18.06 Problem Set 8 Solution
Due Wednesday, 22 April 2009 at 4 pm in 2106.
Total: 160 points.
Problem 1:
If
A
is realsymmetric, it has real eigenvalues. What can you say about
the eigenvalues if
A
is real and antisymmetric (
A
=

A
T
)? Give both a general
explanation for any
n
×
n A
(similar to what we did in class and in the book) and
check by finding the eigenvalues a 2
×
2 antisymmetric example matrix.
Solution
(15 points = 10(proof) + 5(example) )
If
λ
is an eigenvalue of
A
with a nonzero eigenvector
v
, that is
Av
=
λv
. Then,
on one hand, we have
v
H
Av
=
v
H
λv
=
λ
k
v
k
2
, and on the other hand,
v
H
Av
= (

v
H
A
T
)
v
=

(
Av
)
H
v
=

(
λv
)
H
v
=

¯
λ
k
v
k
2
.
Since
v
is nonzero,
k
v
k
2
>
0.
We conclude that
λ
=

¯
λ
.
This implies that
λ
is
purely imaginary, that is the real part of
λ
is zero.
For example, we take
A
=
0
1

1
0
. (Since
A
is antisymmetric, its diagonal
entries must be zero.) We then solve det(
A

λI
) =
λ
2
+1 = 0 to get
λ
1
=
i, λ
2
=

i
.
They are purely imaginary numbers.
Problem 2:
Find an orthogonal matrix
Q
that diagonalizes
A
=

2
6
6
7
, i.e. so
that
Q
T
AQ
= Λ where Λ is diagonal. What is Λ?
Solution
(10 points)
Since
A
is realsymmetric, we should be able to get orthonormal eigenvectors,
and then
Q
is just the matrix whose columns are the eigenvectors (as in class and
the textbook), and Λ is the diagonal matrix of eigenvalues. So, we just solve for the
eigenvalues and eigenvectors of
A
. To get the eigenvalues, we solve det(
A

λI
) =
0 =
λ
2

5
λ

50, obtaining
λ
1
= 10 and
λ
2
=

5. Since the eigenvalues are distinct,
the eigenvectors are automatically orthogonal, and we just need to normalize them
to have length 1:
λ
1
= 10
,
A

λ
1
I
=

12
6
6

3
,
v
1
=
1
2
,
q
1
=
1
/
√
5
2
/
√
5
λ
2
=

5
,
A

λ
2
I
=
3
6
6
12
,
v
2
=

2
1
,
q
2
=

2
/
√
5
1
/
√
5
.
1
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Hence, we have
Q
=
1
√
5

2
√
5
2
√
5
1
√
5
!
,
Λ =
10
0
0

5
Problem 3:
Even if the real matrix
A
is rectangular, the block matrix
B
=
0
A
A
T
0
is symmetric. An eigenvector
~x
of
B
satisfies
B~x
=
λ~x
with:
~x
=
~
y
~
z
,
0
A
A
T
0
~
y
~
z
=
λ
~
y
~
z
,
and thus
A~
z
=
λ~
y
and
A
T
~
y
=
λ~
z
.
(a) Show that

λ
is also an eigenvalue of
B
, with the eigenvector (
~
y,

~
z
)
T
.
(b) Show that
A
T
A~
z
=
λ
2
~
z
, so that
λ
2
is an eigenvalue of
A
T
A
.
(c) Show that
λ
2
is also an eigenvalue of
AA
T
by finding a corresponding eigen
vector.
(d) If
A
=
I
(2
×
2), find all four eigenvalues and eigenvectors of
B
.
Solution
(25 points = 5+5+5+10)
(a) We check this by direct computation.
B
~
y

~
z
=

A~
z
A
T
~
y
=

λ~
y
λ~
z
=

λ
~
y

~
z
.
Hence

λ
is also an eigenvalue of
B
, with the eigenvector
~
y

~
z
.
(b) Again, we check by direct computation.
A
T
A~
z
=
A
T
(
λ~
y
) =
λA
T
~
y
=
λ
(
λ~
z
) =
λ
2
~
z.
Hence,
λ
2
is an eigenvalue of
A
T
A
with eigenvector
~
z
.
(c) By “symmetry”, it is not hard to guess that
~
y
may be an eigenvector of
AA
T
.
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 Spring '08
 Strang
 Matrices, Eigenvalues, Singular value decomposition, Orthogonal matrix, AZ

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