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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

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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