We review here the basics of computing eigenvalues and eigenvectors. Eigenvalues and eigenvectors play a prominent role in the study of ordinary differential equations and in many applications in the physical sciences. Expect to see them come up in a variety of contexts!

Definitions

Let A be an n ×n matrix. The number l is an eigenvalue of A if there exists a non-zero vector v such that

Computing Eigenvalues and Eigenvectors

We can rewrite the condition Av = lv as (A- lI)v = 0. where I is the n ×n identity matrix. Now, in order for a non-zero vector v to satisfy this equation, A -lI must not be invertible. That is, the determinant of A - lI must equal 0. We call p(l) = det(A - lI) the characteristic polynomial of A. The eigenvalues of A are simply the roots of the characteristic polynomial of A.

Otherwise, if A - lI has an inverse, (A - lI)-1(A - lI)v = (A - l I)-10 v = 0. But we are looking for a non-zero vector v.

Example

Let A =

é ê ê ë

2 -4 -1 -1

ù ú ú û

. Then

p(l)	=	det é ê ê ë 2-l -4 -1 -1-l ù ú ú û
	=	(2-l)(-1-l)-(-4)(-1)
	=	l2 -l-6
	=	(l-3)(l+2).

Thus, l₁ = 3 and l₂ = -2 are the eigenvalues of A.

To find eigenvectors v =

é ê ê ê ê ê ë

v1 v2 : vn

ù ú ú ú ú ú û

corresponding to an eigenvalue l, we simply solve the system of linear equations given by

Example

The matrix A =

é ê ê ë

2 -4 -1 -1

ù ú ú û

of the previous example has eigenvalues l₁ = 3 and l₂ = -2. Let's find the eigenvectors corresponding to l₁ = 3. Let v = [ v₁ v₂ ]^T. Then (A-3I)v = 0 gives us

Repeating this process with l₂ = -2, we find that

In the following example, we see a two-dimensional eigenspace.

Example

Let A =

é ê ê ê ë

5 8 16 4 1 8 -4 -4 -11

ù ú ú ú û

. Then p(l) = det

é ê ê ê ë

5-l 8 16 4 1-l 8 -4 -4 -11-l

ù ú ú ú û

=

(l-1)(l+3)² after some algebra! Thus, l₁ = 1 and l₂ = -3 are the eigenvalues of A.

Eigenvectors v =

é ê ê ê ê ë

v1 v2 v3

ù ú ú ú ú û

corresponding to l1 = 1 must satisfy

[ -2 -1 1 ]T is a basis for the eigenspace corresponding to l1 = 1.

Letting v₃ = t, we find from the second equation that v₁ = -2t, and then v₂ = -t. All eigenvectors corresponding to l₁ = 1 are multiples of

é ê ê ê ë

-2 -1 1

ù ú ú ú û

and so the eigenspace corresponding to l1 = 1 is given by the span of

é ê ê ê ë

-2 -1 1

ù ú ú ú û

.

Eigenvectors corresponding to l₂ = -3 must satisfy

The equations here are just multiples of each other! If we let v₃ = t and v₂ = s, then v₁ = -s -2t. Eigenvectors corresponding to l₂ = -3 have the form

{[ -1 1 0]T, [ -2 0 1]T} is a basis for the eigenspace corresponding to l2 = -3.

Thus, the eigenspace corresponding to l₂ = -3 is two-dimensional and is spanned by

é ê ê ê ë

-1 1 0

ù ú ú ú û

and

é ê ê ê ë

-2 0 1

ù ú ú ú û

.

• Eigenvalues and eigenvectors can be complex-valued as well as real-valued.

• The dimension of the eigenspace corresponding to an eigenvalue is less than or equal to the multiplicity of that eigenvalue.

• The techniques used here are practical for 2 ×2 and 3×3 matrices. Eigenvalues and eigenvectors of larger matrices are often found using other techniques, such as iterative methods.

Let A be an n ×n matrix. The eigenvalues of A are the roots of the characteristic polynomial