linearly independent associated eigenvectors. Nevertheless, it is a (somewhat disparaging)
fact that eigenvalues can have fewer linearly independent eigenvectors than their multiplicity
suggests.
Example
:
Find the eigenvalues and associated eigenvectors of the matrix
A
=
2

1
1
2
.
We compute
det(
A

λ
I
)
=
2

λ

1
1
2

λ
=
(
λ

2)
2
+ 1
=
λ
2

4
λ
+ 5
.
The roots of this polynomial are
λ
1
= 2+
i
and
λ
2
= 2

i
; that is, the eigenvalues
are not real numbers.
This is a common occurrence, and we can press on to
find the eigenvectors just as we have in the past with real eigenvalues. To find
eigenvectors associated with
λ
1
= 2 +
i
, we look for
x
satisfying
(
A

(2 +
i
)
I
)
x
=
0
⇒

i

1
1

i
x
1
x
2
=
0
0
⇒

ix
1

x
2
x
1

ix
2
=
0
0
⇒
x
1
=
ix
2
.
Thus all eigenvectors associated with
λ
1
= 2+
i
are scalar multiples of
u
1
= (
i,
1).
Proceeding with
λ
2
= 2

i
, we have
(
A

(2

i
)
I
)
x
=
0
⇒
i

1
1
i
x
1
x
2
=
0
0
⇒
ix
1

x
2
x
1
+
ix
2
=
0
0
⇒
x
1
=

ix
2
,
which shows all eigenvectors associated with
λ
2
= 2

i
to be scalar multiples of
u
2
= (

i,
1).
Notice that
u
2
, the eigenvector associated with the eigenvalue
λ
2
= 2

i
in the last
example, is the complex conjugate of
u
1
, the eigenvector associated with the eigenvalue
λ
1
= 2 +
i
. It is indeed a fact that, if
A
∈ M
n
×
n
(
R
) has a nonreal eigenvalue
λ
1
=
λ
+
iμ
with corresponding eigenvector
ξ
1
, then it also has eigenvalue
λ
2
=
λ

iμ
with corresponding
eigenvector
ξ
2
=
¯
ξ
1
.
5
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 Spring '13
 MRR
 Math, Differential Equations, Linear Algebra, Eigenvectors, Equations, Vectors, Eigenvalue, eigenvector and eigenspace, equation Ax

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