Math 2270, Fall 2015
Instructor: Thomas Goller
8 December 2015
Test #5 Solutions
1
Inventions (10 points)
(a) Invent a unit vector in R4 with no zero entries. (2 points)
Possible solution:
2 3
1
7
16
617
2 415
1
2
3
3
(b) Invent a vector with no zero entries that is orthogonal to 4 7 5. (2 points)
2
Possible solution:
2 3
1
415
5
(c) Invent vectors ~x1 , ~x2 in R2 such that the orthogonal complement of Span{~x1 , ~x2 } is a
line. (2 points)
Possible solution:

1
~x1 =
,
0
~x2 =

2
0
(d) Invent a symmetric 2 ⇥ 2 matrix of rank 1. (2 points)
Possible solution:

1 2
2 4
(e) Invent a matrix that is diagonalizable but not orthogonally diagonalizable. (2 points)
Possible solution:

2 1
0 3
Math 2270, Fall 2015
2
Instructor: Thomas Goller
8 December 2015
Orthogonal Projection (10 points)
8 2 3 2 39
9 =
< 2
4
5
4
1 , 15 of R3 .
Consider the subspace W = Span
:
;
1
5
(a) Compute an orthogonal basis B = {~u1 , ~u2 } of W . (3 points)
Possible solution:
2 3
2
~u1 = 415 ,
1
2 3
9
~u2 = 415
5
2 3 2 3
2
9
18 + 1 + 5 4 5 4 5
1 = 1
4+1+1
1
5
2 3 2 3
2
1
4 415 = 4 35
1
1
2
3
9
(b) Let ~y = 4 2 5. Compute projW ~y . (4 points)
4
Solution:
projW ~y =
2 3
2 3
2
1
18 + 2 + 4 4 5
9 6+44 5
1 +
3 =
6
1+9+1
1
1
2 3
2
2 415
1
2
3 2 3
1
5
4 35 = 4 1 5
1
3
(c) What is the closest point in W to ~y ? (1 point)
Solution:
2
3
5
415
3
(d) Compute the distance from ~y to W . (2 points)
Solution:
k~y
2
3
9
projW ~y k = 4 2 5
4
2
3
2 3
5
4
p
p
4 1 5 = 4 1 5 = 16 + 1 + 49 = 66
3
7
Math 2270, Fall 2015
3
Instructor: Thomas Goller
8 December 2015
Least Squares Approximation (10 points)
2
3
2 3
1 0
4
~
~
4
5
4
Consider the equation A~x = b, where A = 3 1 and b = 15.
0 1
0
(a) Determine whether A~x = ~b has a solution. (2 points)
Solution:
2
3
2
1 0 4
1 0
4 3 1 15 ! 4 0 1
0 1 0
0 1
3
2
4
1 0
5
4
11 ! 0 1
0
0 0
3
4
0 5
11
Since the last row is all zeros except a nonzero number in the augmented column, the
system has no solution.
(b) Compute the unique solution ~x0 of AT A~x = AT~b. (5 points)
Solution:
2
3
2 3



1 0
4
1 3 0 4
10
3
1
3
0
T
T
415 = 7
3 15 =
A A=
,
A ~b =
0 1 1
3 2
0 1 1
1
0 1
0




1 2
1 11
3 7
1
T
1 T~
=
=
~x0 = (A A) A b =
3 10 1
11
1
11
11

(c) Compute kA~x0
~bk. (2 points)
Solution:
A~x0
2
3
2 3 2 3 2 3 2 3
1 0 
4
1
4
3
~b = 43 15 1
415 = 4 2 5 415 = 4 1 5
1
0 1
0
1
0
1
2 3
3
p
p
4 1 5 = 9 + 1 + 1 = 11
1
(d) What can you say about ~x0 to relate it to the original equation A~x = ~b? (1 point)
Solution: ~x0 is the vector such that A~x0 is the best approximation to ~b in Col A.
Math 2270, Fall 2015
4
Instructor: Thomas Goller
8 December 2015
Orthogonal Diagonalization (10 points)
Let A =

4
2
2
.
7
(a) Why can you immediately tell that A is diagonalizable? (1 point)
Solution: A is symmetric.
(b) Compute the eigenvalues of A. (3 points)
Solution:
4
2
2
so
1
= 8,
2
7
=
2
11 + 24 = (
8)(
3),
= 3.
(c) Compute a basis {~v1 , ~v2 } of R2 consisting of orthonormal eigenvectors of A. (4 points)
Solution:
A
A
8I =
3I =



2
2 1
!
,
1
0 0

2
1
2
!
,
4
0 0
4
2
1
2
1
~v1 = p
5

1
2

1 2
~v2 = p
5 1
(d) Write down the diagonalization A = P DP T , where the columns of P are orthonormal
eigenvectors and D is diagonal. (2 points)
Solution:
1
A= p
5

1 2
2 1

8 0 1
p
0 3
5

1 2
2 1
Math 2270, Fall 2015
5
Let
2
2
44
0
Instructor: Thomas Goller
8 December 2015
Singular Value Decomposition (10 points)

1 2
1
1 2 1
3 2 p
4 0
1/p5
8 05 = 42/ 5
0 2
0
A =
and assume that
p 32
0
2/p 5
10
5
4
0
0 1/ 5
0
1
0
an orthogonal diagonalization of AT A is given by
p
32 p
3
0 0
1/ 5 2/ 5 0
2 05 4 0p
0p 15.
0 0
2/ 5 1/ 5 0
(a) Compute the singular value decomposition A = U ⌃V T . (Hint: V and ⌃ can easily be
obtained from the given information!) (7 points)
2 p
1/p5 0
4
V = 2/ 5 0
0
1
p 3
p
2/p 5
10 p0 0
5
⌃=
1/ 5 ,
0
2 0
0
2 3



1
1
1 1 2
1 5
1 1
1 4 5 1
2 p = p
~u1 = A~v1 = p
=p
10 1 2 1
5
5 2 5
2 1
1
0
2 3


0
1
1 1 2
1
1 4 5
1
0 =p
~u2 = A~v2 = p
1
2
1
1
2
2
2
1
p
2 p
3


p
1/ 5 2/ 5 0
1 1
1 2
1
1
10 p0 0 4
0p
0p 15
=p
1 2 1
0
2 0
2 1 1
2/ 5 1/ 5 0
(b) What is the rank r of A? (1 point)
Solution: r = 2
(c) Compute the best rank 1 approximation of A. (2 points)
Solution:
u1~v1T
1~
=
p


⇤ 1
1 1 ⇥
1 2 0
1 2 0 p =
10 p
1 2 0
2 1
5