math111_logo Eigenvalue and Eigenvector

2. Computation

A linear transformation T: RnRn is given by an n by n matrix A. The eigenvalue λ and the eigenvector v of T are defined by

Av = λv.

Equivalently, v is a nonzero vector in the null space nul(A - λI). The number λ and the vector v are also called the eigenvalue and the eignevctor of A.

The following suggests how to find eigenvalues.

λ is an eigenvalue of A
Av = λv for some v0 (definition of eigenvalue)
⇔ (A - λI)x = 0 has x = v as a nontrivial solution
A - λI is not invertible (invertibility criterion for square matrices)
det(A - λI) = 0. (this relation between det and invertiblity)

The characteristic polynomial of a square matrix A is det(A - λI).

Thus the eigenvalues and the eigenvectors can be computed as follows.

Step 1: Solve the characteristic equation det(A - λI) = 0 and get the eigenvalues λ1, λ2, ... .

Step 2: For each eigenvalue λi, solve the homogeneuous system (A - λiI)x = 0 and get the eigenvectors with λi as the eigenvalue.

In the second step, the answer is actually presented as a basis of nul(A - λiI), called the eigenspace of A with eigenvalue λi.

Example Consider the matrix

A = [ 13 -4 ]
-4 7

of the linear transformation T(x1, x2) = (13x1 - 4x2, - 4x1 + 7x2) on R2. The characteristic polynomial is

det(A - λI) = [ 13 - λ -4 ] = (13 - λ)(7 - λ) - (-4)(-4) = 75 - 20λ + λ2.
-4 7 - λ

Since 75 - 20λ + λ2 = (5 - λ)(15 - λ), we get two eigenvalues λ1 = 5, λ2 = 15.

For λ1 = 5, we have

A - 5I = [ 13 - 5 -4 ] = [ 8 -4 ].
-4 7 - 5 -4 2

The solutions of (A - 5I)x = 0 are of the form x = c(1, 2), c arbitrary. We get an eigenvector v1 = (1, 2), which is really a basis of the eigenspace nul(A - 5I).

For λ2 = 15, we have

A - 15I = [ 13 - 15 -4 ] = [ -2 -4 ].
-4 7 - 15 -4 -8

The solutions of (A - 15I)x = 0 are of the form x = c(-2, 1), c arbitrary. We get an eigenvector v2 = (-2, 1), which is really a basis of the eigenspace nul(A - 15I).

Example Consider the matrix

A = [ 1 3 -3 ].
-3 7 -3
-6 6 -2

The characteristic polynomial has been computed in an earlier example: det(A - λI) = (4 - λ)2(- 2 - λ). We have two eigenvalues λ1 = 4, λ2 = -2.

For λ1 = 4, we have

A - 4I = [ 1 - 4 3 -3 ] = [ -3 3 -3 ].
-3 7 - 4 -3 -3 3 -3
-6 6 -2 - 4 -6 6 -6

The solutions of (A - 4I)x = 0 are of the form x = x2(1, 1, 0) + x3(1, 0, -1), x2 and x3 arbitrary. We get two eigenvectors v1 = (1, 1, 0), v2 = (1, 0, -1), which form a basis of the eigenspace nul(A - 4I).

For λ2 = -2, we have

A + 2I = [ 1 + 2 3 -3 ] = [ 3 3 -3 ].
-3 7 + 2 -3 -3 9 -3
-6 6 -2 + 2 -6 6 0

The solutions of (A + 2I)x = 0 are of the form x = c(1, 1, 2), c arbitrary. We get an eigenvector v3 = (1, 1, 2), which is really a basis of the eigenspace nul(A + 2I).

We may verify that v1, v2, v3 actually form a basis of R3. Geometrically, we have complete undestanding of the linear transformation given by A.

From the examples, we saw that for an n by n matrix A, det(A - λI) is a polynomial of degree n. The general reason can be found here.

If A is the matrix of a linear transformation T: VV on a finite dimensional vector space with respect to one basis, then according to the discussion here, the matrix of T with respect to another basis is B = PAP-1. By this property, we have

det(B - λI) = det(PAP-1 - λI) = det(P(A - λI)P-1) = det(P-1P(A - λI)) = det(A - λI).

Thus the characteristic polynomial of a linear transformation does not depend on the choice of the basis. We also call det(A - λI) the characteristic polynomial of T.


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