Complex Diagonalization Complex DiagonalizationBy the Fundamental Theorem of Algebra, any (real or complex) polynomial have roots, as long as complex numbers are allowed. In fact, if we count the repeated roots, a polynomial of degree n has n roots.
Since the eigenvalues were computed as the roots of the characteristic polynomial, we need to consider the possibility of complex eigenvalues, even for real matrices.
| A = [ | 1 | 5 | ]. |
| -2 | 3 |
The characteristic polynomial is det(A - λI) = (1 - λ)(3 - λ) - (-2)5 = 13 - 4λ + λ2, and A has two eigenvalues λ1 = 2 + 3i, λ2 = 2 - 3i.
For λ1 = 2 + 3i, we have
| A - (2 + 3i)I = [ | -1 - 3i | 5 | ]. |
| -2 | 1 - 3i |
The two equations (-1 - 3i)x1 + 5x2 = 0 and -2x1 + (1 - 3i)x2 = 0 are equivalent (you may verify that (1 - 3i)[equation 1] = 2[equation 2]). The solutions of (A - (2 + 3i)I)x = 0 are of the form x = c(1 - 3i, 2), c arbitrary (complex number). We get an eigenvector v1 = (1 - 3i, 2), which is really a basis of the complex eigenspace nul(A - (2 + 3i)I) (a subspace of C2).
For λ1 = 2 - 3i, we have
| A - (2 - 3i)I = [ | -1 + 3i | 5 | ]. |
| -2 | 1 + 3i |
The solutions of (A - (2 - 3i)I)x = 0 are of the form x = c(1 + 3i, 2), c arbitrary. We get an eigenvector v2 = (1 + 3i, 2), which is really a basis of the eigenspace nul(A - (2 - 3i)I).
We conclude that A is diagonalizable, with diagonalization
| A = [ | 1 + 3i | 1 - 3i | ] [ | 2 + 3i | 0 | ] [ | 1 + 3i | 1 - 3i | ]-1. |
| 2 | 2 | 0 | 2 - 3i | 2 | 2 |
We can also try to diagonalize complex matrices.
Example The matrix
| A = [ | i | -i | 1 | ] |
| -1 | 1 | i | ||
| 0 | 0 | 1 + i |
has the characteristic polynomial is det(A - λI) = [(i - λ)(1 - λ) - i](1 + i - λ) = - λ(1 + i - λ)2. We get two eigenvalues λ1 = 0, λ2 = 1 + i.
For λ1 = 0, we solve Ax = 0 and find solutions x = x2(1, 1, 0), x2 arbitrary. Thus v1 = (1, 1, 0) form a basis of the eigenspace nulA.
For λ2 = 1 + i, we have
| A - (1 + i)I = [ | -1 | -i | 1 | ]. |
| -1 | -i | i | ||
| 0 | 0 | 0 |
The solutions of (A - (1 + i)I)x = 0 are of the form x = x2(-i, 1, 0), x2 arbitrary. Thus v2 = (-i, 1, 0) form a basis of the eigenspace nul(A - (1 + i)I).
The matrix is not diagonalizable.