math111_logo Complex Diagonalization

3. Complex eigenvalue in real diagonalization

In the previous part, we studied the geometric meaning of one conjugate pair of complex eigenvalues and eigenvectors. In this part, we consider several such pairs.

We illustrate the situation by considering a real linear transformation T: R5R5. We assume T has a complex basis (for C5) of eigenvectors (see the earlier result)

v1, v2 = u2 + iw2, v2 = u2 - iw2, v3 = u3 + iw3, v3 = u3 - iw3,

with corresponding eigenvalues

λ1, λ2 = μ2 + 2, λ2 = μ2 - 2, λ3 = μ3 + 3, λ3 = μ3 - 3.

First we prove that the real vectors v1, u2, w2, u3, w3 form a basis of R5. In fact, for real numbers c1, a2, b2, a3, b3, we have

c1v1 + a2u2 + b2w2 + a3u3 + b3w3 = c1v1 + c2v2 + c2v2 + c3v3 + c3v3,

where c2 = (a2 - ib2)/2, c3 = (a3 - ib3)/2. Then we have

c1v1 + a2u2 + b2w2 + a3u3 + b3w3 = 0
c1v1 + c2v2 + c2v2 + c3v3 + c3v3 = 0
c1 = c2 = c3 = 0 (v1, v2, v2, v3, v3 are complex linearly independent)
c1 = a2 = b2 = a3 = b3 = 0 (complex number = 0 ⇔ real and imaginary parts = 0)

This means that v1, u2, w2, u3, w3 are linear independent vectors. By this result, they must form a basis of R5.

Similar to the computation in the example in the previous part (also see the remark after the example), we have

T(v1) = λ1v1,
T(u2) = μ2u2 - ν2w2,
T(w2) = ν2u2 + μ2w2,
T(u3) = μ3u3 - ν3w3,
T(w3) = ν3u3 + μ3w3.

Note that the basis {v1, u2, w2, u3, w3} gives us a direct sum (see later part of the proof of this result) span{v1}⊕span{u2, w2}⊕span{u3, w3} = R5. This gives the geometric interpretation below.

Let V be a finite dimensional real vector space. Let T: VV be complex diagonalizable. Then V is decomposed as a direct sum of 1-dimensional subspaces span{vi} and 2-dimensional subspaces span{uj, wj}, such that

The scalar multiplication on span{vi} is by the real eigenvalue associated to vi. The rotation and scalar multiplication on span{uj, wj} is determined by the polar expressions of the complex eigenvalue associated to uj + iwj.

Finally let us consider the discussion from the matrix viewpoint. Let the 5 by 5 matrix A be the usual matrix of T: R5R5 (with respect to the standard basis). Then we have A = PDP-1, with

P = [v1 u2 w2 u3 w3], D = [ λ1 0 0 0 0 ].
0 μ2 ν2 0 0
0 -ν2 μ2 0 0
0 0 0 μ3 ν3
0 0 0 -ν3 μ3

Note that D is a block diagonal matrix, obtained by replacing the conjugate pairs of complex eigenvalues in the usual complex diagonal matrix by special 2 by 2 matrices.


[exercise]
[previous topic] [part 1] [part 2] [part 3] [next topic]