MATH 115A - Assignment Seven Extra Solutions
Paul Skoufranis
February 26, 2012
For Question 3 of Section 5.1, for each matrix
A
∈ M
n
×
n
(
F
)
, determine all eigenvalues of
A
and for each
eigenvalue
λ
of
A
, find the set of eigenvectors corresponding to
λ
. If possible, find a basis of
F
n
consisting
of eigenvectors of
A
and determine an invertible matrix
Q
and a diagonal matrix
D
such that
Q
-
1
AQ
=
D
.
§
5.1 Question 3a)
A
=
1
2
3
2
for
F
=
R
.
Solution
:
We desire to find the eigenvalues and eigenvector of
A
.
First we compute the eigenvalues.
To find the eigenvalues, we compute the characteristic polynomial:
χ
A
(
λ
) = det(
λI
-
A
) = det
λ
-
1
-
2
-
3
λ
-
2
= (
λ
-
1)(
λ
-
2)
-
6 =
λ
2
-
3
λ
-
4 = (
λ
-
4)(
λ
+ 1)
Thus the eigenvalues of
A
are 4 and
-
1.
Now we shall compute the eigenspaces. We notice that
E
4
=
ker
(4
I
-
A
) =
ker
3
-
2
-
3
2
=
ker
3
-
2
0
0
=
span
(
{
(2
,
3)
}
)
where the third equality comes from a simple row reduction. Therefore
{
(2
,
3)
}
is a basis for
E
4
.
Next we notice that
E
-
1
=
ker
(
-
I
-
A
) =
ker
-
2
-
2
-
3
-
3
=
ker
1
1
0
0
=
span
(
{
(1
,
-
1)
}
)
where the third equality comes from a simple row reduction. Therefore
{
(1
,
-
1)
}
is a basis for
E
-
1
.
Combining these two bases, we see that
β
=
{
(2
,
3)
,
(1
,
-
1)
}
is an eigenbasis for
A
(and thus a basis for
R
2
consisting of eigenvectors).
Finally, to find the desired matrices
Q
and
D
, we notice that, if
γ
is the
standard basis for
R
2
, then
Q
-
1
AQ
=
D
where
D
=
4
0
0
-
1
and
Q
is the change of basis matrix that takes
β
-coordinates to
γ
-coordinates.
Therefore, by using the
definition of the change of basis matrix
Q
=
2
1
3
-
1
which completes the problem.
§
5.1 Question 3b)
A
=
0
-
2
-
3
-
1
1
-
1
2
2
5
for
F
=
R
.
Solution
:
We desire to find the eigenvalues and eigenvector of
A
.
First we compute the eigenvalues.
To find the eigenvalues, we compute the characteristic polynomial:
χ
A
(
λ
) = det
λ
2
3
1
λ
-
1
1
-
2
-
2
λ
-
5
=
λ
(
λ
-
1)(
λ
-
5) + (2)(1)(
-
2) + (3)(1)(
-
2)
-
(3)(
λ
-
1)(
-
2)
-
(2)(1)(
λ
-
5)
-
λ
(1)(
-
2)
= (
λ
3
-
6
λ
2
+ 5
λ
)
-
4
-
6 + (6
λ
-
6)
-
(2
λ
-
10) + 2
λ
=
λ
3
-
6
λ
2
+ 11
λ
-
6
Therefore, to find the eigenvalues of
A
, we need to factor
χ
A
.
One way is to notice that
χ
A
(1) = 0 (we
would try 1 as 1 divides
-
6 (see the Rational Roots Theorem)) and then use long division of polynomials.
Another way is to notice that
χ
A
(
λ
) = (
λ
3
-
6
λ
2
+ 5
λ
) + (6
λ
-
6)
=
λ
(
λ
-
5)(
λ
-
1) + 6(
λ
-
1)
= (
λ
-
1)(
λ
(
λ
-
5) + 6)
= (
λ
-
1)(
λ
2
-
5
λ
+ 6) = (
λ
-
1)(
λ
-
2)(
λ
-
3)
Thus the eigenvalues of
A
are 1, 2, and 3.
Now we shall compute the eigenspaces. We notice that
E
3
=
ker
3
2
3
1
2
1
-
2
-
2
-
2
=
ker
1
0
1
0
1
0
0
0
0
=
span
(
{
(1
,
0
,
-
1)
}
)
where the second equality comes from a simple row reduction. Therefore
{
(1
,
0
,
-
1)
}
is a basis for
E
3
.
Next we notice that
E
2
=
ker
2
2
3
1
1
1
-
2
-
2
-
3
=
ker
1
1
0
0
0
1
0
0
0
=
span
(
{
(1
,
-
1
,
0)
}
)
where the second equality comes from a simple row reduction. Therefore
{
(1
,
-
1
,
0)
}
is a basis for
E
2
.
