Eigenvalue and Eigenvector Eigenvalue and EigenvectorA linear transformation T: Rn → Rn 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 v ≠ 0 (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)
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.
| 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).
| 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: V → V 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.