# Find P and C such that $A=PCP^{-1}$

1. Apr 5, 2014

### Zondrina

1. The problem statement, all variables and given/known data

Find an invertible matrix $P$ and a matrix $C$ of the form $\left( \begin{array}{cc} a & -b \\ b & a \end{array} \right)$ such that $A=PCP^{-1}$ when $A = \left( \begin{array}{cc} 1 & -2 \\ 1 & 3 \end{array} \right)$.

2. Relevant equations

Eigenvalues for A:

$\lambda_1 = 2+i$
$\lambda_2 = 2-i$

Eigenvectors for eigenvalues:

$v_1 = [-1 + i, 1]^t$
$Re(v_1) = [-1, 1]^t$
$Im(v_1) = [1, 0]^t$

$v_2 = [-1 - i, 1]^t$
$Re(v_2) = [-1, 1]^t$
$Im(v_2) = [-1, 0]^t$

A theorem:

Let $A$ be a real 2x2 matrix with a complex eigenvalue $\lambda = a - bi, (b≠0)$ and an associated eigenvector $v$.

Then $A=PCP^{-1}$ where $P = [ Re(v), Im(v) ]$ and $C = \left( \begin{array}{cc} a & -b \\ b & a \end{array} \right)$

3. The attempt at a solution

I've basically solved this, but I had a question.

Does the theorem state that I only care about the eigenvalue $a - bi$? In this case it would be $\lambda_2 = 2-i$. If that's the case, $\lambda_1$ is useless in finding the required matrices.

Granted that $\lambda_1$ and $v_1$ have nothing to do with the problem, the matrices required are:

$C = \left( \begin{array}{cc} 2 & -1 \\ 1 & 2 \end{array} \right)$
$P = \left( \begin{array}{cc} -1 & -1 \\ 1 & 0 \end{array} \right)$
$P^{-1} = \left( \begin{array}{cc} 0 & 1 \\ -1 & -1 \end{array} \right)$

Where I have plugged the magnitudes of $a$ and $b$ into $C$.

My real curiosity lies in why I only care about the eigenvalue $a - bi$?

2. Apr 5, 2014

### Staff: Mentor

Complex eigenvalues come in pairs, so if you know one of them, the other is uniquely determined. Also notice that the first components of your eigenvectors are conjugates of one another. Notice also that the real parts of your two eigenvectors are the same, and the imaginary parts are scalar multiples of one another.

3. Apr 5, 2014

### Zondrina

This much I have already seen. I was more curious as to why the eigenvalue $a+bi$ is being neglected in the theorem. Is it solely due to the definition of $C$?

4. Apr 5, 2014

### Staff: Mentor

I'll bet the theorem could be rewritten so that it uses a + bi instead of a - bi.

5. Apr 5, 2014

### Zondrina

Indeed, I conjecture that:

Let $A$ be a real 2x2 matrix with a complex eigenvalue $\lambda = a + bi, (b≠0)$ and an associated eigenvector $v$.

Then $A=PCP^{-1}$ where $P = [ Re(v), Im(v) ]$ and $C = \left( \begin{array}{cc} a & b \\ -b & a \end{array} \right)$

I have verified the application of this theorem. I'll go prove it in a more general setting now.