Homework 7 Solutions

Joshua Hernandez
November 16, 2009
5.2 - Diagonalizability
2. For each of the following matrices A M
nn
(R), test A for diagonalizability, and if A is diagonalizable,
nd an invertible matrix Q and a diagonal matrix D such that Q
-1
AQ = D.
b. A =
_
1 3
3 1
_
Solution: Computing eigenvalues:

A
() = det
_
1 3
3 1
_
= (1 )
2
3
2
This polynomial has roots 13 = 4, -2. Two distinct eigenvalues mean that A is diagonalizable.
Then
E
4
= N
_
1 4 3
3 1 4
_
= N
_
-3 3
3 -3
_
= span
__
1
1
__
E
-2
= N
_
1 (-2) 3
3 1 (-2)
_
= N
_
3 3
3 3
_
= span
__
1
-1
__
.
Our diagonalization is therefore
A =
_
1 1
1 -1
__
4 0
0 -2
__
1 1
1 -1
_
-1
=: QDQ
-1
.
d. A =
_
_
7 -4 0
8 -5 0
6 -6 3
_
_
Solution: Computing eigenvalues:

A
() = det
_
_
7 -4 0
8 -5 0
6 -6 3
_
_
= (7 )(-5 )(3 ) (-4 8(3 ))
= (3 )(
2
2 3) = (3 )( 3)( + 1).
This polynomial has roots 3 and -1 (a repeated root means that A might not be diagonalizable).
E
3
= N
_
_
7 3 -4 0
8 -5 3 0
6 -6 3 3
_
_
= N
_
_
4 -4 0
8 -8 0
6 -6 0
_
_
= span
_
_
_
_
_
1
1
0
_
_
,
_
_
0
0
1
_
_
_
_
_
1
E
-1
= N
_
_
7 (-1) -4 0
8 -5 (-1) 0
6 -6 3 (-1)
_
_
= N
_
_
8 -4 0
8 -4 0
6 -6 4
_
_
= span
_
_
_
_
_
2
4
3
_
_
_
_
_
.
The two eigenspaces have a total dimension of 3, so A is diagonalizable. Our diagonalization is
therefore
A =
_
_
1 0 2
1 0 4
0 1 3
_
_
_
_
3 0 0
0 3 0
0 0 -1
_
_
_
_
1 0 2
1 0 4
0 1 3
_
_
-1
=: QDQ
-1
.
f. A =
_
_
1 1 0
0 1 2
0 0 3
_
_
Solution: Since A is an upper-triangular matrix, we can read its eigenvalues o the diagonal:
= 1, 3. Then
E
1
= N
_
_
1 1 1 0
0 1 1 2
0 0 3 1
_
_
= N
_
_
0 1 0
0 0 2
0 0 2
_
_
= span
_
_
_
_
_
1
0
0
_
_
_
_
_
.
We neednt bother to compute E
3
. The dimension of E
1
is 1, although the root = 1 has
multiplicity 2. Therefore A is not diagonalizable.
3b. Let V = P
2
(R). Dene T : V V by the mapping T(ax
2
+ bx + c) = cx
2
+ bx + a. If diagonalizable,
nd a basis for V such that [T]

is a diagonal matrix.
Solution: If = 1, x, x
2
is the standard basis on P
2
(R), then
A := [T]

=
_
_
0 0 1
0 1 0
1 0 0
_
_
.
Computing eigenvalues of A:

A
() = det
_
_
0 0 1
0 1 0
1 0 0
_
_
= (0 )
2
(1 ) (1 ) = (1 )(
2
1)
= -(1 )
2
(1 +).
The roots of this polynomial are = 1. Now,
E
1
= N
_
_
0 1 0 1
0 1 1 0
1 0 0 1
_
_
= N
_
_
-1 0 1
0 0 0
1 0 -1
_
_
= span
_
_
_
_
_
1
0
1
_
_
,
_
_
0
1
0
_
_
_
_
_
E
-1
= N
_
_
0 (-1) 0 1
0 1 (-1) 0
1 0 0 (-1)
_
_
= N
_
_
1 0 1
0 2 0
1 0 1
_
_
= span
_
_
_
_
_
1
0
-1
_
_
_
_
_
.
2
Thus = (1, 0, 1), (0, 1, 0), (1, 0, -1) is a diagonalizing basis of L
A
. Noting that
[L
A
]

= [L
[T]

= [

T
-1

= [T]

-1

()
, (1)
we know that
-1

() = 1 +x
2
, x, 1 x
2
is a diagonalizing basis of T (equation (1) justies the
obvious nal step of converting the vectors of into their corresponding polynomials).
8. Suppose that A M
nn
n(F) has two distinct eigenvalues,
1
and
2
, and that dim(E

1
) = n 1. Prove
that A is diagonalizable.
Solution: Distinct eigenspaces intersect trivially, and any eigenspace has dimension 1, so
dim(E

1
+E

2
) = dim(E

1
E

2
) = dim(E

1
) + dim(E

2
) (n 1) + 1 = n.
The eigenspaces of A span F
n
, and so A is diagonalizable.
11. Let A be an nn matrix that is similar to an upper triangular matrix, and has the distinct eigenvalues

1
,
2
, . . . ,
k
with corresponding multiplicities m
1
, m
2
, . . . , m
k
. Prove the following statements.
Lemma: If A and B are similar matrices, then
A
() =
B
().
Let A = Q
-1
BQ for some matrix Q M
nn
(F). By the multiplicative property of determinants,

A
() = det(AI) = det(Q
-1
BQI) = det(Q
-1
(B I)Q) = det(B I) =
B
().
Solution: Let M be an upper-triangular matrix such that A = QMQ
-1
. It was proved (5.4:9)
that the eigenvalues of M coincide with the diagonal elements M
ii
. Since similar transforma-
tions have the same characteristic polynomials (lemma, above), they share eigenvalues
i
and
multiplicities m
i
.
a. trace(A) =

k
i=1
m
i

i
Solution: By (2.5:10),
trace(A) = trace(M) =
n

i=1
M
ii
=
k

i=1
m
i

i
. (2)
b. det(A) = (
1
)
m
1
(
2
)
m
2
(
n
)
m
n
.
Solution: The determinant of an upper-triangular matrix is the product of its diagonal entries
M
ii
(determinant property 4). By the multiplicative property of determinants,
det(A) = det(QMQ
-1
) = det(M) =
n

i=1
M
ii
=
n

i=1

m
i
i
. (3)
Finally, one can show (the proof is a little too complicated to show here, but see pp.370,385 in the
text) that every matrix is similar, over some eld, to an upper-triangular matrix. The identities (2)
and (3) above are therefore universal properties of matrices.
3
14a. Find the general solution to the system of dierential equations
x

= x +y, y

= 3x y. (4)
Solution: Let V = (

(R, R
2
) be the space of smooth curves in R
2
. We can consider the
derivative as a linear operator D : V V. Then
D
_
x
y
_
=
_
x

_
=
_
x +y
3x y
_
=
_
1 1
3 -1
__
x
y
_
=: A
_
x
y
_
.
We diagonalize A in the usual fashion:

A
() = det
_
1 1
3 -1
_
= (1 )(-1 ) 3 =
2
4.
This has roots = 2. Computing eigenspaces:
E
2
= N
_
1 2 1
3 -1 2
_
= N
_
-1 1
3 -3
_
= span
__
1
1
__
E
-2
= N
_
1 (-2) 1
3 -1 (-2)
_
= N
_
3 1
3 1
_
= span
__
1
-3
__
.
We have a basis of eigenvectors = (
1
1
) , (
1
-3
). Let be the standard basis of R
2
. Changing
basis,
_
x
y
_
= [I]

_
f
1
f
2
_
=
_
1 1
1 -3
__
f
1
f
2
_
,
where and f
1
(t) and f
2
(t) satisfy
f

1
(t) = 2f
1
(t) and f

2
(t) = -2f
2
(t).
These dierential equations have solutions f
1
(t) = c
1
e
2t
and f
2
(t) = c
2
e
-2t
. Observe, then,
_
x(t)
y(t)
_
=
_
1 1
1 -3
__
f
1
(t)
f
2
(t)
_
=
_
c
1
e
2t
+c
2
e
-2t
c
1
e
2t
3c
2
e
-2t
_
.
18. Two linear operators T and U on an n-dimensional vector space V are called simultaneously diago-
nalizable if there exists some basis of V such that [T]

and [U]

are diagonal matrices.

k=1
(D
1
)
ik
(D
2
)
kj
=
n

k=1

ik
(D
1
)
ik

kj
(D
2
)
kj
=
ij
(D
1
)
ii
(D
2
)
ii
=
ij
(D
2
)
ii
(D
1
)
ii
= (D
2
D
1
)
ij
.
4
Solution: Let be a basis of V that diagonalizes both T and U. Since diagonal matrices
commute with each other (lemma, above),
[TU]

= [T]

[U]

= [U]

[T]

= [UT]

.
Now we can relate the two operators in the same way:
TU =
-1

L
[TU]

=
-1

L
[UT]

= UT.
6.1 - Inner Products and Norms
5. In C
2
, show that x, y) = xAy

is an inner product, where

= (x
1
A+cx
2
A)y

= x
1
Ay

+x
2
Ay

= x
1
, y) +c x
2
, y)
2. Symmetry: Observe that
A

=
_
1 i
-i 2
_

=
_

1

-i

i

2
_
=
_
1 i
-i 2
_
= A.
Thus (observing that the Hermitian adjoint of a scalar is just its complex conjugate),
y, x) = yAx

= ((x

= (xAy

= xAy

= x, y).
3. Coercivity:
(x
1
, x
2
), (x
1
, x
2
)) =
_
x
1
x
2
_
_
1 i
-i 2
__
x
1
x
2
_
=
_
x
1
x
2
_
_
x
1
+i x
2
- x
1
+ 2 x
2
_
= [x
1
[
2
+ 2[x
2
[
2
If (x
1
, x
2
) ,= (0, 0), then the RHS is a positive real number.
Compute x, y) for x = (1 i, 2 + 3i) and y = (2 +i, 3 2i).
Solution:
x, y) =
_
1 i 2 + 3i
_
_
1 i
-i 2
__
2 +i
3 2i
_
=
_
1 i 2 + 3i
_
_
1 i
-i 2
__
2 i
3 + 2i
_
=
_
1 i 2 + 3i
_
_
(2 i) +i(3 + 2i)
-i(2 i) + 2(3 + 2i)
_
=
_
1 i 2 + 3i
_
_
2i
5 + 2i
_
= (1 i)(2i) + (2 + 3i)(5 + 2i) = (2 + 2i) + (4 + 19i) = 6 + 21i.
9. Let be a basis for a nite-dimensional inner product space.
a. Prove that if x, z) = 0 for all z , then x = 0.
5
Solution: Since is a spanning set, we may write x =

n
i=1
c
i
z
i
, with all z
i
. Then
x, x) =
_
x,

n
i=1
c
i
z
i
_
=
n

i=1
c
i
x, z) = 0.
Coercivity of the inner product implies that x = 0.
b. Prove that if x, z) = y, z) for all z , then x = y.
Solution: If x, z) = y, z), then x y, z) = x, z) y, z) = 0 for all z . By the above,
x y = 0, so x = y.
10. Let V be an inner product space, and suppose that x and y are orthogonal vectors in V. Prove that
|x +y|
2
= |x|
2
+ |y|
2
. Deduce the Pythagorean theorem in R
2
.
Solution:
|x +y|
2
= x +y, x +y) = x, x) + x, y) + y, x) + x, y) .
By orthogonality of x and y,
= x, x) + y, y) = |x|
2
+ |y|
2
.
In R
2
, if two orthogonal vectors x and y are laid head-to-tail, they form two edges of a right
triangle, of which x + y forms the third edge. If we denote the lengths of these edges by a, b,
and c, respectively, then
c
2
= |x +y|
2
= |x|
2
+ |y|
2
= a
2
+b
2
.
Thus we prove the Pythagorean Theorem.
12. Let v
1
, v
2
, . . . , v
k
be an orthogonal set in V, and let a
1
, a
2
, . . . , a
k
be scalars. Prove that
_
_
_
_
_
k

i=1
a
i
v
i
_
_
_
_
_
2
=
k

i=1
[a
i
[
2
|v
i
|
2
.
Solution: Clearly, in the case that k = 1,
_
_
_
_
_
k

i=1
a
i
v
i
_
_
_
_
_
2
= |a
1
v
1
|
2
= [a
1
[
2
|v
1
|
2
=
k

i=1
[a
i
[
2
|v
i
|
2
.
Now, assume the result is proven for sets of k1 or fewer orthogonal vectors. Suppose v
1
, . . . , v
k
are orthogonal, and dene b
k1
=

k1
i=1
a
i
k
i
. Observe that
b
k1
, a
k
) =
_
k1

i=1
a
i
v
i
, a
k
v
k
_
=
k1

i=1
a
i
v
i
, v
k
) = 0,
6
so a
k
and b
k1
are orthogonal. Applying the previous problem,
_
_
_
_
_
k

i=1
a
i
v
i
_
_
_
_
_
2
= |a
k
v
k
+b
k1
|
2
= [a
k
v
k
|
2
+ |b
k
|
2
.
By assumption,
= [a
k
[
2
|v
k
|
2
+
k1

i=1
[a
i
[
2
|v
i
|
2
=
k

i=1
[a
i
[
2
|v
i
|
2
.
17. Let T be a linear operator on an inner product space V, and suppose that |T(x)| = |x| for all x. Prove
that T is one-to-one.
Solution: If v N(T), then |v| = |T(v)| = |0| = 0. By coercivity of the norm, v = 0. Thus
T is one-to-one.
7