# Interesting theorem, complex eigenvalues.

1. Jul 28, 2013

### 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?

Last edited: Jul 28, 2013
2. Jul 29, 2013

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

3. Jul 29, 2013

### 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} &= 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}
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} &= \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}
But both $AP$ and $PC$ are real, hence
$$AP=PC \quad \text{or} \quad A=PCP^{-1}.$$

4. Jul 30, 2013

### bobby2k

Very nice proof!, thank you very much!