Download Final Exam Solutions

1. (5 points each) State whether the following are true or false, and justify all of your
answers.
0 
a. Let A be a 3×3 matrix with eigenvectors v1 = 1  , v2 =
 5 
diagonalizable.
1 
2 , v =
  3
 8 
4
  1 . A is
 
 0 
Solution:
TRUE. These vectors are linearly independent since:
0 1 4 
1 2  1 ~


5 8 0 
1 2  1 1 2  1
0 1
4  ~ 0 1 4   the columns are linearly independent.

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
e. Let A = 2 3 7 . The vector v =


3 6 9
 10 
  1  is in Nul(A).
 
  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
1 0
If A = 
, then the eigenvalues of A are 1 and 2. But A2 = 

 , and its
0 2 
0 4 
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
2. Let u =  2  , v =
 3
0 
4 , w =
 
 2 
5 
 6  be vectors in R3.
 
  7 
a. (5 points) What is the distance from u to w?
Solution:
 4
dist(u, w) = ║u – w║ = ║   4 ║ = 16  16  16 =
 
 4 
48 = 4 3
b. (10 points) Find the component of u orthogonal to v.
Solution:
Let L = Span{v}.
1
1
0   1 
uv
2   




4 = 1.6 
v = 2 –
u – projLu =  2  –
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!
0   5 
 5 
wv
10    


4 = 4 .
v = 6 –
Let v1 = v; v2 = w –
  20    
vv
 2   8
  7 
Then, if W = Span{v, w}:
0
 5   37 21   37 21 
u  v1
u  v2
37
 4  =  190  =  38 
projWu =
v1 +
v2 = .4 +
105
21
105    275   55 
v1  v1
v2  v2
.2
 8  105   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
dist(u, W) = ║u – projWu║ = ║  2  –
 3
 37 21 
 16 21
 38  ║ = ║  4  ║ =
 21 
 21 
  55 21
  8 21 
=
(16) 2  (4) 2  (8) 2
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  up
x  u1
up
u1 + … +
up  up
u1  u1
To show that T is a linear transformation, we must show 2 things:
projW x =
(i) T(x + y) = T(x) + T(y):
( x  y)  u p
( x  y)  u1
up
u1 + … +
up  up
u1  u1
x  up  y  up
x  u1  y  u1
up
=
u1 + … +
up  up
u1  u1
T(x + y) =
(prop’s of dot product)
x  up
y  up
x  u1
y  u1
up +
u p (prop’s of vectors)
u1 + … +
u1 + … +
up  up
up  up
u1  u1
u1  u1
= T(x) + T(y)
=
(ii) T(cx) = cT(x):
(cx)  u p
(cx)  u1
up
u1 + … +
up  up
u1  u1
c( x  u p )
c( x  u1 )
up
=
u1 + … +
up  up
u1  u1
T(cx) =
 x  u1
x  up 

= c

u p  u p 
 u1  u1
= cT(x)
Thus, T is a linear transformation.
(prop’s of dot product)
(prop’s of vectors)
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 
x 
1 
0 
0 
 2 
 
 
 
7x 4  = x2 0  + x4 7  + x5 0  . Therefore a basis is:


 
 
 
 x4 
0 
1 
0 
 x 5 
0 
 0 
1 
  3   0  0  
      
1 0 0 


 0  , 7  , 0  
 0  1  0  
      
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
For  = 0: A – 0I = 
 ~ 
  basis is
 5 25  0 0 
 25  5 1
For  = 26: A – 26I = 
 ~ 
  5  1  0
5
1
 

 1
 basis is  

0
5
1
5
Since these eigenvectors correspond to different eigenvalues, they are orthogonal. So to
get an orthonormal basis we just normalize.
5
u1 =  1

5
P= 1

26

1
26
5
26
26

 1 26 
,
u
=
 2  5

26 
 26 
26

2
2
 and the new quadratic form is 0y1 + 26y2 .

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
For  = 7: A – 7I =  2  1 2  ~  0
0 0 ~ 0

 4
2  4  0
0 0 0
 1 1  


 basis is  2 , 0 
  0  1  
    
1
2
0
0
 1
0 
0 
 5  2 4 1  4  1 1  4  1 1 0
1  ~ 0
For  = -2: A + 2I =  2 8 2 ~ 5  2 4  ~ 0 1
2
  1


5  0 0
0  0 0
 4
2 5 4 2
  1  
 2 
 1  


 basis is    2   OR    1  
 1  
 2  


  
1
1 
2

0 
The basis that we found for the eigenspace of  = 7 is not orthogonal, so we use GramSchmidt to make it orthogonal:
  1
v1 =  2  ; v2 =
 0 
1 
 1  4 5 
0  –  1  2  =  2 
5
 
5    
1 
 0   1 
 2
So {v1, v2} is an orthogonal basis for the eigenspace of  = 7, and {v1, v2,   1  } is an
 
 2 
 2
3
orthogonal basis for R since   1  corresponds to a different eigenvalue than v1 and v2,
 
 2 
which implies that it is orthogonal to both v1 and v2.
Now we just have to normalize:
4
 1 5 

u1 =  2 5  , u2 =  2
5
 0 


 2 3 

, u3 =  1 3  .
45 

 2 3 
45 
45
Thus A = PDPT for
 1 5

P=  2 5
 0

4
45
2
45
5
45
 23

1
3  and D =
2 
3 
7 0 0 
0 7 0  .


0 0  2
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 >
= <u, u> – <u, v> – <v, u> + <v, v>
= <u, u> – 2<u, v> + <v, v>
= ║u║2 – 2<u, v> + ║v║2
(by the def of ║u║2)
(by prop’s of inner product)
(by prop’s of inner product)
(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
a 
c. (10 points) Let u =   and v =
b 
 a2  b2
1
ab

.
Show
that



1
 2 

 2
2
cow-chee

 .

Solution:
We use Cauchy-Schwarz!
║u║ =
<u, v> = a + b
a2  b2
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

2
22
a2  b2
ab

.
 
2
 2 
2
║v║ = 1  1  2