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Math 355 Take Home Final by gddmZl

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									1. (5 points each) State whether the following are true or false, and justify all of your
   answers.

                                                0           1         4
a. Let A be a 3×3 matrix with eigenvectors v1 = 1  , v2 =
                                                 
                                                              2 , v =
                                                                3
                                                                           1 . A is
                                                                           
                                                5 
                                                            8 
                                                                        0
                                                                           
   diagonalizable.

Solution:

TRUE. These vectors are linearly independent since:

0 1 4         1 2  1 1 2  1
1 2  1 ~     0 1   4  ~ 0 1 4   the columns are linearly independent.
                               
5 8 0 
              0  2 5  0 0 13 
                                 

Thus we have 3 linearly independent eigenvectors for a 3×3 matrix, which implies that A
is diagonalizable by The Diagonalization Theorem.

b. det (ATA) ≥ 0.

Solution:

TRUE.

det (ATA) = det (AT)(det A) since det(AB) = (det A)(det B)
          = (det A)(det A) since det A = det (AT)
          = (det A)2 ≥ 0    since det A  R

c. It is possible for a 3×3 matrix with real entries to have the eigenvalues 2, -3, and i.

Solution:

FALSE.

Complex eigenvalues of a real-valued matrix come in conjugate pairs, so -i would also
need to be an eigenvalue.
1. (continued)

d. If J is the Jordan canonical form of a matrix A, then A is similar to J.

Solution:

TRUE.

If J is the Jordan canonical form of a matrix A, then A = PJP-1, which implies that A is
similar to J.

           1 2 3                        10 
e. Let A = 2 3 7  . The vector v =
                 
                                           1  is in Nul(A).
                                          
           3 6 9 
                                        2
                                          

Solution:

FALSE.

A vector v is in Nul(A) if Av = 0.
            1 2 3  10   2 
Here, Av = 2 3 7    1  =  3   0. Thus v  Nul(A).
                         
            3 6 9    2   6 
                         

f. Each eigenvalue of A is also an eigenvalue of A2.

Solution:

FALSE.

       1 0                                                 2   1 0
If A =       , then the eigenvalues of A are 1 and 2. But A = 0 4 , and its
       0 2                                                         
eigenvalues are 1 and 4. Thus the eigenvalue 2 of A is not an eigenvalue of A2.

In general, if  is an eigenvalue of A, then 2 is an eigenvalue of A2.
           1           0          5 
2. Let u =  2  , v =
            
                         4 , w =
                          
                                      6  be vectors in R3.
                                      
             3
                       2
                                    7 
                                      

a. (5 points) What is the distance from u to w?

Solution:

                          4
dist(u, w) = ║u – w║ = ║   4  ║ = 16  16  16 =
                                                             48 = 4 3
                          4 
                          


b. (10 points) Find the component of u orthogonal to v.

Solution:

Let L = Span{v}.

             1               1       0   1 
              2  – u  v v =  2  – 2  4  =  1 .6 
u – projLu =  
                     vv         20                
               3
                               3
                                        2    3 .2 
                                                     


c. (10 points) Find the projection of u onto the subspace of R3 spanned by v and w.
   [NOTE: To use the nice dot product formula, what kind of set/basis must you have??]

Solution:

We must have an orthogonal basis for Span{v, w} – we use Gram-Schmidt!

                              5       0   5 
                     wv      6  – 10  4  =  4  .
Let v1 = v; v2 = w –     v =  
                     vv             20    
                              7 
                                       2    8
                                           
Then, if W = Span{v, w}:

                                    0         5   37 21   37 21 
         u  v1       u  v2
                              v 2 = .4  +
                                            37    190   38 
                 v1 +
projWu =
         v1  v1      v2  v2         105  4  =  105  =  21 
                                    .2 
                                               8  275105   55 21
                                                                     
d. (10 points) Find the distance from u to the subspace of R3 spanned by v and w.

Solution:

Let W = Span{v, w}.

                              1        37 21       16 21
                                                                   (16) 2  (4) 2  (8) 2
dist(u, W) = ║u – projWu║ = ║  2  –
                               
                                         38  ║ = ║  4  ║ =
                                         21         21                  21
                                3
                                        55 21
                                                     8 21 
                                                             
                                                                   336   4 21
                                                               =       =
                                                                   21     21
3. (20 points) Let u1, …, up be an orthogonal basis for a subspace W of Rn, and let
   T: Rn→ Rn be defined by T(x) = projW x. Show that T is a linear transformation.

Solution:

Since u1, …, up is an orthogonal basis, we can write:

                           x  u1            x  up
                  projW x =        u1 + … +         up
                          u 1  u1          up  up
To show that T is a linear transformation, we must show 2 things:

(i) T(x + y) = T(x) + T(y):

              ( x  y )  u1          ( x  y)  u p
T(x + y) =                   u1 + … +                up
                 u1  u 1                up  up
              x  u1  y  u1          x  up  y  up
          =                   u1 + … +                 up          (prop’s of dot product)
                  u1  u1                  up  up
             x  u1            x  up       y  u1            y  up
          =          u1 + … +         up +          u1 + … +         u p (prop’s of vectors)
            u 1  u1          up  up      u 1  u1          up  up
          = T(x) + T(y)

(ii) T(cx) = cT(x):

          (c x )  u1          (cx)  u p
T(cx) =               u1 + … +            up
           u1  u1              up  up
          c( x  u1 )          c(x  u p )
      =               u1 + … +             up       (prop’s of dot product)
           u1  u1              up  up
          x  u1       x  up 
      = c                             (prop’s of vectors)
          u1  u1
                      up  up 
                               
      = cT(x)

Thus, T is a linear transformation.
4. Let a subspace U of R5 be defined by: U = {(x1, x2, x3, x4, x5): x1 = 3x2, x3 = 7x4}

a. (15 points) Find a basis for U.

Solution:

Vectors in U look like:

3 x 2      3        0        0                               3  0   0  
x          1        0        0                                   
 2                                                         1 0 0 
                                                                                       
7x 4  = x2  0  + x4  7  + x5  0  . Therefore a basis is:    0  , 7  , 0  
                                                            0  1  0  
 x4        0        1        0                                   
 x5        0        0        1                             0 0 1 
                                                                 




b. (5 points) What is the dimension of U?

Solution:

The dimension of U is the number of elements in the basis, which is 3.
5. Given the quadratic form on R3: Q(x) = x12 + 25x22 – 10x1x2

a. (5 points) What is the matrix of this quadratic form?

Solution:

    1  5
A=        
    5 25 

b. (15 points) Make a change of variable that transforms this quadratic form into one with
   no cross- product terms. Tell me what the P matrix is, and write what the new
   quadratic form looks like in the new variable.

Solution:

To make a change of variable to satisfy the conditions that we want, we find a P such that
x = Py, and with yT(PTAP)y a quadratic form with no cross product terms (i.e. with PTAP
a diagonal matrix).

So, we want to orthogonally diagonalize A! Meaning we find an orthonormal basis for
R2 such that PTAP is diagonal, where P is the matrix whose columns are the orthonormal
basis.

The eigenvalues of A are the roots of the polynomial:
       (1 – )(25 – ) – 25 = 2 – 26 = ( – 26)

        eigenvalues are 0 and 26

Now we find bases for the eigenspaces:

                     1  5 1  5                                5
For  = 0: A – 0I =         ~     basis is                     1
                     5 25  0 0                                  

                       25  5 1                     1
                                                                      1
For  = 26: A – 26I =                                     basis is  5 
                                                            5
                               ~ 
                        5  1 0                     0             

Since these eigenvectors correspond to different eigenvalues, they are orthogonal. So to
get an orthonormal basis we just normalize.

                                     5    26          1 
                                u1 =  1       , u2 =  5 26 
                                          26          26 
    5              1
                               
P= 1      26
                 5
                          26                                      2      2
                                and the new quadratic form is 0y1 + 26y2 .
           26           26    
6. (20 points) Suppose that T1, …, Tn are injective linear transformations such that
   T1◦…◦ Tn makes sense (i.e. the codomain of Ti+1 is the domain of Ti.) Prove that
   T1◦…◦ Tn is injective.
   [Recall: T1◦…◦ Tn(x) = T1(…(Tn(x))…) = composition of functions]

Solution:

We can prove this in a few ways… I’ll do 2:

(i) A transformation T is one-to-one if an only if ker(T) = {0}

Since T1, …, Tn are injective, ker(T1) = {0}, …, ker(Tn) = {0}.

So now, what is in the kernel of T1◦…◦ Tn?

Assume T1◦…◦ Tn(x) = T1(…(Tn(x))…) = 0. We show that x = 0.

T1 injective  T2(…(Tn(x))…) = 0.
T2 injective  T3(…(Tn(x))…) = 0.
        ………….
Tn-1 injective  Tn(x) = 0.
Tn injective  x = 0.

Thus, T1◦…◦ Tn(x) = 0  x = 0, which means that ker(T1◦…◦ Tn) = {0}, and T1◦…◦ Tn is
injective.

(ii) A transformation T is one-to-one if and only if [T(x) = T(y)  x = y]

So Ti(x) = Ti(y)  x = y for all i = 1, …, n.

Assume T1◦…◦ Tn(x) = T1◦…◦ Tn(y). We show that x = y.

T1 injective  T2(…(Tn(x))…) = T2(…(Tn(y))…)
T2 injective  T3(…(Tn(x))…) = T3(…(Tn(y))…)
        ………….
Tn-1 injective  Tn(x) = Tn(y)
Tn injective  x = y

Thus T1◦…◦ Tn(x) = T1◦…◦ Tn(y)  x = y , which means that T1◦…◦ Tn is injective.
7.
a. (5 points) What is the definition of orthogonally diagonalizable?

Solution:

A matrix A is orthogonally diagonalizable if there exist an orthogonal matrix P and a
diagonal matrix D such that A = PDP-1 = PDPT.

                                                        3  2 4
b. (20 points) Orthogonally diagonalize the matrix A =  2 6 2 , given that its
                                                               
                                                       4
                                                           2 3
   eigenvalues are 7 and -2.

Solution:

We want to find an orthonormal basis for R3.

Since we have the eigenvalues already, we find bases for the eigenspaces:

                     4  2 4                  4  2 4 1     1
                                                                       2    1
For  = 7: A – 7I =  2  1 2  ~
                                     
                                                0
                                                     0 0  ~ 0
                                                                 0       0 
                     4
                           2  4             0
                                                     0 0  0
                                                                 0       0 
                             1 1  
                                        
                basis is   2  , 0  
                               
                            0  1  
                              

                      5  2 4 1  4  1 1  4  1 1 0                      1
For  = -2: A + 2I =  2 8 2 ~ 5  2 4  ~ 0 1
                                                    
                                                             1  ~ 0
                                                              2
                                                                  1
                                                                                  1 
                                                                                   2
                                                                                     
                     4
                            2 5 4 2
                                                  5  0 0
                                                           0  0 0
                                                                                0 
                            1        2 
                           1                 
                basis is    2   OR   1  
                                            
                           1          2  
                                       

The basis that we found for the eigenspace of  = 7 is not orthogonal, so we use Gram-
Schmidt to make it orthogonal:

      1          1          1  4 5 
                            1   2 
v1 =  2  ; v2 =
      
                    0  –
                                2 = 5
                            5    
     0
                  1 
                              0  1
                                  
                                                                              2
So {v1, v2} is an orthogonal basis for the eigenspace of  = 7, and {v1, v2,   1  } is an
                                                                              
                                                                              2 
                                                                              
                                2
orthogonal basis for R since   1  corresponds to a different eigenvalue than v1 and v2,
                        3
                                
                                2 
                                
which implies that it is orthogonal to both v1 and v2.

Now we just have to normalize:

                             1 5        4    45
                                                             2 3 
                                                   
                       u1 =  2 5  , u2 =  2
                                               45 
                                                      , u3 =  1 3  .
                                                                   
                             0 
                                         5                23 
                                                                   
                                                45 



Thus A = PDPT for

                   1 5     4
                                 45
                                       23            7 0 0 
                                                     0 7 0  .
               P=  2 5      2
                                 45
                                       1
                                         3  and D =
                                                              
                   0        5         2              0 0  2
                                                              
                                45      3 
8. (15 points) Show that two vectors u and v of an inner product space V are orthogonal if
   and only if ║u – v║2 = ║u║2 + ║v║2.

Solution:

First, we note that:

║u – v║2 = <u – v, u – v >                          (by the def of ║u║2)
         = <u, u> – <u, v> – <v, u> + <v, v>        (by prop’s of inner product)
         = <u, u> – 2<u, v> + <v, v>                (by prop’s of inner product)
         = ║u║2 – 2<u, v> + ║v║2                    (by the def of ║u║2)

Then ║u – v║2 = ║u║2 + ║v║2  -2<u, v> = 0
                             <u, v> = 0
                             u and v are orthogonal.
9.
a. (5 points) State the Cauchy-Schwarz Inequality.

Solution:

For all u, v in a vector space V, <u, v>  ║u║║v║.



b. (5 points) Circle one: “Cauchy” is pronounced cow-shee coh-shee        cow-chee




                       a           1             ab   a2  b2 
                                                          2

c. (10 points) Let u =   and v =   1 . Show that        2 .
                                                                   
                       b                          2            

Solution:

We use Cauchy-Schwarz!

<u, v> = a + b                ║u║ =      a2  b2              ║v║ = 1  1  2

Then Cauchy-Schwarz   a + b  ( a 2  b 2 )( 2 )

                        (a + b)2  (a2 + b2)(2)

                           (a  b) 2
                                     a2 + b2
                              2

                           (a  b) 2   a2  b2
                                    
                              22          2

                        ab   a2  b2
                                     2

                                    .
                         2       2

								
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