19. Algebraic and Geometric Multiplicity. May 20, 2013
19.1. Starting Example
Find eigenvalues and eigenvectors for
0 −1
A=
2 3
The characteristic polynomial is
−λ −1 = (−λ)(3 − λ) − (−1)2 = λ2 − 3λ + 2 = (λ − 1)(λ − 2).
det(A − λI2 ) = 2 3 − λ
It has roots λ1 = 1 and λ2 = 2. Find eigenvectors corresponding to λ1 = 1:
−λ1
−1
x1
0
=
2 3 − λ 1 x2
0
We have the system with augmented matrix
−1 −1 0
1 1 0
⇒
2
2 0
0 0 0
Therefore, we have: x1 +x2 = 0. Any nonzero vectors which satisfies this is an eigenvector for λ1 = 1.
For example,
1
−2
,
, etc.
−1
2
We need only one of these eigenvectors, because all others are proportional to it. Let us find eigenvector corresponding to λ2 = 2:
−λ2
−1
x1
0
=
2 3 − λ 2 x2
0
We have the system with augmented matrix
−2 −1 0
2 1 0
⇒
2
1 0
0 0 0
Therefore, we have: 2x1 + x2 = 0. Any nonzero vectors which satisfies this is an eigenvector for
λ2 = 2. For example,
1/2
−1
, etc.
,
2
−1
The answer should be like this:
−1
There are two eigenvalues, 1 and 2. Eigenvector corresponding to 1:
Eigenvector corre1
−1
sponding to 2:
.
2
19.2. Example of a Defective Matrix
Let
1 1
A=
0 1
1
Let us find its eigenvalues and eigenvectors. Characteristic polynomial:
1 − λ
1 det(A − λI2 ) = = (1 − λ)2 ,
0
1 − λ
so λ = 1 is a double root. (Double means that it has power
corresponding to this eigenvalue:
1−λ
1
x1
0
0
=
⇒
0
1 − λ x2
0
0
two in this polynomial.) Eigenvectors
1
0
x1
0
=
x2
0
1
So x2 = 0, and x1 can be anything. There is only one linearly independent vector:
0
Consider another example: the identity matrix,
1 0
A=
0 1
Then, by the same token, λ = 1 is a double root of the characteristic polynomial and the only
2
eigenvalue. But any nonzero vector can serve as its eigenvector,because
for
any x ∈ R we have:
1
0
Ax = x = 1 · x. So it has two linearly independent eigenvectors:
and
0
1
19.3. Definitions
For an eigenvalue λ, its algebraic multiplicity is the mulitplicity of λ as a root of the characteristic
polynomial. Its geometric multiplicity is the maximal number of linearly independent eigenvectors
corresponding to it.
Here, for both matrices λ = 1 is the
only eigenvalue
with algebraic multiplicity two. But its
1 1
1 0
geometric multiplicity is one for
, and two for
.
0 1
0 1
2