Let
A be an
n x n matrix and consider the set
E = {
xε
R
n
:
A
x = λ
x}. If
x ε
E, then so is
t
x for any scalar
t, since
Furthermore, if
x1 and
x2 are in
E, then
These calculations show that
E is closed under scalar multiplication and vector addition, so
E is a subspace of
R
n
. Clearly, the zero vector belongs to
E; but more notably,
the nonzero elements in E are precisely the eigenvectors of A corresponding to the eigenvalue λ. When the zero vector is adjoined to the collection of eigenvectors corresponding to a particular eigenvalue, the resulting collection,
forms a vector space called the
eigenspace of
A correspondign to the eigenvalue λ. Since it depends on both
A and the selection of one of its eigenvalues, the notation
will be used to denote this space. Since the equation
A
x = λ
x is equivalent to (
A − λ
I)
x = 0, the eigenspace
Eλ(
A) can also be characterized as the nullspace of
A − λ
I:
This observation provides an immediate proof that
Eλ(
A) is a subspace of
R
n
.
Consider the matrix
The determination of the eigenvectors of
A shows that its eigenspaces are
and
E−1(
A) is the line in
R2 through the origin and the point (1, 1), and
E−2(
A) is the line through the origin and the point (2, 3). Both of these eigenspaces are 1-dimensional subspaces of
R2.
Example 1: Determine the eigenspaces of the matrix
First, form the matrix
The determinant will be computed by performing a Laplace expansion along the second row:
The roots of the characteristic equation,
are clearly λ = −1 and 3, with 3 being a double root; these are the eigenvalues of
B. The associated eigenvectors can now be found. Substituting λ = −1 into the matrix
B − λ
I in (*) gives
which is the coefficient matrix for the equation (
B − λ
I)
x = 0 with λ = −1, which determines the eigenvectors corresponding to the eigenvalue λ = −1. These eigenvectors are the nonzero solutions of
The identical first and third equations imply that x1 + x3 = 0— that is,
x3 = −
x1—and the second equation says
x2 = 0. Therefore, the eigenvectors of
B associated with the eigenvalue λ = −1 are all vectors of the form (
x1, 0, −
x1)T =
x1(1, 0, −1)T for
x1 ≠ 0. Removing the restriction that the scalar multiple be nonzero includes the zero vector and gives the full eigenspace:
Now, since
the eigenvectors corresponding to the eigenvalue λ = 3 are the nonzero solutions of
These equations imply that
x3 =
x1, and since there is no restriction on
x2, this component is arbitrary. Therefore, the eigenvectors of
B associated with λ = 3 are all nonzero vectors of the form (x1,x2,x1)T = x1(1,0,1)T + x2(0,1,0)T The inclusion of the zero vector gives the eigenspace:
Note that dim
E−1(
B) = 1 and dim
E3(
B) = 2.
Vector Algebra
Eigenvalues and Eigenvectors
