# Linear Algebra Lecture Note _2

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```					Lecture 15 Linear Transformation & Eigenvalues and Eigenvectors
Last Time
- Introduction to Linear Transformations - The Kernel and Range of a Linear Transformation

Elementary Linear Algebra R. Larsen et al. (5 Edition)

TKUEE翁慶昌-NTUEE SCC_01_2008

Lecture 15: Linear Transformation
Today  Matrices for Linear Transformations  Transition Matrix and Similarity  Eigenvalues and Eigenvectors Reading Assignment: Secs 6.3-7.1 Final Exam 2:20 – 4:20 Scope: Sections 4.7-7.1 70% Sections 1.1 – 4.6 30% Tip: Practice your homework problems and really understand Makeup Lecture  Diagonalization  Symmetric Matrices and Orthogonal Diagonalization  Applications Reading Assignment: Secs 7.2-7.4

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What Have You Actually Learned about LT So Far?

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Keywords in Section 6.2:
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kernel of a linear transformation T: 線性轉換T的核空間 range of a linear transformation T: 線性轉換T的值域 rank of a linear transformation T: 線性轉換T的秩 nullity of a linear transformation T: 線性轉換T的核次數 one-to-one: 一對一 onto: 映成 isomorphism(one-to-one and onto): 同構 isomorphic space: 同構的空間

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Today Matrices for Linear Transformations Transition Matrix and Similarity  Eigenvalues and Eigenvectors
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6.3 Matrices for Linear Transformations
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Two representations of the linear transformation T:R3→R3 :
(1)T ( x1 , x2 , x3 )  (2 x1  x2  x3 , x1  3x2  2 x3 ,3x2  4 x3 )

 2 1  1   x1  (2)T (x)  Ax   1 3  2  x2     0 3 4   x3  
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Three reasons for matrix representation of a linear transformation:
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It is simpler to write. It is simpler to read.

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It is more easily adapted for computer use.

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Thm 6.10: (Standard matrix for a linear transformation)
Let T : R n  R m be a linear trtansformati such that on

 a11   a12   a1n   a21   a22   a2 n  T (e1 )   , T (e2 )   , , T (en )   ,          am1   am 2  amn       
Then the m n matrix who se n columns correspond to T (ei )  a11 a12  a1n   a21 a22  a2 n  A  T (e1 ) T (e2 )  T (en )          am1 am 2  amn   

is such that T ( v )  Av for every v in R n . A is called the standard matrix for T .
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Pf:
 v1   v2  v     v1e1  v2 e2    vn en  vn   
T is a L.T.  T ( v)  T (v1e1  v2e2    vn en )  T (v1e1 )  T (v2e2 )    T (vn en )  v1T (e1 )  v2T (e2 )    vnT (en )

 a11  a21 Av     am1 

a12 a22  am 2

 a1n   v1   a11v1  a12 v2    a1n vn   a2 n  v2   a21v1  a22 v2    a2 n vn                amn  vn  am1v1  am 2 v2    amnvn     
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 a11   a12   a1n   a21   a22   a2 n   v1    v2      vn            am1   am 2  amn         v1T (e1 )  v2T (e2 )    vnT (en )
Therefore, T ( v)  Av for each v in R n

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Ex 1: (Finding the standard matrix of a linear transformation)
Find the standard matrix for the L.T. T : R 3  R 2 define by

T ( x, y, z )  ( x  2 y, 2 x  y)

Sol:

Vector Notation
T (e1 )  T (1, 0, 0)  (1, 2)

T (e2 )  T (0, 1, 0)  (2, 1)
T (e3 )  T (0, 0, 1)  (0, 0)
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Matrix Notation 1 0  )  1  T (e1 )  T (    2 0    0    2 T (e2 )  T ( 1 )      1  0  0  0  T (e3 )  T ( 0 )      0  1

A  T (e1 ) T (e2 ) T (e3 ) 1  2 0   2 1 0  Check:  x  y   1  2 A   2 1 z  

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 x 0    x  2 y    y   2 x  y  0  z  

i.e. T ( x, y, z )  ( x  2 y,2 x  y)
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Note:
1  2 0  1x  2 y  0 z A  2 1 0  2 x  1 y  0 z 
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Ex 2: (Finding the standard matrix of a linear transformation) The linear transformation T : R 2  R 2 is given by projecting
each point in R 2 onto the x - axis. Find the standard matrix for T .

Sol:
T ( x, y)  ( x, 0)
1 0 A  T (e1 ) T (e2 )  T (1, 0) T (0, 1)   0 0    Notes: (1) The standard matrix for the zero transformation from Rn into Rm is the mn zero matrix. (2) The standard matrix for the zero transformation from Rn into Rn is the nn identity matrix In
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Composition of T1:Rn→Rm with T2:Rm→Rp :
T ( v)  T2 (T1 ( v)), v  R n

T  T2  T1 , domain of T  domain of T1
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Thm 6.11: (Composition of linear transformations)
Let T1 : R n  R m and T2 : R m  R p be L.T. with standard matrices A1 and A2 , then

(1) The compositio n T : R n  R p , defined by T ( v)  T2 (T1 ( v)), is a L.T.

(2) The standard matrix A for T is given by the matrix product A  A2 A1

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Pf:
(1)( T is a L.T.) Let u and v be vectorsin R n and let c be any scalar the n T (u  v)  T2 (T1 (u  v))  T2 (T1 (u)  T1 ( v))  T2 (T1 (u))  T2 (T1 ( v))  T (u)  T ( v)
T (cv)  T2 (T1 (cv))  T2 (cT1 ( v))  cT2 (T1 ( v))  cT ( v)

(2)( A2 A1 is the standard matrix for T ) T ( v)  T2 (T1 ( v))  T2 ( A1v)  A2 A1v  ( A2 A1 ) v
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Note:
T1  T2  T2  T1
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Ex 3: (The standard matrix of a composition) Let T1 and T2 be L.T. from R 3 into R 3 s.t.
T1 ( x, y, z )  (2 x  y, 0, x  z ) T2 ( x, y, z )  ( x  y, z, y) Find the standard matrices for the compositio ns T  T2  T1 and T '  T1  T2 ,

Sol:

2 A1  0  1  1 A2  0  0 

1 0 0 0 (standard matrix for T1 )  0 1   1 0 0 1 (standard matrix for T2 )  1 0 
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The standard matrix for T  T2  T1

1  1 0 2 1 0 2 1 0 A  A2 A1  0 0 1 0 0 0  1 0 1      0 1 0  1 0 1   0 0 0  
The standard matrix for T '  T1  T2

2 1 0 1  1 0 2  2 1 A'  A1 A2  0 0 0 0 0 1  0 0 0      1 0 1  0 1 0   1 0 0      

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Inverse linear transformation:
If T1 : R n  R n and T2 : R n  R n are L.T. s.t. for every v in R n

T2 (T1 ( v))  v and T1 (T2 ( v))  v
Then T2 is called the inverse of T1 and T1 is said to be invertible
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Note:
If the transformation T is invertible, then the inverse is unique and denoted by T–1 .

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Ex 6.12: (Existence of an inverse transformation)
Let T : R n  R n be a L.T. with standard matrix A, Then thefollowing condition are equivalent.

(1) T is invertible. (2) T is an isomorphism.

(3) A is invertible.
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Note: If T is invertible with standard matrix A, then the standard matrix for T–1 is A–1 .

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Ex 4: (Finding the inverse of a linear transformation)
The L.T. T： R 3  R 3 is defined by

T ( x1 , x2 , x3 )  (2 x1  3x2  x3 , 3x1  3x2  x3 , 2 x1  4 x2  x3 )

Sol: The standard matrix for T
2 3 1 A  3 3 1   2 4 1  

Show that T is invertible, and find its inverse.

 2 x1  3x2  x3  3x1  3x2  x3  2 x1  4 x2  x3

 2 3 1 1 0 0  A I 3    3 3 1 0 1 0   2 4 1 0 0 1 
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0 1 0 0  1 1 G . J . E  0 1 0  1 0   1  I   0 0 1 6  2  3 

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A1

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ThereforeT is invertible and the standard matrix for T 1 is A1
0  1 1 A 1   1 0 1   6  2  3  0   x1    x1  x2   1 1 T 1 ( v )  A1 v   1 0 1   x2     x1  x3       6  2  3  x3  6 x1  2 x2  3 x3   In other word s,

T 1 ( x1 , x2 , x3 )  ( x1  x2 ,  x1  x3 , 6 x1  2 x2  3x3 )
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the matrix of T relative to the bases B and B':
T :V  W B  {v1 , v2 ,, vn } (a L.T.) (a basis for V )

B'  {w1 , w2 ,, wn } (a basis for W )

Thus, the matrix of T relative to the bases B and B' is
A  T (v1 )B ' , T (v2 )B ' ,, T (vn )B '  M mn

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Transformation matrix for nonstandard bases:
Let V and W be finite - dimensiona l vector spaces with basis B and B' , respectively, where B  {v1 , v2 ,, vn }

If T : V  W is a L.T.s.t.

T (v1 )B '

 a11   a12   a1n   a21   a22   a2 n    , T (v2 )B '   ,, T (vn )B '             am1   am 2  amn       

then the m  n matrix who se n columns correspond to T (vi )B '

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 a11  a21 A  T (e1 ) T (e2 )  T (en )     am1 

a12 a22  am 2

 a1n   a2 n       amn  

is such that T ( v)B '  A[ v]B for every v in V .

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Ex 5: (Finding a matrix relative to nonstandard bases)
Let T： R 2  R 2 be a L.T. defined by T ( x1 , x2 )  ( x1  x2 , 2 x1  x2 )

Find the matrix of T relative to the basis B  {(1, 2), (1, 1)} and B'  {(1, 0), (0, 1)} Sol: T (1, 2)  (3, 0)  3(1, 0)  0(0, 1) T (1, 1)  (0,  3)  0(1, 0)  3(0, 1) 3 T (1, 2)B '  0, T (1, 1)B '  03        

the matrix for T relative to B and B' 3 0  A  T (1, 2)B ' T (1, 2)B '    0  3 
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Ex 6: For theL.T.T： R 2  R 2 given in Example 5, use the matrix A to find T ( v), where v  (2, 1) Sol: v  (2, 1)  1(1, 2)  1(1, 1) B  {(1, 2), (1, 1)}
1  v B     1 3 0   1  3  T ( v)B '  Av B     1  3 0  3      T ( v)  3(1, 0)  3(0, 1)  (3, 3) B' {(1, 0), (0, 1)}

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Check: T (2, 1)  (2  1, 2(2)  1)  (3, 3)
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Notes:
(1)In thespecial case whereV  W and B  B' , thematrix A is alled the matrix of T relative to the basis B

(2)T : V  V : theidentity transformat ion B  {v1 , v2 ,  , vn } : a basis for V  the matrix of T relative to the basis B 1 0  0 0 1  0    In A  T (v1 )B , T (v2 )B ,  , T (vn )B          0 0  1 

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Keywords in Section 6.3:
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standard matrix for T: T 的標準矩陣 composition of linear transformations: 線性轉換的合成 inverse linear transformation: 反線性轉換 matrix of T relative to the bases B and B' : T對應於基底B到 B'的矩陣 matrix of T relative to the basis B: T對應於基底B的矩陣

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Today Matrices for Linear Transformations Transition Matrix and Similarity  Eigenvalues and Eigenvectors
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6.4 Transition Matrices and Similarity
T :V  V B  {v1 , v2 ,, vn } ( a L.T.) ( a basis of V )
( matrix of T relative to B) (matrix of T relative to B' )

A  T (v1 )B , T (v2 )B ,, T (vn )B  P  w1 B , w2 B ,, wn B 

B'  {w1 , w2 ,, wn } (a basis of V )

A'  T ( w1 )B ' , T ( w2 )B ' ,, T ( wn )B ' 

( transitio n matrix from B' to B )
( transitio n matrix from B to B ' )

P 1  v1 B ' , v2 B ' ,  , vn B ' 

 v B  Pv B ' , v B '  P 1 v B

T ( v)B  AvB T ( v)B '  A' vB '
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Two ways to get from v B ' to T ( v)B ' :
(1)( direct ) A'[ v]B '  [T ( v)] B '

indirect

(2)(indirect) P 1 AP[ v]B '  [T ( v)]B '
 A'  P AP
1

direct

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Ex 1: (Finding a matrix for a linear transformation)
Find the matrix A' for T： R 2  R 2
T ( x1 , x2 )  (2 x1  2 x2 ,  x1  3x2 )

reletive to the basis B'  {(1, 0), (1, 1)}

Sol: (I) A'  T (1, 0)B '

T (1, 1)B ' 

3 T (1, 0)  (2,  1)  3(1, 0)  1(1, 1)  T (1, 0)B '     1   2 T (1, 1)  (0, 2)  2(1, 0)  2(1, 1)  T (1, 1)B '    2  3  2  A'  T (1, 0)B ' T (1, 1)B '     1 2  
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(II) standard matrix for T ( matrix of T relative to B  {(1, 0), (0, 1)})
 2  2 A  T (1, 0) T (0, 1)    1 3   transition matrix from B' to B

1 1 P  (1, 0)B (1, 1)B    0 1   transition matrix from B to B'

1  1 P  0 1   matrix of T relative B'
1

1  1  2  2 1 1  3  2 A'  P AP     1 3  0 1   1 2  0 1      
1

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Ex 2: (Finding a matrix for a linear transformation)
Let B  {(3, 2), (4,  2)} and B'  {(1, 2), (2,  2)} be basis for R 2 ,  2 7  and let A   be the matrix for T : R 2  R 2 relative to B.  3 7   Find the matrix of T relative to B'.

3  2 (2,  2)B    2  1     1 2 1 transition matrix from B to B': P  (3, 2)B ' (4,  2)B '     2 3   matrix of T relative to B':
  1 2  2 7  3  2  2 1 A'  P AP      3 7  2  1    1 3   2 3     
1

Sol: transition matrix from B' to B : P  (1, 2)B

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Ex 3: (Finding a matrix for a linear transformation) For thelinear transformation T : R 2  R 2 given in Ex.2, find v B 、 T ( v)B and T ( v)B ' , for the vector v whosecoordinate matrix is
 3 vB '      1 Sol: 3  2  3  7  v B  PvB '  2  1   1    5       2 7   7   21 T ( v)B  Av B    3 7   5   14         1 2  21  7  1 T ( v)B '  P T ( v)B   2 3  14    0      

or T ( v)B '  A' vB '

 2 1  3  7     1   0   1 3    
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

Similar matrix: For square matrices A and A‘ of order n, A‘ is said to be similar to A if there exist an invertible matrix P s.t. A'  P 1 AP



Thm 6.13: (Properties of similar matrices) Let A, B, and C be square matrices of order n. Then the following properties are true. (1) A is similar to A. (2) If A is similar to B, then B is similar to A. (3) If A is similar to B and B is similar to C, then A is similar to C. Pf:
(1) A  I n AI n

(2) A  P 1BP  PAP1  P( P 1BP) P 1 PAP1  B  Q 1 AQ  B (Q  P 1 )
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

Ex 4: (Similar matrices)
 2  2  3  2 (a) A    and A'   1 2  are similar  1 3    1 1 1 because A'  P AP, whereP   0 1  

 2 7   2 1 (b) A    and A'   1 3 are similar   3 7  

3  2 because A'  P AP, whereP   2  1  
1

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

Ex 5: (A comparison of two matrices for a linear transformation) 1 3 0  Suppose A  3 1 0  is the matrix for T : R 3  R 3 relative   0 0  2   
to the standard basis. Find the matrix for T relative to the basis B'  {(1, 1, 0), (1,  1, 0), (0, 0, 1)} Sol: The transition matrix from B' to the standard matrix

P  (1, 1, 0)B

(1,  1, 0)B
0 0  1

1 1 2 2 1 1  P  2  1 2  0 0

1 1 0 (0, 0, 1)B   1  1 0   0 0 1   

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matrix of T relative to B': 1 1 2 2 A'  P 1 AP   1  1 2 2 0 0  0 4 0  0  2 0     0 0  2   0 1 3 0  1 1 0 0 3 1 0  1  1 0    1  0 0  2  0 0 1    

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

Notes: Computational advantages of diagonal matrices:
d1k 0 k (1) D    0 
(2) DT  D

0 k d2  0

 0  0     k  dn  

d1 0 D  0 

0 d2  0

 0  0     dn  

 d11 0 (3) D 1    0 

0
1 d2

 0

   

0 0 , d i  0  1  dn 
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Today Matrices for Linear Transformations Transition Matrix and Similarity  Eigenvalues and Eigenvectors
 

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Chapter 7 Eigenvalues and Eigenvectors
7.1 Eigenvalues and Eigenvectors 7.2 Diagonalization 7.3 Symmetric Matrices and Orthogonal Diagonalization

Elementary Linear Algebra R. Larsen et al. (5 Edition)

TKUEE翁慶昌-NTUEE SCC_01_2008

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7.1 Eigenvalues and Eigenvectors


Eigenvalue problem: If A is an nn matrix, do there exist nonzero vectors x in Rn such that Ax is a scalar multiple of x？



Eigenvalue and eigenvector: A：an nn matrix ：a scalar x： a nonzero vector in Rn
Eigenvalue
Geometrical Interpretation



Ax  x
Eigenvector
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

Ex 1: (Verifying eigenvalues and eigenvectors)

2 0  A 0  1  

1 0  x1    x2    0  1 
Eigenvalue

2 0  1 2 1 Ax1    0  0  2 0  2 x1 0  1      
Eigenvector
Eigenvalue

 2 0  0   0  0  Ax2     1   (1) x2 0  1 1  1      1
Eigenvector
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

Thm 7.1: (The eigenspace of A corresponding to ) If A is an nn matrix with an eigenvalue , then the set of all eigenvectors of  together with the zero vector is a subspace of Rn. This subspace is called the eigenspace of  .

Pf:

x1 and x2 are eigenvectors corresponding to 
(i.e. Ax1  x1 , Ax2  x2 )

(1) A( x1  x2 )  Ax1  Ax2  x1  x2   ( x1  x2 ) (i.e. x1  x2 is an eigenvector correspond to λ) ing (2) A(cx1 )  c( Ax1 )  c(x1 )   (cx1 ) (i.e. cx1 is an eigenvector correspond to  ) ing
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

Ex 3: (An example of eigenspaces in the plane) Find the eigenvalues and corresponding eigenspaces of

 1 0 A 0 1  
Sol: If
v  ( x, y)

 1 0  x   x Av     y   y   0 1    
For a vector on the x-axis

Eigenvalue

1  1

 1 0  x  x  x  0 1 0   0   10       
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For a vector on the y-axis

Eigenvalue

2  1

 1 0  0   0   0   0 1  y    y   1 y        
Geometrically, multiplying a vector (x, y) in R2 by the matrix A corresponds to a reflection in the y-axis. The eigenspace corresponding to 1  1 is the x-axis. The eigenspace corresponding to 2  1 is the y-axis.

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

Thm 7.2: (Finding eigenvalues and eigenvectors of a matrix AMnn ) Let A is an nn matrix. (1) An eigenvalue of A is a scalar  such that det(I  A)  0 . (2) The eigenvectors of A corresponding to  are the nonzero solutions of (I  A) x  0 .


Note: Ax  x  (I  A) x  0 (homogeneous system)
If (I  A) x  0 has nonzero solutions iff det(I  A)  0.



Characteristic polynomial of AMnn:

det(I  A)  (I  A)  n  cn 1n 1    c1  c0


Characteristic equation of A:
det(I  A)  0
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

Ex 4: (Finding eigenvalues and eigenvectors)

2  12 A 1 5   
Sol: Characteristic equation:
(I  A) 

 2
1

12

 5

 2  3  2  (  1)(  2)  0

   1,  2

Eigenvalue: 1  1, 2  2

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(1)1  1

 3 12  x1  0  (1I  A) x     x   0   1 4  2     x1  4t  4  x    t   t 1  , t  0  2      4 12  x1  0  (2 I  A) x     x   0   1 3  2     x1  3t  3  x    t   t 1, t  0  2    

(2)2  2

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

Ex 5: (Finding eigenvalues and eigenvectors) Find the eigenvalues and corresponding eigenvectors for the matrix A. What is the dimension of the eigenspace of

each eigenvalue?
2 1 0 A  0 2 0    0 0 2 

Sol: Characteristic equation:   2 1 0 I  A  0  2 0  (  2) 3  0 0 0  2
Eigenvalue:   2
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The eigenspace of A corresponding to   2 : 0  1 0  x1  0 (I  A) x  0 0 0  x2   0      0 0 0  x3  0 
 x1   s  1 0  x2   0   s 0   t 0  , s , t  0         x3   t   0 1

 1 0        ing s 0  t 0 s, t  R  : the eigenspace of A correspond to   2  0 1       
Thus, the dimension of its eigenspace is 2.
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

Notes:

(1) If an eigenvalue 1 occurs as a multiple root (k times) for
the characteristic polynominal, then 1 has multiplicity k. (2) The multiplicity of an eigenvalue is greater than or equal to the dimension of its eigenspace.

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

Ex 6：Find the eigenvalues of the matrix A and find a basis for each of the corresponding eigenspaces.
0  1 0 0 0 1 5  10 A  1 0 2 0   3  1 0 0   Sol: Characteristic equation:  1 0 0 0 0  1  5 10 I  A  1 0  2 0 1 0 0  3

 (  1) 2 (  2)(  3)  0 Eigenvalue:1  1, 2  2, 3  3
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(1)1  1

0 0  (1I  A) x    1   1

0   x1  0 0  5 10   x2  0      0  1 0   x3  0     0 0  2  x4  0 0 0

 x1   2t  0  2  x   s  1  0   2     s    t   , s, t  0  x3   2t  0  2          x4   t  0  1    0    2         1   0    ,  is a basis for the eigenspace 0  2   of A corresponding to   1  0   1       
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(2)2  2

0   x1  0 1 0 0  0 1  5 10   x  0  2      (2 I  A) x    1 0 0 0   x3  0       1 0 0  1  x4  0  x1   0  0  x  5t  5  2      t  , t  0  x3   t  1        x4   0  0
 0       5      is a basis for the eigenspace 1  of A corresponding to   2  0     
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(3)3  3

 2 0 0 0   x1  0  0 2  5 10  x  0  2      (3I  A) x    1 0 1 0   x3  0       1 0 0 0   x4  0   x1   0   0   x   5t   5  2     t  , t  0  x3   0   0        x4   t   1  

 0         5    is a basis for the eigenspace  0   of A corresponding to   3  1     
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

Thm 7.3: (Eigenvalues for triangular matrices) If A is an nn triangular matrix, then its eigenvalues are the entries on its main diagonal. Ex 7: (Finding eigenvalues for diagonal and triangular matrices)  1 0 0 0 0 2 0 0  0 2 0 0 0 (a) A   1 1 0  (b) A   0 0 0 0 0    0 0 0  4 0  5 3  3    0 0 0 0 3    2 0 0 Sol: ( a ) I  A  1  1 0  (  2)(  1)(  3) 5 3  3 1  2, 2  1, 3  3
(b) 1  1, 2  2, 3  0, 4  4, 5  3
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



Eigenvalues and eigenvectors of linear transformations:
A number  is called an eigenvalue of a linear transformation T : V  V if there is a nonzero vector x such thatT (x)  x. The vector x is called an eigenvector of T correspond to  , ing and the setof all eigenvectors of  (with thezero vector)is called the eigenspace of .

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

Ex 8: (Finding eigenvalues and eigenspaces)

Find the eigenvalue s and corresponding eigenspace s 1 3 0  A  3 1 0 .   0 0  2    Sol: 0    1  3 I  A    3   1 0   (  2) 2 (  4)   0   2  0  
eigenvalues 1  4, 2  2

The eigenspaces for thesetwoeigenvalues are as follows. B1  {(1, 1, 0)} Basis for 1  4 B2  {(1,  1, 0), (0, 0, 1)}
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Basis for 2  2

Notes: (1) Let T:R 3  R 3 be the linear transformatio n whose standard matrix


is A in Ex. 8, and let B' be the basis of R 3 made up of three linear independent eigenvecto rs found in Ex. 8. Then A' , the matrix of T relative to the basis B' , is diagonal. B'  {(1, 1, 0), (1,  1, 0), (0, 0, 1)}
0 4 0 A'  0  2 0    0 0  2 

Eigenvecto rs of A

Eigenvalue s of A

(2) The main diagonal entries of the matrix A' are the eigenvalue s of A.

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Keywords in Section 7.1:


eigenvalue problem: 特徵值問題 eigenvalue: 特徵值 eigenvector: 特徵向量 characteristic polynomial: 特徵多項式 characteristic equation: 特徵方程式











eigenspace: 特徵空間
multiplicity: 重數



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Lecture 15: Linear Transformation
Today Matrices for Linear Transformations Transition Matrix and Similarity Eigenvalues and Eigenvectors Reading Assignment: Secs 6.3-7.1
  

Final Exam 2:20 – 4:20 Scope: Sections 4.7-7.1: 70%, Sections 1.1 – 4.6: 30% Tip: Practice your homework problems and really understand Makeup Lecture Diagonalization Symmetric Matrices and Orthogonal Diagonalization Applications Reading Assignment: Secs 7.2-7.4
  

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