Math 2270, Fall 2014
Instructor: Thomas Goller
25 November 2014
Test #5 Solutions
1
Singular Value Decomposition (12 points)

2
1
. Compute the singular value decomposition A = U ⌃V T of A. Show all
4
2
of your work! (Hint: start by computing AT A.)
Let A =
Solution: First, we compute

2
A A=
1
T
4
2


1
20 10
=
.
2
10 5
2
4
The eigenvalues of AT A are the roots of the polynomial
20
10
so
1
10
5
2
=
25 + 100
100 = (
25),
p
= 25 and 2 = 0 and 1 = 25 = 5. Now we compute unit eigenvectors of AT A:
✓
◆
✓
◆
✓ ◆

20 25
10
5 10
2
2
1
N
=N
= span
,
so ~v1 = p5
;
10
5 25
10
20
1
1
✓
◆
✓
◆
✓
◆

20 0 10
20 10
1
1
1
N
=N
= span
,
so ~v2 = p5
.
10
5 0
10 5
2
2
From ~v1 , we get
~u1 =
1
1
A~v1 =
We get ~u2 by computing
T
N (A ) = N
✓
2
1
4
2
1
5

◆
2
4
1
2
= span
p1
5
✓
2
1

2
=
1
◆
Thus the singular value decomposition of A is



2
1
1
2
5 0
1
= p5
4
2
2 1 0 0
p1
5
,

1
.
2
so ~u2 =
p1
5

2 1
.
1 2
p1
5

2
.
1
Math 2270, Fall 2014
2
Instructor: Thomas Goller
25 November 2014
Inventions (10 points)
(a) Invent a 2 ⇥ 2 matrix A that rotates every vector in R2 by 180 . (2 points)
Solution: A =

1
0
0
1
(b) Invent a 3 ⇥ 3 matrix J with eigenvalues 7, 7, 7 and only one independent eigenvector.
(2 points)
2
3
7 1 0
Possible solution: 40 7 15
0 0 7
(c) Let A =

1
3
0
. Invent a matrix B that is similar to A but not equal to A. (2 points)
1
Possible solution:

1
0
0
1
(d) Invent a basis B = {~v1 , ~v2 } for R such that the B-coordinates of the vector

1
. (2 points)
0
2
Possible solution: B =
⇢
Possible solution: T
v1
v2
1000
are
90

1000
1
,
90
0
(e) Invent a linear transformation R2
✓

◆
=

v1
0
T
/
R2 that has a non-trivial kernel. (2 points)
Math 2270, Fall 2014
3
Instructor: Thomas Goller
25 November 2014
Linear Transformations in R2 (10 points)
◆ 
2
1
/ R is linear and that T
(a) Suppose you know that R
=
. Compute each
3
4
of the following if you can, or state that not enough information is given: (4 points)
✓ ◆

0
0
(i) T
Solution:
0
0
✓
◆

4
2
(ii) T
Solution:
6
8
✓ ◆
3
(iii) T
Solution: Not enough information.
2
✓ ◆
1
(iv) T
Solution: Not enough information.
4

1 1
T / 2
(b) Let A =
and consider R2
R defined by T (~v ) = A~v .
0
1

1
(i) Draw the line spanned by
and draw what you get when T transforms that line.
1
(2 points)
2
T
✓
2

1
is the line through the origin
1
✓ ◆

1
0
with slope 1. T transforms that line to the line spanned by T
=
,
1
1
which is the y-axis.
Description of solution: The line spanned by




0
1
0
1
(ii) Draw the square with vertices
,
,
,
and draw what you get when T
0
1
1
0
transforms that square. (4 points)
Description

 of solution:

T transforms the square into the parallelogram with ver0
1
1
0
tices
,
,
,
.
0
0
1
1
Math 2270, Fall 2014
4
Instructor: Thomas Goller
25 November 2014
Change of Basis (10 points)
(a) Consider the basis B =
following vectors in R2 :

4
(i)
Solution:
1

2
(ii)
Solution:
0

0
(iii)
Solution:
1
⇢

4
2
,
of R2 . Compute the B-coordinates of each of the
1
0
(3 points)

1
0

0
1

1
2
(b) What is the change of basis matrix M that converts B-coordinates to standard coordinates? Check your M by multiplying each of your answers in (a) by M . (4 points)

4 2
1 0



4 2 1
4
=
,
1 0 0
1
Solution: M =

4 2
1 0


0
2
=
,
1
0

4 2
1 0


1
0
=
.
2
1
(c) What is the change
of basis

 matrix N that converts standard coordinates to B-coordinates?
4
2
Compute N
and N
. You should get your answers in (a). (3 points)
1
0
Solution: N = M
1
1
2
=

0
1
2


0
2
0
= 1
1 4
2


1
4
1
=
,
2 1
0
1
.
2

0
1
2
1
2


2
0
=
.
0
1
Math 2270, Fall 2014
5
Instructor: Thomas Goller
25 November 2014
Matrix for the Second Derivative (8 points)
In this problem, let V denote the vector space of all polynomials in x of degree  3. Let W
denote the vector space of all polynomials in x of degree  1. Let B = {1, x, x2 , x3 } be a
basis for V and let B 0 = {1, x} be a basis for W .
T
/ W be the linear transformation that takes the second derivative. For exam(a) Let V
ple, T (x2 2x3 ) = 2 12x. Compute the B-coordinates of x2 2x3 and the B 0 -coordinates
of 2 12x. (2 points)
Solution: [x
2
2
3
0
607
7
2x3 ]B = 6
4 1 5,
2
[2
12x]B0 =

2
.
12
(b) Compute the B 0 -coordinates of T (1), T (x), T (x2 ), and T (x3 ). Construct the 2⇥4 matrix
A that represents the second derivative in these bases B and B 0 . (5 points)
Solution:
[T (1)]B0 = [0]B0
[T (x2 )]B0 = [2]B0


0
0
=
,
[T (x)]B0 = [0]B0 =
,
0
0


2
0
3
=
,
[T (x )]B0 = [6x]B0 =
0
6

0 0 2 0
A=
0 0 0 6
(c) Check your A is correct by multiplying A by the B-coordinates of x2
(a). You should get the B 0 coordinates of 2 12x. (1 point)
Solution:
2
3
0


7
0 0 2 0 6
0
6 7= 2 .
0 0 0 6 415
12
2
2x3 you found in