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Interesting theorem, complex eigenvalues.

  1. Jul 28, 2013 #1
    Take a look at this theorem.


    Is it a way to show this theorem? I would like to show it using the standard way of diagonalizing a matrix.

    I mean if P = [v1 v2] and D =
    [lambda1 0
    0 lambda D]

    We have that AP = PD even for complex eigenvectors and eigenvalues.

    But the P matrix in this theorem is real, and so is the C matrix. I think they have used that v1 and v2 are conjugates, and so is lambda 1 and lambda 2.

    How would you show this theorem? Can you use ordinary diagonolisation to show it?
    Last edited: Jul 28, 2013
  2. jcsd
  3. Jul 29, 2013 #2
    I would suggest: check that AP= PC writing A with some matrix elements (say p,q,r,s) and using the definitions.
  4. Jul 29, 2013 #3
    Before showing the theorem, I'll first establish the lemma that if you have two real 2x2 matrices, and if you multiply them on the right by [itex]\begin{bmatrix} 1 \\ i \end{bmatrix}[/itex] and you get the same result for both, then the two original matrices are the same:

    Let [itex] X= \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix} [/itex] and [itex] Y= \begin{bmatrix} y_1 & y_2 \\ y_3 & y_4 \end{bmatrix} [/itex]. Then if
    [tex] \begin{align}X\begin{bmatrix} 1 \\ i \end{bmatrix} &= Y\begin{bmatrix} 1 \\ i \end{bmatrix}, \\
    \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix}\begin{bmatrix} 1 \\ i \end{bmatrix} &= \begin{bmatrix} y_1 & y_2 \\ y_3 & y_4 \end{bmatrix}\begin{bmatrix} 1 \\ i \end{bmatrix} \end{align} \\
    \begin{bmatrix} x_1 + x_2i \\ x_3 + x_4i \end{bmatrix} = \begin{bmatrix} y_1 + y_2i \\ y_3 + y_4i \end{bmatrix} [/tex]
    Hence [itex]x_1=y_1[/itex], [itex]x_2=y_2[/itex], [itex]x_3=y_3[/itex], [itex]x_4=y_4[/itex], and hence [itex]X=Y[/itex].

    Now for the theorem itself:
    Notice that [itex]P \begin{bmatrix} 1 \\ i \end{bmatrix} = {\bf v}[/itex] and that
    [tex]C\begin{bmatrix} 1 \\ i \end{bmatrix}=\begin{bmatrix} a-bi \\ b+ai \end{bmatrix}=\begin{bmatrix} \lambda \\ \lambda i \end{bmatrix} = \lambda\begin{bmatrix} 1 \\ i \end{bmatrix}.[/tex]
    Then, for any real matrix [itex]A[/itex] with complex eigenvalues, we have
    [tex]\begin{align}A{\bf v} &= \lambda {\bf v} \\
    AP\begin{bmatrix} 1 \\ i \end{bmatrix} &= \lambda P\begin{bmatrix} 1 \\ i \end{bmatrix} \\ &= P \lambda \begin{bmatrix} 1 \\ i \end{bmatrix} \\ &= PC \begin{bmatrix} 1 \\ i \end{bmatrix}\end{align}[/tex]
    But both [itex]AP[/itex] and [itex]PC[/itex] are real, hence
    [tex]AP=PC \quad \text{or} \quad A=PCP^{-1}.[/tex]
  5. Jul 30, 2013 #4
    Very nice proof!, thank you very much!
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