Selected Solutions, Leon
§
6.1
6.1.1
Find the eigenvalues and associated eigenspaces of each of the following matrices.
(e)
A
=
±
1 1

2 3
²
. The characteristic polynomial is
p
(
λ
) =
λ
2

4
λ
+ 5, with roots
λ
1
= 2

i
and =
λ
2
= 2 +
i
. We know that the associated eigenvectors will come
in conjugate pairs, so our work is cut in half. We ﬁnd that
A

λ
1
I
=
±

1 +
i
1

2
1 +
i
²
,
and verify that the matrix is singular (row 2 is (1+
i
) times row 1). If the eigenvector
we seek has the form (
x
1
,x
2
)
T
, then setting
x
2
=
s
we have 2
x
1
= (1 +
i
)
s
, or
x
1
=
(
1+
i
2
)
s
. If we choose
s
= 2, we have
x
= (1+
i,
2)
T
. The associated eigenspace
is Span(
x
). The eigenspace associated with
λ
2
, then, is Span
(
(1

i,
2)
T
)
.
(f)
A
=
0 1 0
0 0 1
0 0 0
. The characteristic polynomial is
p
(
λ
) =

λ
3
. Setting
p
(
λ
) = 0,
we ﬁnd that
λ
= 0 is an eigenvalue of algebraic multiplicity 3. But
N
(
A

0
I
) =
N
(
A
) = Span
(
(1
,
0
,
0)
T
)
, a 1dimensional subspace of
R
3
.
(g)
A
=
1 1 1
0 2 1
0 0 1
. The characteristic polynomial is
p
(
λ
) = (1

λ
)
2
(2

λ
). Setting
p
(
λ
) = 0, we ﬁnd
λ
1
=
λ
2
= 1 and
λ
3
= 2. It remains to be seen whether we can
ﬁnd two linearly independent eigenvectors associated with the repeated eigenvalue.
We compute
A

I
=
0 1 1
0 1 1
0 0 0
.
The free variables are
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 Spring '08
 Anshelvich
 Linear Algebra, Algebra, Eigenvectors, Vectors, Matrices, Characteristic polynomial, λ, eigenvector

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