples of Finding Eigenvalues
1                 0
The spanning eigenvectors v2 =    0 , v3 =          1   are linearly independent. So, {v2 , v3 } is
1                 0
a basis for E1 and the geometric multiplicity of    λ = 1 is 2.

NOTE If A = [aij ] is a triangular matrix, then so is A − λI. Hence, in this case
det(A − λI) = (a11 − λ)(a22 − λ) · · · (ann − λ)
We conclude that the eigenvalues of a triangular matrix are the diagonal entries.

1   −1        0
Example A =      0   −4        2    .
0    0       −2

A is triangular, so the eigenvalues are the diagonal entries 1, −2, −4. By row reducing
[A − 1I : 0] , [A − (−2)I : 0], and [A − (−4)I : 0] we get
1                         1/3                         1/5
E1 = Span        0        , E−2 = Span      1       , E−4 = Span        1
0                          1                           0
The spanning eigenvectors deﬁne bases for the corresponding eigenspaces.

Lecture 3 / @ Copyright: George Nakos                                                      126
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7.5 Diagonalization

Matrix arithmetic with diagonal matrices is easier than with any other matri-
ces. This is most notable in matrix multiplication. For example, a diagonal
matrix D does not mix the components of x in the product Dx.
2 0         a        2a
=
0 3         b        3b
Also, it does not mix rows of A in a product DA (or columns in AD).
2 0      a b c               2a 2b 2c
=
0 3      d e f               3d 3e 3f
Moreover, it is very easy to compute the powers Dk .
k
2 0             2k   0
=
0 3             0    3k

We study matrices that can be transformed to diagonal matrices and take
advantage of the easy arithmetic. We use eigenvalues to develop criteria that
identify these matrices and we explore their basic properties.
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7.5 Diagonalization

Deﬁnition Let A and B be two n × n matrices. We say that B is similar to
A if there exists an invertible matrix P such that

B = P −1 AP

Deﬁnition If a n × n matrix A is similar to a diagonal matrix D, then it is
called diagonalizable. We also say that A can be diagonalized. This means
that there exists an invertible n × n matrix P such that P −1 AP is a diagonal
matrix D.
P −1 AP = D
The process of ﬁnding matrices P and D is called diagonalization. We say
that P and D diagonalize A.

The answer of how to diagonalize a matrix is provided in the next theorem.

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7.5 Diagonalization
Theorem Let A be an n × n matrix.

1. A is diagonalizable if and only if it has n linearly independent eigenvectors.

2. If A is diagonalizable with P −1 AP = D, then the columns of P are eigen-
vectors of A and the diagonal entries of D are the corresponding eigen-
values.

3. If {v1 , . . . , vn} are linearly independent eigenvectors of A with correspond-
ing eigenvalues λ1 , . . . , λn, then A can be diagonalized by
             
λ1 · · ·  0
P = [v1 v2 · · · vn] and D =  . . . .
.
.        . 
.
.
0 · · · λn

Theorem Let A be a n × n matrix. The following are equivalent.

1. A is diagonalizable.

2. Rn has a basis of eigenvectors of A.

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7.5 Diagonalization
           
0 0 1
Example A =  0 1 0 .
0 0 1
Solution: We found before       that λ1 = 0, λ2 = λ3 = 1 and
                                
         1                  1       0 
E0 = Span          0  , E1 = Span  0  ,  1 
         0                  1       0 
A has 3 linearly independent      eigenvectors so it is diagonalizable. We may
take
                              
1      1 0         0 0 0
P = 0       0 1  , D= 0 1 0 
0      1 0         0 0 1
We may check this by
         −1                       
1 1 0        0 0 1    1 1 0     0 0 0
−1
P AP =  0 0 1   0 1 0   0 0 1  =  0 1 0  = D
0 1 0        0 0 1    0 1 0     0 0 1

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7.5 Diagonalization

Example
         
1 −1  0
A =  0 −4  2 .
0  0 −2

Solution: We have found that λ1 = 1, λ2 = −2, λ3 = −4 and
                     1                1 
 1                     3                  5 
E1 = Span  0  , E−2 = Span  1  , E−4 = Span  1 
 0                                            
1                   0
A has 3 linearly independent eigenvectors so it is diagonalizable. We may
take
1 1 1
                              
3   5            1    0    0
P =  0 1 1  , D =  0 −2         0 
0 1 0              0    0 −4

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7.5 Diagonalization
Theorem Let λ1 , . . . , λl be any distinct eigenvalues of an n × n matrix A.

1. Then any corresponding eigenvectors v1 , . . . , vl are linearly independent.

2. If B1 , . . . , Bl are bases for the corresponding eigenspaces, then
B = B1 ∪ · · · ∪ Bl
is linearly independent.

3. Let l be the number of all distinct eigenvalues of A. Then A is diago-
nalizable, if and only if B in part 2 has exactly n elements.
                
1    0    3
Example Is A =  1 −1            2  diagonalizable?
−1     1 −2
Solution: We have found that λ1 = λ2 = 0, λ3 = −2 and
−3                        −1
E0 = Span      −1      , E−2 = Span      −1
1                        1
This time A has at most 2 (< 3) linearly independent eigenvectors, so it is not diagonalizable,
by part 2 of the theorem.
Lecture 3 / @ Copyright: George Nakos                                                  132
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7.5 Powers of Diagonalizable Matrices
Let A be diagonalizable n × n matrix, diagonalized by P and D, so A = P DP −1 . We have
A2 = (P DP −1 )(P DP −1 ) = P D2 P −1 . We iterate to get
Ak = P Dk P −1

Example Find a formula for Ak , k = 0, 1, 2, . . . , where
1   0     1
A=    0   2     0
3   0     3

Solution: A has eigenvalues 0, 2, 4 and the corresponding basic eigenvectors (−1, 0, 1),
(0, 1, 0), (1, 0, 3) are linearly independent. Hence,
k                            −1
−1    0    1     0   0     0            −1       0       1
Ak   =       0    1    0     0   2     0             0       1       0
1    0    3     0   0     4             1       0       3
                        
0    0         0           −3       0       1
−1    0    1                                 4               4
=       0    1    0     0   2k         0           0       1       0   
1    0    3     0    0        4k            1
0       1
4               4

4k−1      0     4k−1
=          0      2k       0
3 · 4k−1   0    3 · 4k−1

Lecture 3 / @ Copyright: George Nakos                                                                133
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7.5 An Important Change of Variables

Let us now discuss an idea that is in the core of most applications of di-
agonalization. Let A be diagonalizable, diagonalized by P and D. Often a
matrix-vector equation f (A, x) = 0 can be substantially simpliﬁed, if we re-
place x by the new vector y such that
x = P y or y = P −1x                      (16)
and replace A with P DP −1 to get an equation of the form g(D, y) = 0 that
involves the diagonal matrix D and the new vector y.
To illustrate suppose we have a linear system Ax = b. Then we can convert
this system into a diagonal system as follows. We consider the new variable
vector y deﬁned by y = P x. We have
Ax = b ⇔ P Ax = P b
⇔ P AP −1 y = P b
⇔ Dy = P b
The last equation deﬁnes a diagonal system.
Lecture 3 / @ Copyright: George Nakos                                 134
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7.3 Orthogonal Matrices

Deﬁnition A square matrix A is called orthogonal if it is has orthonormal
columns. This means that every pair of columns is orthogonal and each
column is a unit vector.
Note that a nonsquare matrix with orthonormal columns is not called orthog-
onal. (Perhaps a better name for orthogonal matrix would be “orthonormal”.)
Orthogonal matrices are invertible, because they are square with linearly
independent columns. We have the following important theorem.
Theorem Let A be a square matrix. The following are equivalent.

1. A is orthogonal.

2. AT A = I

3. A−1 = AT
Lecture 3 / @ Copyright: George Nakos                                135
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7.3 Examples of Orthogonal Matrices
Example Show that the rotation matrix A is orthogonal
cos θ − sin θ
A=
sin θ  cos θ
and ﬁnd its inverse
Solution: A is orthogonal because
T      cos2 θ + sin2 θ        0                  1 0
AA =                                            =
0        cos2 θ + sin2 θ           0 1

Note that the inverse of an orthogonal matrix is its transpose. Hence,
cos θ sin θ
A−1 = AT =
− sin θ cos θ

Example It is easy to see that B is also orthogonal.
 2 1        2 
3     3    3
 2          2    1
B =  −3

3    3   
1     2
3     3
−2
3

Lecture 3 / @ Copyright: George Nakos                                    136
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7.3 Orthogonal Matrices
Theorem For A a n × n matrix, the following statements are equivalent.

1. A is orthogonal.

2. Au · Av = u · v for any n-vectors u and v (Preservation of dot products).

3.   Av = v for any n-vector v                          (Preservation of lengths).

REMARK The matrix transformation T (x) = Ax deﬁned by an orthogonal
matrix A is also called orthogonal. By the last theorem we see that orthogonal
matrix transformations preserve dot products. Hence, they preserve lengths
and angles.
Theorem If λ is an eigenvalue of an orthogonal matrix A, then |λ| = 1.
Proof: If v an eigenvector of A, then by part 3 of the last theorem
v = Av = λv = |λ| v
Hence, |λ| = 1, since v = 0.

This theorem also holds for complex eigenvalues of A.
Lecture 3 / @ Copyright: George Nakos                                       137
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7.3 Eigenvalues of Symmetric Matrices
Theorem

1.   A real symmetric matrix has only real eigenvalues.

2. A real skew-symmetric matrix eigenvalues that are either pure imaginary
or zero.

Example The symmetric matrix A
                    
1 1 2
A=  1 2 1 
2 1 1
has real eigenvalues: 1, 4, −1.
The skew-symmetric matrix B
          
0    2 1
B =  −2     0 1 
−1 −1 0
√     √
has pure imaginary or zero eigenvalues: 0, i 6, −i 6.
Lecture 3 / @ Copyright: George Nakos                               138
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7.4 Hermitian, Skew-Hermitian,
and Unitary Matrices
Deﬁnitions Let A be a square complex matrix. Then
T
1. A is called Hermitian, if A = A.
T
2. A is called skew-Hermitian, if A = −A.
T
3. A is called unitary, if A = A−1 .

Example Show that matrix A is Hermitian, matrix B is skew-Hermitian, and
matrix C is unitary.
√
1            3
4  2+i                         0    2−i                  2
−    2
i
A=                    ,   B=                         ,   C=       √
2−i  0                        −2 − i −4i              −    3
i       1
2          2
Solution: That A is Hermitian because
T                     T
T       4  2+i                     4  2−i             4  2+i
A =                         =                     =                    =A
2−i  0                     2+i  0             2−i  0

Lecture 3 / @ Copyright: George Nakos                                                139
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7.4 Hermitian, Skew-Hermitian,
and Unitary Matrices
B is skew-Hermitian, because
T                                        T
T        0    2−i                        0    2+i                                 0  −2 + i
B =                              =                                        =                    = −B
−2 − i −4i                      −2 + i  4i                                2+i  4i

T
To show that C is unitary, it suﬃces to check that C C = I. We have
√     T                           √
1
T            2
− 23 i               1
2
−        2
3
i
C C=           √                                √                       =
3           1                    3              1
−   2
i         2
−   2
i            2
√                           √
1           3               1             3
2          2
i             2
−    2
i               1 0
=    √                            √                           =         = I2
3
i       1
−    3
i        1                   0 1
2          2                2           2

Lecture 3 / @ Copyright: George Nakos                                                              140
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7.4 Hermitian, Skew-Hermitian,
and Unitary Matrices
REMARKS

1. For a real skew-Hermitian matrix A, we have A = −A. Such a matrix is
called skew-symmetric.

2. For a real unitary matrix A, we have A = A−1 . Hence, a real unitary
matrix is orthogonal.

3. The main diagonal of a Hermitian matrix consists of real numbers.

4. The main diagonal of a skew-Hermitian matrix consists of 0s, or pure
imaginary numbers.

T
5. Equivalent statements for A being a unitary matrix are: A A = I and
also by taking the transpose
AT A = I

Lecture 3 / @ Copyright: George Nakos                               141
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7.4 Hermitian, Skew-Hermitian,
and Unitary Matrices
Theorem Let A be a complex square matrix. Then

1. If A is Hermitian, then its eigenvalues are real. (Thus, this holds for
symmetric matrices.)

2. If A is skew-Hermitian, then its eigenvalues are pure imaginary, or 0.
(Thus, this holds for skew-symmetric matrices.)

3. If A is unitary, then its eigenvalues have absolute value 1. (Thus, this
holds for real orthogonal matrices.)

Note Let A be a n × n unitary matrix. Then for any complex n-vectors u
and v, we have with respect to the complex dot product:

1. Au · Av = u · v              (Preservation of the complex dot product)

2.   Av = v                                  (Preservation of complex norm)

Note The complex dot product is given by u · v = uT v = u1 v1 + · · · + unvn
Lecture 3 / @ Copyright: George Nakos                                  142
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Lecture 4 in Engineering
Mathematics
George Nakos

Engineering Mathematics
The Johns Hopkins University

Fall 2004

143
Engineering Mathematics / The Johns Hopkins University

Part 2: Partial Diﬀerential Equations

1. Orthogonal Sets of Functions

2. Generalized Fourier Series

3. Euler’s Formula

4. Review of Homogenous Diﬀerential Equations with Constant
Coeﬃcients

5. Sturm-Liouville Theory

144
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Some Trigonometric Identities
1. sin (a + b) = sin a cos b + cos a sin b

2. cos (a + b) = cos a cos b − sin a sin b

3. sin (a − b) = sin a cos b − cos a sin b

4. cos (a − b) = cos a cos b + sin a sin b

5. sin (2a) = 2 sin a cos a

6. cos (2a) = 2 cos2 a − 1

1 + cos 2a
7. cos2 (a) =
2

1 − cos 2a
8. sin2 (a) =
2

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Some Trigonometric Identities (Cont.)

1              1
9. sin a cos b =     sin (a + b) + sin (a − b)
2              2

1              1
10. sin a sin b =     cos (a − b) − cos (a + b)
2              2

1              1
11. cos a cos b =      cos (a − b) + cos (a + b)
2              2

12. cos (kπ) = (−1)k , k integer.

13. sin (kπ) = 0, k integer.

π
14. cos (2k − 1)         = 0, k integer.
2

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4.7 Orthogonal Sets of Functions
We consider continuous real-valued functions deﬁned on an interval [a, b] . So
they are in the set C [a, b] .
Deﬁnitions

1. We say that the distinct functions gm (x) and gn (x) are orthogonal on
[a, b] , if their integral inner product is zero. I.e., if
b
gm , gn =           gm (x) gn (x) dx = 0,           for m = n
a

2. We say that the sequence of distinct functions g1 (x) , g2 (x) , . . . , gn (x) , . . .
is an orthogonal set on [a, b] , if all functions are pairwise orthogonal.
I.e., if
gm , gn = 0,      for all m = n

Recall that the norm or length of each gm on [a, b] under this inner product
is
b                             b
2
gm =      gm , gm =                    gm (x) gm (x) dx =            gm (x) dx
a                             a

Lecture 4 / @ Copyright: George Nakos                                                         147
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4.7 Orthonormal Sets of Functions
Deﬁnition We say that the sequence of distinct functions g1 (x) , g2 (x) , . . . , gn (x) , . . .
is an orthonormal set on [a, b] , if

1. The set is orthogonal: gm , gn = 0 for m = n, and

2. All functions are unit:           gm = 1.

2
Because gm = 1 is equivalent to gm                  = 1, the above deﬁnition is equivalent
to saying
b
0,     for all m = n
gm , gn =           gm (x) gn (x) dx =
a
1,     for all m = n

Note: From an orthogonal set we may obtain an orthonormal set by dividing
1
each function by its own norm. So we replace gm with    gm .
gm

1             1             1
This is because if gm = 1, 0, then                 gm =             gm =         gm = 1
gm            gm            gm
Lecture 4 / @ Copyright: George Nakos                                              148
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4.7 Assumptions

Standing assumptions: From now on we assume that

a. all functions we discuss are bounded on [a, b],

b. their integrals over [a, b] are ﬁnite, and

c. their norms are nonzero.

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4.7 Examples of Orthogonal Sets
Example 1 Let gm (x) = sin mx, m = 1, 2, . . .

1. Show that gm (x), m = 1, 2, . . . forms an orthogonal set on [−π, π] .

2. Find each norm and the corresponding orthonormal set.

Solution:

1. If m = n, then
π
gm , gn   =           sin (mx) sin (nx) dx
−π
π
1
=                [cos ((m − n) x) − cos ((m + n) x)] dx
2       −π
1                             1
=           sin ((m − n) x)|π −
−π            sin ((m + n) x)|π
−π
2 (m − n)                     2 (m + n)
= 0+0
= 0
Lecture 4 / @ Copyright: George Nakos                                      150
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4.7 Examples of Orthogonal Sets
2. Each norm is computed from
π
2
gm       =         sin2 (mx) dx
−π
1 π
=        (1 − cos (2mx)) dx
2 −π
π
1      sin (2mx)
=       x−
2          2m     −π
1
=     (2π)
2
=   π
So
√
sin (mx) =          π,   for m = 1, 2, . . .
So, the corresponding orthonormal set is

sin x sin (2x) sin (3x)       sin (mx)
√ , √        , √       ,...,    √     ,...
π       π        π              π
Lecture 4 / @ Copyright: George Nakos                                    151
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4.7 Examples of Orthogonal Sets
Example 2 Consider the set
1, cos x, sin x, cos (2x) , sin (2x) , . . . , cos (mx) , sin (mx) , . . .

1. Show that this set forms an orthogonal set on [−π, π] .

2. Find each norm and the corresponding orthonormal set.

Solution:

1. We have
a.
π                                  π
sin (mx)
1, cos (mx) =    (1) cos (mx) dx =                               =0
−π                       m                    −π

b.
π                                      π
cos (mx)
1, sin (mx) =    (1) sin (mx) dx = −                             =0
−π                        m                   −π

Lecture 4 / @ Copyright: George Nakos                                                     152
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4.7 Examples of Orthogonal Sets
c. If m = n, then sin (mx) , sin (nx) = 0. This was proved before.

d. If m = n, then
π
cos (mx) , sin (nx)   =           cos (mx) sin (nx) dx
−π
π
1
=                (sin ((m + n) x) − sin ((m − n) x)) dx
2       −π
−1                           1
=           cos (m + n) x|π +
−π            cos (m − n) x|π
−π
2 (m + n)                   2 (m − n)
= 0+0=0

e.
π
cos (mx) , sin (mx)          =           cos (mx) sin (mx) dx
−π
π
1
=                sin (2mx) dx
2       −π
π
1 cos 2mx
=   −                       =0
2 2m             −π
Lecture 4 / @ Copyright: George Nakos                                                 153
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4.7 Examples of Orthogonal Sets
f. If m = n, then
π
cos (mx) , cos (nx)    =            cos (mx) cos (nx) dx
−π
π
1
=                 (cos ((m − n) x) + cos ((m + n) x)) dx
2       −π
1                           1
=           sin (m − n) x|π +
−π            sin (m + n) x|π
−π
2 (m + n)                   2 (m − n)
= 0+0
= 0

2. The norms are computed from
a.
π
2
1        =          1dx = x|π = 2π
−π
−π

Lecture 4 / @ Copyright: George Nakos                                             154
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4.7 Examples of Orthogonal Sets
b.
π
2
cos (mx)         =         cos2 (mx) dx
−π
π
1
=              (1 + cos (2mx)) dx
2       −π
π
1          sin (2mx)
=         x+
2              2m         −π
= π

c.    sin (mx) 2 = π was proved in the last example. So the norms are
√                    √                   √
1 = 2π,       cos (mx) = π,        sin (mx) = π, for m = 1, 2, . . .
So the orthonormal set is
1 cos x sin x cos (2x) sin (2x)        cos (mx) sin (mx)
√ , √ , √ ,,     √     , √       ,,...,   √     ,   √     ,...
2π  π      π       π        π               π        π

Lecture 4 / @ Copyright: George Nakos                                       155
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4.8 Generalized Fourier Series
Orthogonal sets are very important because if f (x) is a given
function deﬁned on [a, b] and g1 (x) , g2 (x) , . . . , gn (x) , . . . orthog-
onal on [a, b] , then in general f (x) can be represented as a con-
vergent series of the gn (x) .
∞
f (x) =         angn (x) = a1g1 (x) + · · · + angn (x) + · · ·    (GFS)
n=1
where the an are constants that depend on the function f (x) .

Equation (GFS) is called the generalized Fourier series of f (x)
with respect to the orthogonal set gn (x) n = 1, 2, . . . . The
coeﬃcients an are the generalized Fourier coeﬃcients of f (x) .

Under some general conditions the series in (GFS) converges to
the function f (x) .
Lecture 4 / @ Copyright: George Nakos                                  156
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4.8 Generalized Fourier Series

Under the convergence conditions the generalized Fourier coef-
ﬁcients can be computed as follows.
∞                      ∞
f, gn =               amgm, gn   =         am gm, gn = an gn, gn
m=1                     m=1
because gm, gn = 0 for m = n, by orthogonality. Therefore,
f, gn   f, gn
an =          =                       (GFC1)
gn, gn   gn 2
Or by the deﬁnition of the integral inner product
1        b              b
an =           f (x) gn (x) dx = a f (x) gn (x) dx           (GFC2)
2 a                       b 2
gn                          a gn (x) dx

Lecture 4 / @ Copyright: George Nakos                                     157
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4.8 Example The (Classical) Fourier Series
Consider the orthogonal functions on [−π, π] of Example 2: 1, cos x, sin x, cos (2x) , sin (2
If a function f (x) is deﬁned on [−π, π] then the special generalized Fourier
series
∞
f (x) = a0 +         (an cos (nx) + bn sin (nx))               (FS)
n=1
is called the (classical) Fourier series of f (x) . The coeﬃcients are computed
by using (GFC2) to get
π
1
a0 =            f (x) dx
2π −π
1 π
an =          f (x) cos (nx) dx
π −π
1 π
bn =         f (x) sin (nx) dx
π −π
because we have already seen in Example 2 that
√                     √                     √
1 = 2π,       cos (mx) = π,         sin (mx) = π,               for m = 1, 2, . . .

Relations (FC) are the (classical) Fourier coeﬃcients of f (x) .
Lecture 4 / @ Copyright: George Nakos                                             158
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4.8 Orthogonality with Respect to a Weight
Function
Let be a positive function deﬁned on [a, b] . I.e., p (x) > 0 for all x in [a, b].
The assignment
b
(f, g) → f, g =                         p (x) f (x) g (x) dx
a
deﬁnes as inner product on the vector space of all real-valued continuous
functions C [a, b] deﬁned on [a, b] . The norm deﬁned by this inner product is
b
f =               p (x) f 2 (x) dx
a

Deﬁnition Let p (x) be a positive function deﬁned on [a.b] . The sequence of
functions g1 (x) , g2 (x) , . . . is an orthogonal set on [a, b] with respect to the
weight function p (x) , if the inner product with weight p (x) is zero. I.e., if
b
gm , gn =           p (x) gm (x) gn (x) dx = 0,                 for m = n
a

Lecture 4 / @ Copyright: George Nakos                                                      159
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4.8 Orthogonality with Respect to a Weight
Function
If each function in an orthogonal set has norm 1 with respect to the weight
function p (x) , then the set is orthonormal with respect to the weight
function p (x) .
Notes

1. Orthogonality is the same as orthogonality with respect to the weight
function p (x) = 1, for all x in [a, b] .

2. If the g1 (x) , g2 (x) , . . . is orthogonal with respect to weight p (x) and we
set hn (x) = p (x)gn (x) , then by the weighted orthogonality we get
b                            b
hm (x) hn (x) dx =             p (x)gm (x)    p (x)gn (x) dx
a                            a
b
=            p (x) gm (x) gn (x) dx
a
= 0
So the functions h1 (x) , h2 (x) , . . . are orthogonal in the usual sense.

Lecture 4 / @ Copyright: George Nakos                                                  160
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Euler’s Formula
Euler’s Formula relates the complex exponential function with the trigono-
metric sines and cosines. If t is a real number, then
eit = cos t + i sin t
√
where i =       −1 is the complex unit such that
i2 = −1

Example We have
eiπ = −1,     eiπ/2 = i,      ei2π = 1,     e2+3i = e2 (cos 3 + i sin 3)
because
eiπ = cos π + i sin π = −1 + i0 = −1
eiπ/2 = cos (π/2) + i sin (π/2) = 0 + i = i
ei2π = cos (2π) + i sin (2π) = 1 + i0 = 1
e2+3i = e2 e3i = e2 (cos 3 + i sin 3)       −7. 315 1 + 1. 042 7i

Lecture 4 / @ Copyright: George Nakos                                                161
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Review
Linear Homogeneous with Constant Coeﬃcients

Let y = y (x) be an unknown function of the independent variable x.

A nth order diﬀerential equation with constant coeﬃcients is of the form
dny       dn−1 y           dy
an n + an−1 n−1 + · · · + a1    + a0 y = f (x)         (N)
dx        dx               dx
where all ai are constants and f (x) is a given function.

If f (x) is nonzero (N) is called nonhomogeneous.

If the function f (x) is the zero function, then we have a homogeneous
diﬀerential equation:
dny      dn−1 y            dy
an n + an−1 n−1 + · · · + a1    + a0 y = 0           (H)
dx       dx                dx

(H) is called the associated homogeneous equation of (N).

Lecture 4 / @ Copyright: George Nakos                                 162
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Review
Linear Homogeneous with Constant Coeﬃcients

In order to solve (H), we seek solutions of the form y = erx , where r is a
constant. Given that
dk
(erx ) = r k erx
dxk
substitution into (H) yields
an (rnerx ) + an−1 r n−1 erx + · · · + a1 (rerx ) + a0 (erx ) = 0
⇒ erx anr n + an−1 rn−1 + · · · + a1 r + a0 = 0
⇒ anrn + an−1 rn−1 + · · · + a1 r + a0 = 0
So to ﬁnd solutions of the form y = erx it suﬃces to solve the auxiliary
polynomial equation
anrn + an−1 rn−1 + · · · + a1 r + a0 = 0                   (A)
The roots of (A) can be real, or complex, or repeated.

Lecture 4 / @ Copyright: George Nakos                                           163
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Review
Linear Homogeneous with Constant Coeﬃcients

For the special case of a second order equation the general solution is dis-
cussed the following theorem.

Theorem Let y = y(x) be an unknown function
d2 y  dy
a 2 +b    + cy = 0                         (H2)
dx    dx
with auxiliary
ar 2 + br + c = 0                       (A2)

1. If (A2) has two distict real roots r1 and r2 , then the general real solution
of (H2) is given by
y (x) = c1 er1 x + c2 er2 x

for any constants c1 and c2 .

Lecture 4 / @ Copyright: George Nakos                                    164
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Review
Linear Homogeneous with Constant Coeﬃcients

2. If (A2) has a double real root r, then the general real solution
of (H2) is given by

y (x) = c1erx + c2xerx
for any constants c1 and c2.

3. If (A2) has a complex conjugate pair of roots a ± ib, then
the general real solution of (H2) is given by

y (x) = c1eax cos (bx) + c2eax sin (bx)
for any constants c1 and c2.
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Review
Linear Homogeneous with Constant Coeﬃcients

Example Solve 2y + 5y − 3y = 0.

Solution: We have 2r2 − 7r + 3 = 0, so r = 1 , 3. Hence,
2

y (x) = c1 ex/2 + c2 e3x

Example Solve y − 8y + 16y = 0.

Solution: We have r2 − 8r + 16, so r = 4, 4. Hence,
y (x) = c1 e4x + c2 xe4x

Example y − 8y + 20y = 0.

Solution: We have r2 − 8r + 20 = 0, so r = 4 − 2i, 4 + 2i. Therefore,
y (x) = c1 e4x cos (2x) + c1 e4x sin (2x)

Lecture 4 / @ Copyright: George Nakos                                   166
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Review
Linear Homogeneous with Constant Coeﬃcients

An initial value problem (IVP) is a set of diﬀerential equations
and initial conditions (IC’s).

Example Solve the IVP.
y + 16y = 0,             y (π ) = 3,      y (π ) = −5

Solution: We have r2 + 16 = 0. So, r = ±4i. Hence, y (x) =
c1 cos (4x) + c2 sin (4x) . Diﬀerentiate to get y = −4c1 sin 4x +
4c2 cos 4x. Now y (π ) = 3 yields c1 = 3 and y (π ) = −5 yields
c2 = − 5 . So the solution is
4
5
y (x) = 3 cos (4x) −         sin (4x)
4
Lecture 4 / @ Copyright: George Nakos                                  167
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4.7 Sturm-Liouville Theory
A Sturm-Liouville problem (S-L) deﬁned on [a, b] is a boundary value problem
for a second order homogeneous diﬀerential equation in unknown function
y = y (x) that can be written in the form

r (x) y   + [q (x) + λp (x)] y = 0
with two boundary conditions of the form
k1 y (a) + k2 y (a) = 0
l1 y (b) + l2 y (b) = 0
where the constants k1 , k2 are not both zero and the constants l1 , l2 are also
not both zero. The number λ is called the parameter of the S-L problem.

Note that a S-L problem has always the trivial solution y (x) = 0 for all x in
[a, b] . If λ is a scalar such that the S-L problem has a nontrivial solution y (x),
λ is called an eigenvalue of the the problem and the nontrivial y (x) is called
an eigenfunction corresponding to λ.

Lecture 4 / @ Copyright: George Nakos                                        168
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4.7 Sturm-Liouville Theory
Example      Find the eigenvalues and eigenfunctions of the S-L problem.
y + λy = 0,         y (0) = 0,        y (π) = 0

Solution: We have the following cases:

Case 1 Let λ < 0, say λ = −v 2 for v > 0. Then we have
y − v 2 y = 0 ⇒ r 2 − v 2 = 0 ⇒ r = ±v
This is a case of two real roots. So

y (x) = c1 evx + c2 e−vx
Using the boundary conditions: The ﬁrst yields y (0) = 0 = c1 + c2 , so
c2 = −c1 . The second condition yields y (π) = c1 (evπ − e−vπ ) = 0. So, c1 = 0.
Hence, c2 = 0. We only get the trivial solution.

Case 2 Let λ = 0. Then y (x) = 0. Hence, y (x) = c1 x + c2 , by integration.
Using the boundary conditions we get y (0) = 0 = c1 . Hence, y (x) = c2 . Now
y (π) = c2 = 0. Thus, we again get the trivial solution.

Lecture 4 / @ Copyright: George Nakos                                    169
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4.7 Sturm-Liouville Theory

Case 3 Let λ > 0, say λ = v 2 for v > 0. Then we have
y + v 2 y = 0 ⇒ r2 + v 2 = 0 ⇒ r = ±vi
This is a case of two complex conjugate roots. So

y (x) = c1 cos (vx) + c2 sin (vx)
Using the boundary conditions we get y (0) = 0 = c1 . So y (x) = c2 sin (vx) .
Now y (π) = c2 sin (vπ) = 0. If c2 = 0, we get the trivial solution. If c2 = 0,
then sin (vπ) = 0. Hence, vπ = nπ, where n is any integer. Therefore, v = n
is an integer. So there are inﬁnitely many eigenvalues
λn = n2
with corresponding eigenfunctions
yn (x) = sin (nx) ,   n = 1, 2, 3, . . .

Exercise Find the eigenvalues and eigenfunctions of the S-L problem.
y + λy = 0,      y (π) = y (−π)      y (π) = y (−π)

Lecture 4 / @ Copyright: George Nakos                                   170
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4.8 Orthogonality of Eigenfunctions
Theorem 1 (Orthogonality of Eigenfunctions) If p (x) , q (x) , r (x) , and
r (x) are real-valued continuous functions deﬁned on [a, b] for an S-L problem.

r (x) y   + [q (x) + λp (x)] y = 0
k1 y (a) + k2 y (a) = 0
l1 y (b) + l2 y (b) = 0
Let ym (x) and yn (x) be two eigenfunctions corresponding to diﬀerent eigen-
vlues λm and λn. Then ym (x) and yn (x) are orthogonal with respect to weight
function p (x) . Furthermore:

1. If r (a) = 0, then the ﬁrst boundary condition can be dropped.

2. If r (b) = 0, then the second boundary condition can be dropped.

3. If r (a) = r (b) , then the two boundary conditions can be replaced by
y (a) = y (b) ,   y (a) = y (b)

Lecture 4 / @ Copyright: George Nakos                                   171
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4.8 Reality of Eigenvalues

Theorem 2 (Real Eigenvalues)            If p (x) , q (x) , r (x) , and
r (x) are real-valued continuous functions deﬁned on [a, b] for

r (x ) y   + [q (x) + λp (x)] y = 0
k 1 y (a ) + k 2 y (a ) = 0
l1y (b) + l2y (b) = 0
and p (x) is either positive in the entire interval [a, b], or nega-
tive in the entire interval [a, b] , then all the eigenvalues are real
numbers.

Lecture 4 / @ Copyright: George Nakos                           172
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4.7 Sturm-Liouville Example:
Periodic Boundary Conditions
Example 2      Find the eigenvalues and eigenfunctions of the S-L problem.

y + λy = 0,      y (0) = y (2π) ,   y (0) = y (2π)
Solution:    We have

Case 1 Let λ < 0, say λ = −v 2 . Then the auxiliary is r2 − v 2 = 0. We get
r = ±v. So, y (x) = c1 evx + c2 e−vx . Using the boundary conditions we have
y = vc1 evx − vc2 e−vx . Hence,
y (0) = c1 + c2 = y (2π) = c1 e2πv + c2 e−2πv
y (0) = vc1 − vc2 = y (2π) = vc1 e2πv − vc2 e−2πv
Thus, we get the homogoenuous linear system in c1 and c2
c1 1 − e2πv + c2 1 − e−2πv        = 0
c1 1 − e2πv + c2 e−2πv − 1        = 0

Lecture 4 / @ Copyright: George Nakos                                 173
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4.7 Sturm-Liouville Example:
Periodic Boundary Conditions
The coeﬃcient matrix has determinant
1 − e2πv   1 − e−2πv                                         2
= 2e−2πv + 2e2πv − 4 = 2 eπv − e−πv       =0
1 − e2πv   e−2πv − 1
So the system has only the trivial solution.
Case 2 Let λ = 0. Then y (x) = 0. Hence, y (x) = c1 x + c2 , by integration.
Using the boundary conditions we get
y (0) = c2 = y (2π) = 2πc1 + c2
y (0) = c1 = y (2π) = c1
Hence, c1 = 0. However, there is no restriction on c2 . So y (x) can be any
constant. Say, y (x) = a0 .
Case 3 Let λ < 0, say λ = −v 2 . Then the auxiliary is r2 + v 2 = 0. We get
r = ±iv. So, y (x) = c1 cos (vx) + c2 sin (vx) . Using the boundary conditions
we have y = −vc1 sin vx + vc2 cos vx. Hence,
y (0) = c1 = y (2π) = c1 cos (2πv) + c2 sin (2πv)
y (0) = vc2 = y (2π) = −vc1 sin (2πv) + vc2 cos (2πv)

Lecture 4 / @ Copyright: George Nakos                                          174
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4.7 Sturm-Liouville Example:
Periodic Boundary Conditions
Thus, we get the homogoenuous linear system in c1 and c2
c1 (1 − cos (2πv)) + c2 (− sin (2πv)) = 0
c1 (sin (2πv)) + c2 (1 − cos (2πv)) = 0

For nontrivial solutions the coeﬃcient determinant must be zero.

1 − cos (2πv)  − sin (2πv)
= 2 − 2 cos (2πv) = 4 sin2 (πv) = 0
sin (2πv)   1 − cos (2πv)

Therefore, πv = nπ, where n is an integer. So the system has eigenfunctions
c1 cos (nx) + c2 sin (nx), n is any integer.

So all eigenfunctions are nontrivial linear combinations of the eigenfunctions
1, cos x, cos (2x) , cos (3x) , . . . , sin x, sin (2x) , sin (3x) , . . .

Lecture 4 / @ Copyright: George Nakos                                                    175
Engineering Mathematics / The Johns Hopkins University

Lecture 5 in Engineering
Mathematics
George Nakos

Engineering Mathematics
The Johns Hopkins University

Fall 2004

176
Engineering Mathematics / The Johns Hopkins University

11.2 Modeling the Vibrating String

We consider a string of length L attached to ﬁxed points with x-coordinates 0
and L. Let u (x, t) be the deﬂection or displacement (signed vertical distance
from the x-axis) of the string at location x at time t.

Goal: Calculate u (x, t) , given (a) the ends of the strings are ﬁxed and (b)
initial displacement u (x, 0) and initial velocity ut (x, 0) .

We need assumptions to simplify the partial diﬀerential equation for u (x, t) .
This is because PDEs are very hard or impossible to solve exactly.

Lecture 5 / @ Copyright: George Nakos                                   177
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11.2 Modeling the Vibrating String

Assumptions:

1. The mass of the string per unit length is constant (homogeneous string).
The string is elastic and does not resist to bending.

2. The tension caused by stretching is much greater that gravity. So, gravity
is not a factor here.

3. The string performs a small transverse motion in the vertical plane, so
that both the deﬂection u (x, t) and its slope ux (x, t) are small.

Lecture 5 / @ Copyright: George Nakos                                  178
Engineering Mathematics / The Johns Hopkins University

11.2 Modeling the Vibrating String

Forces:
Consider forces acting on small portions of the string.
Since there is no resistance to bending, the tension is tangential to the curve
of the string at each point.
Let T1 and T2 be the tensions at P and Q.
Horizontal direction: There is no motion in the horizontal direction, so the
horizontal component must be constant, say T . So
T1 cos α = T2 cos β = T                      (1)

Lecture 5 / @ Copyright: George Nakos                                   179
Engineering Mathematics / The Johns Hopkins University

11.2 Modeling the Vibrating String

Vertical direction: In the vertical direction we have two forces, the vertical
components −T1 sin α and T2 sin β.

Let ρ be the linear mass density of the string, i.e., mass per unit length. By
∂ 2u
Newton’s second law the resultant force is mass ρ∆x times acceleration
∂t2
evaluated at some point between x and x + ∆x.

∂ 2u
T2 sin β − T1 sin β = ρ∆x 2
∂t

Lecture 5 / @ Copyright: George Nakos                                   180
Engineering Mathematics / The Johns Hopkins University

11.2 Modeling the Vibrating String
Using (1) we get
T2 sin β    T1 sin α                     ∆x ∂ 2 u
=           = tan β − tan α = ρ
T2 cos β    T1 cos α                     T ∂t2
∂u                ∂u
But tan α =             , tan β =              are the slopes at x and x + ∆x. So
∂x x              ∂x x+∆x
we have

1      ∂u              ∂u         ρ ∂ 2u
−            =
∆x     ∂x   x+∆x       ∂x   x     T ∂t2

Now we take the limit as ∆x → 0 to get
∂ 2u      2
2∂ u
=c                              (W-1)
∂t2      ∂x2
T
where c2 =     .
ρ
Equation (W-1) is the one-dimensional wave equation.
Lecture 5 / @ Copyright: George Nakos                                     181
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11.3 Solving the 1-D Wave Equation
We solve the one-dimensional wave equation

∂ 2u      2
2∂ u
=c                            (W-1)
∂t2      ∂x2

subject to the ﬁxed-ends boundary conditions
u (0, t) = 0,      u (L, t) = 0,        t≥0         (BC)

and initial conditions specifying an initial deﬂection f (x) and initial velocity
g (x) , for x such that 0 ≤ x ≤ L.
∂u
u (x, 0) = f (x) ,                 = g (x) ,   0≤x≤L        (IC)
∂t   t=0

Method of solution

Lecture 5 / @ Copyright: George Nakos                                     182
Engineering Mathematics / The Johns Hopkins University

11.3 Solving the 1-D Wave Equation
Stage 1: Separation of Variables

First we seek nontrivial solutions of the system (W-1), (BC). Notice that the
trivial solution is already a solution.To solve (W-1), (BC) we use the method
of separation of variables. I.e., we seek solutions of the form.
u (x, t) = X (x) T (t)
where X = X (x) is a function of x only and T = T (t) is a function of t only.
Substitution into (W-1) yields
XT = c2 X T
where by X we mean dX and by T we mean
dx
dT
dt
.   Now we separate the variables
by dividing both sides by c2 XT to get
T     X
=
c2 T    X
Now x and t are completely independent variables, one being location and
one being time. So the only way the functions cTT of t and X of x is if they
2           X
are both the same constant, say −λ. So,
Lecture 5 / @ Copyright: George Nakos                                         183
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11.3 Solving the 1-D Wave Equation
T     X
2T
=     = −λ
c      X
Therefore, we get a system of two ordinary diﬀerential equations homoge-
neous with constant coeﬃcients: one in X only and only in T only.
T + c2 λT = 0,        X + λX = 0
These can be readily solved, provided we know the constant λ.

We use X + λX = 0 and the boundary conditions (BC) to ﬁnd λ and X (x) .
The boundary conditions (BC) are written in terms of X and T. For all t ≥ 0
u (0, t) = X (0) T (t) = 0,        u (L, t) = X (L) T (t) = 0
T cannot be identically zero (T (t) = 0, for all t), or else u (x, t) would be
zero for all x and t, hence we would get the trivial solution. So we must have
X (0) = 0 and X (L) = 0. We get the Sturm-Liouville problem
X + λX = 0,            X (0) = 0,     X (L) = 0
which we have essentially solved before. We have the following cases:
Lecture 5 / @ Copyright: George Nakos                                       184
Engineering Mathematics / The Johns Hopkins University

11.3 Solving the 1-D Wave Equation
Case 1 Let λ < 0, say λ = −v 2 for v > 0. Then we have
X − v 2 X = 0 ⇒ r 2 − v 2 = 0 ⇒ r = ±v
This is a case of two real roots. So

X (x) = c1 evx + c2 e−vx
Using the boundary conditions we get X (0) = 0 = c1 + c2 and X (L) =
c1 evL + c2 e−vL = 0. So c2 = −c1 . Hence, X (L) = c1 evL − e−vL = 0. Thus,
c1 = 0. So, c2 = 0 and we get the trivial solution.
Case 2 Let λ = 0. Then X (x) = 0. Hence, X (x) = c1 x + c2 , by integration.
Using the boundary conditions we get X (0) = 0 = c2 . Hence, X (x) = c1 x.
Now X (L) = c1 L = 0. So, c1 = 0. Thus, we again get the trivial solution.
Case 3 Let λ > 0, say λ = v 2 for v > 0. Then we have
X + v 2 X = 0 ⇒ r2 + v 2 = 0 ⇒ r = ±vi
This is a case of two complex conjugate roots. So

X (x) = c1 cos (vx) + c2 sin (vx)

Lecture 5 / @ Copyright: George Nakos                                185
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11.3 Solving the 1-D Wave Equation
Using the boundary conditions we get y (0) = 0 = c1 . So X (x) = c2 sin (vx) .
Now X (L) = c2 sin (vπ) = 0. If c2 = 0, we get the trivial solution. If c2 = 0,
then sin (vL) = 0. Hence, vL = nπ, where n is any integer. Therefore,
v = nπ/L. So there are inﬁnitely many eigenvalues
nπ 2
λn =     , n = 1, 2, 3, . . .
L
with corresponding eigenfunctions
nπ
Xn (x) = sin     x , n = 1, 2, 3, . . .
L

Note that since sin (−x) = − sin (x) and sin (0) = 0, so we need not keep any
negative integer values for n. These signs can be absorbed by the constant
coeﬃcients.

Now that we know X and λ we turn to T. The equation T + c2 λT = 0 takes
the form

Lecture 5 / @ Copyright: George Nakos                                   186
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11.3 Solving the 1-D Wave Equation
cnπ         2
Tn +                 Tn = 0
L
cnπ 2
which can be solved right away, becuase                    L
> 0. The auxiliary is r2 +
cnπ 2                    cnπ
L
= 0. So, r = ±    L
i. We have for any constants an and bn
cnπ             cnπ
Tn = an cos   t + bn sin      t , n = 1, 2, 3, . . .
L               L
Hence, u = XnTn becomes
cnπ             cnπ        nπ
un (x, t) = an cos     t + bn sin      t sin     x ,    n = 1, 2, 3, . . .
L               L         L

Note that since the system (W-1), (BC) is homogeneous, any ﬁnite sum of
solutions un is also a solution.
k
u (x, t) =            un (x, t)
n=1

Lecture 5 / @ Copyright: George Nakos                                                  187
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11.3 Solving the 1-D Wave Equation
Under certain conidtions an inﬁnite sum of solutions is also a solution
∞
u (x, t) =          un (x, t)
n=1
So we may have a general solution of the form
∞
cnπ            cnπ                    nπ
u (x, t) =         an cos       t + bn sin     t            sin      x   (W-1-Sol)
n=1
L              L                     L

Stage 2: Fourier Analysis
Solution (W-1-sol) is the kind of solution this method produces, provided
we know the coeﬃcients an and bn. These can be computed by using the
boundary conditions u (x, 0) = f (x) and ∂u t=0 = g (x) .
∂t

Using the ﬁrst condition and (W-1-sol) with t = 0 yields
∞
nπ
u (x, 0) =         an sin       x = f (x)
n=1
L

Lecture 5 / @ Copyright: George Nakos                                              188
Engineering Mathematics / The Johns Hopkins University

11.3 Solving the 1-D Wave Equation
Now recall that the functions sn (x) = sin nπ x were eigenfunctions to the
L
S-L problem on [0, L] . Therefore, by Theorem 1 on S-L problems, these
eigenfunctions must be orthogonal. So we can use the generic formula an =
f, sn / sn, sn to ﬁnd an. We have
L                                      L
f, sn         0
f (x) sin nπ x dx
L                        0
f (x) sin nπ x dx
L
an =              =       L
=       L
sn , s n
0
sin2 nπ x dx
L
1
2   0
1 − cos 2nπ x dx
L
L
0
f   (x) sin nπ x dx
L          2 L                        nπ
=                                 =            f (x) sin      x dx
L/2                   L   0               L

Using the second condition and (W-1-sol) with t = 0 yields
∞
∂u                        cnπ     nπ
=         bn       sin    x = g (x)
∂t   t=0       n=1
L      L

The functions sn (x) = sin nπ x are orthogonal, So we can use the generic
L
formula bn cnπ = g, sn / sn, sn to ﬁnd bn. We have
L

Lecture 5 / @ Copyright: George Nakos                                                       189
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11.3 Solving the 1-D Wave Equation
L                             L
L f, sn      L             0
g (x) sin nπ x dx
L           L    0
g (x) sin   nπ
L
x   dx
bn =                =                   L
=
cnπ sn, sn   cnπ
0
sin2 nπ x dx
L
cnπ           L/2
L
2                         nπ
=                  g (x) sin       x dx
cnπ   0                    L

So the method of separation of variables yields the following solution to the
one-dimensional wave equation.

∞
cnπ            cnπ            nπ
u (x, t) =                 an cos       t + bn sin     t    sin      x
n=1
L              L             L
2 L              nπ
an =               f (x) sin       x dx,         n = 1, 2, . . .
L 0               L
L
2                 nπ
bn =                   g (x) sin     x dx,         n = 1, 2, . . .
cnπ 0               L

Lecture 5 / @ Copyright: George Nakos                                                       190

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