Interesting theorem, complex eigenvalues.

[bobby2k]

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?

 

[arkajad]

I would suggest: check that AP= PC writing A with some matrix elements (say p,q,r,s) and using the definitions.

 

[robotopia]

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 \begin{bmatrix} 1 \\ i \end{bmatrix} and you get the same result for both, then the two original matrices are the same:

Let X= \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix} and Y= \begin{bmatrix} y_1 & y_2 \\ y_3 & y_4 \end{bmatrix}. Then if
\begin{align}X\begin{bmatrix} 1 \\ i \end{bmatrix} &amp;= Y\begin{bmatrix} 1 \\ i \end{bmatrix}, \\<br /> \begin{bmatrix} x_1 &amp; x_2 \\ x_3 &amp; x_4 \end{bmatrix}\begin{bmatrix} 1 \\ i \end{bmatrix} &amp;= \begin{bmatrix} y_1 &amp; y_2 \\ y_3 &amp; y_4 \end{bmatrix}\begin{bmatrix} 1 \\ i \end{bmatrix} \end{align} \\<br /> \begin{bmatrix} x_1 + x_2i \\ x_3 + x_4i \end{bmatrix} = \begin{bmatrix} y_1 + y_2i \\ y_3 + y_4i \end{bmatrix}
Hence x_1=y_1, x_2=y_2, x_3=y_3, x_4=y_4, and hence X=Y.

Now for the theorem itself:
Notice that P \begin{bmatrix} 1 \\ i \end{bmatrix} = {\bf v} and that
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}.
Then, for any real matrix A with complex eigenvalues, we have
\begin{align}A{\bf v} &amp;= \lambda {\bf v} \\<br /> AP\begin{bmatrix} 1 \\ i \end{bmatrix} &amp;= \lambda P\begin{bmatrix} 1 \\ i \end{bmatrix} \\ &amp;= P \lambda \begin{bmatrix} 1 \\ i \end{bmatrix} \\ &amp;= PC \begin{bmatrix} 1 \\ i \end{bmatrix}\end{align}
But both AP and PC are real, hence
AP=PC \quad \text{or} \quad A=PCP^{-1}.

 

[bobby2k]

Very nice proof!, thank you very much!