Eigen Methods
Math 246, Fall 2009, Professor David Levermore
Eigenpairs.
Let
A
be a real
n
×
n
matrix. A number
λ
(possibly complex) is an
eigenvalue
of
A
if there exists a nonzero vector
v
(possibly complex) such that
(1)
Av
=
λ
v
.
Each such vector is an
eigenvector
associated with
λ
, and (
λ,
v
) is an
eigenpair
of
A
.
Fact 1:
If (
λ,
v
) is an eigenpair of
A
then so is (
λ, α
v
) for every complex
α
negationslash
= 0. In
other words, if
v
is an eigenvector associated with an eigenvalue
λ
of
A
then so is
α
v
for every complex
α
negationslash
= 0. In particular, eigenvectors are not unique.
Reason.
Because (
λ,
v
) is an eigenpair of
A
you know that (1) holds. It follows that
A
(
α
v
) =
α
Av
=
αλ
v
=
λ
(
α
v
)
.
Because the scalar
α
and vector
v
are nonzero, the vector
α
v
is also nonzero.
Therefore
(
λ, α
v
) is also an eigenpair of
A
.
square
Finding Eigenvalues.
Recall that the characteristic polynomial of
A
is defined by
(2)
p
A
(
z
) = det(
z
I
−
A
)
.
It has the form
p
A
(
z
) =
z
n
+
π
1
z
n

1
+
π
2
z
n

2
+
· · ·
+
π
n

1
z
+
π
n
,
where the coefficients
π
1
,
π
2
,
· · ·
,
π
n
are real. In other words, it is a real monic polynomial
of degree
n
. One can show that in general
π
1
=
−
tr(
A
)
,
π
n
= (
−
1)
n
det(
A
)
.
In particular, when
n
= 2 one has
p
A
(
z
) =
z
2
−
tr(
A
)
z
+ det(
A
)
.
Because det(
z
I
−
A
) = (
−
1)
n
det(
A
−
z
I
), this definition of
p
A
(
z
) coincides with the book’s
definition when
n
is even, and is its negative when
n
is odd. Both conventions are common.
We have chosen the convention that makes
p
A
(
z
) monic. What matters most about
p
A
(
z
)
is its roots and their multiplicity, which are the same for both conventions.
Fact 2:
A number
λ
is an eigenvalue of
A
if and only if
p
A
(
λ
) = 0. In other words, the
eigenvalues of
A
are the roots of
p
A
(
z
).
Reason.
If
λ
is an eigenvalue of
A
then by (1) there exists a nonzero vector
v
such that
(
λ
I
−
A
)
v
=
λ
v
−
Av
= 0
.
It follows that
p
A
(
λ
) = det(
λ
I
−
A
) = 0.
Conversely, if
p
A
(
λ
) = det(
λ
I
−
A
) = 0 then there exists a nonzero vector
v
such that
(
λ
I
−
A
)
v
= 0. It follows that
λ
v
−
Av
= (
λ
I
−
A
)
v
= 0
,
whereby
λ
and
v
satisfy (1), which implies
λ
is an eigenvalue of
A
.
square
Fact 2 shows that the eigenvalues of a
n
×
n
matrix
A
can be found if you can find all
the roots of the characteristic polynomial of
A
.
Because the degree of this characteristic
polynomial is
n
, and because every polynomical of degree
n
has exactly
n
roots counting
multiplicity, the
n
×
n
matrix
A
therefore must have at least one eigenvalue and at most
n
eigenvalues.
1
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2
Example.
Find the eigenvalues of
A
=
parenleftbigg
3
2
2
3
parenrightbigg
.
Solution.
The characteristic polynomial of
A
is
p
A
(
z
) =
z
2
−
6
z
+ 5 = (
z
−
1)(
z
−
5)
.
By Fact 2 the eigenvalues of
A
are 1 and 5.
Example.
Find the eigenvalues of
A
=
parenleftbigg
3
2
−
2
3
parenrightbigg
.
Solution.
The characteristic polynomial of
A
is
p
A
(
z
) =
z
2
−
6
z
+ 13 = (
z
−
3)
2
+ 4 = (
z
−
3)
2
+ 2
2
.
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 Spring '10
 LEVERMORE
 Linear Algebra, Matrices, Orthogonal matrix, real solutions, real n×n matrix

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