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The Eigenvalue Problem: Examples

Module by: Steven J. Cox. E-mail the author

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We take a look back at our previous examples in light of the results of two previous sections The Spectral Representation and The Partial Fraction Expansion of the Transfer Function. With respect to the rotation matrix B= 01-10 B 01 -10 we recall, see Cauchy's Theorem eqn6, that Rs=1s2+1s1-1s Rs 1 s 2 1 s1 -1s

Rs=1s−ⅈ1/2-ⅈ2ⅈ21/2+1s+ⅈ1/2ⅈ2-ⅈ21/2 Rs 1 s 12 2 2 12 1 s 12 2 2 12 (1)
Rs=1s− λ1 P1 +1s− λ2 P2 Rs 1 s λ1 P1 1 s λ2 P2 and so B= λ1 P1 + λ2 P2 =ⅈ1/2-ⅈ2ⅈ21/2+-ⅈ1/2ⅈ2-ⅈ21/2 B λ1 P1 λ2 P2 12 2 2 12 12 2 2 12 From m1 = m2 =1 m1 m2 1 it follows that ℛ P1 ℛ P1 and ℛ P2 ℛ P2 are actual (as opposed to generalized) eigenspaces. These column spaces are easily determined. In particular, ℛ P1 ℛ P1 is the span of e1 =1ⅈ e1 1 while ℛ P2 ℛ P2 is the span of e2 =1-ⅈ e2 1 To recapitulate, from partial fraction expansion one can read off the projections from which one can read off the eigenvectors. The reverse direction, producing projections from eigenvectors, is equally worthwhile. We laid the groundwork for this step in the discussion of Least Squares. In particular, this Least Squares projection equation stipulates that P1 = e1 e1 T e1 -1 e1 T   and   P2 = e2 e2 T e2 -1 e2 T P1 e1 e1 e1 e1   and   P2 e2 e2 e2 e2 As e1 T e1 = e1 T e1 =0 e1 e1 e1 e1 0 these formulas can not possibly be correct. Returning to the Least Squares discussion we realize that it was, perhaps implicitly, assumed that all quantities were real. At root is the notion of the length of a complex vector. It is not the square root of the sum of squares of its components but rather the square root of the sum of squares of the magnitudes of its components. That is, recalling that the magnitude of a complex quantity zz is zz¯ z z , ∥ e1 ∥2≠ e1 T e1   rather   ∥ e1 ∥2≠ e1 ¯T e1 e1 2 e1 e1   rather   e1 2 e1 e1 Yes, we have had this discussion before, recall complex numbers, vectors, and matrices. The upshot of all of this is that, when dealing with complex vectors and matrices, one should conjugate before every transpose. Matlab (of course) does this automatically, i.e., the ' symbol conjugates and transposes simultaneously. We use xH x to denote `conjugate transpose', i.e., xH≡x¯T x x All this suggests that the desired projections are more likely
P1 = e1 e1 H e1 -1 e1 H   and   P2 = e2 e2 H e2 -1 e2 H P1 e1 e1 e1 e1   and   P2 e2 e2 e2 e2 (2)
Please check that Equation 2 indeed jives with Equation 1.

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