HW 9 Solutions, MATH 110 with Professor Stankova
5.4 #2eg (e) The characteristic polynomial is

λ
(

λ
(1

λ
) + 1) + 1 =

λ
3
+
λ
2

λ
+ 1 =

λ
(
λ
2
+
1) + (
λ
2
+ 1) = (1

λ
)(
λ
2
+ 1). This polynomial has roots
λ
= 1
,
±
i
, so it is diagonalizable.
The eigenvectors are (1
,
0
,
1)
t
,
(1
,

i

1
, i
)
t
,
(1
, i

1
,

i
)
t
, so
Q
=
0
@
1
1
1
0

i

1
i

1
1
i

i
1
A
and
D
=
0
@
1
0
0
0
i
0
0
0

i
1
A
. (g) The characteristic polynomial is (3

λ
)((4

λ
)(1

λ
)+2)

(2(1

λ
)+
2)+(

2+(4

λ
)) = (3

λ
)(4

λ
)(1

λ
)+(4

λ
) = (4

λ
)(
λ
2

4
λ
+4) = (4

λ
)(2

λ
)
2
, so
the roots are
λ
= 4
,
2. However,
A

2
I
=
0
@
1
1
1
2
2
2

1

1
1
1
A
is rank 2, so the 2eigenspace is one
dimensional and we cannot have a basis of eigenvectors, so the matrix is not diagonalizable.
5.4 #3cdf (c) This transformation is not diagonalizable. Any diagonalizable transformation is diagonal
izable when restricted to an invariant subspace, and
T
rotates the
xy
plane by

90 degrees.
(d) This is not diagonalizable. Using the standard basis the matrix is
A
=
0
@
1
0
0
1
1
1
1
1
1
1
A
which
has characteristic polynomial (1

t
)
(

t
); the eigenvalues
t
= 1 has only one eigenvector since
A

I
is rank 2. (f) This matrix is diagonalizable. Its eigenspaces are three 3dimensional
subspace of symmetric matrices (which has eigenvalue 1 under
T
since
T
(
A
) =
A
if
A
is sym
metric) and the onedimensional subspace of skewsymmetric matrices (which has eigenvalue

1).
5.4 #4 Let
β
=
{
b
i
}
be the basis of eigenvectors. Put them into the columns of a matrix
Q
which is
invertible since it is a basis. Then we claim that
Q

1
AQ
is a diagonal matrix. To see this,
note that
Q

1
AQe
i
=
Q

1
Ab
i
=
Q

1
λ
i
b
i
=
λ
i
e
i
.
5.4 #5 (a) If
A
is diagonalizable then det(
A

tI
) splits as a polynomial over
t
. If
A
is diagonalizable,
say
A
=
QDQ

1
, then det(
A

tI
) = det(
QDQ

1

tI
) = det(
Q
(
D

tI
)
Q

1
) = det(
D

tI
) =
Q
(
λ
i

t
).
5.4 #6 (a) A linear transformation is diagonalizable if and only if it has a basis of eigenvectors. Let
E
λ
denote the
λ
eigenspace; this is equivalent to asking that
P
λ
dim(
E
λ
) = dim(
V
). By Theorem
5.7, dim(
E
λ
) is
the multiplicity of the root
λ
in the characteristic polynomial.
Since
eigenspaces for di
↵
erent eigenvalues have zero intersection, if any dim(
E
λ
)
<
the algebraic
multiplicity, then it would be impossible for the dimension of the eigenspaces to be dim(
V
).
Thus, we need equalities across the board, and we need for every root of the characteristic
polynomial to be in the field
F
. (b) Same.
Find
a direct formula for the terms of the sequence:
a
n
+1
= 4
a
n

3
a
n

1
,
a
0
= 0,
a
1
= 1. Solution:
We have
✓
a
n
+1
a
n
◆
=
✓
4

3
1
0
◆ ✓
a
n
a
n

1
◆
=
✓
4

3
1
0
◆
n
✓
a
1
a
0
◆
Denote the matrix
A
. We will diagonalize
A
; it has characteristic polynomial
t
2

4
t
+ 3 =
(
t

3)(
t

1). Its eigenvalues are 1
,
3, with eigenvectors (1
,
1)
t
and (3
,
1)
t
respectively. So,
1
HW 9 Solutions, MATH 110 with Professor Stankova
A
=
QDQ

1
where
Q
=
✓
1
3
1
1
◆
and
D
=
✓
1
0
0
3
◆
. We can compute
Q

1
=
1
2
✓

1
3
1

1
◆
.