# Invertable matrix P and a matrix C

• Nov 23rd 2009, 03:25 PM
statman101
Invertable matrix P and a matrix C
I need to find and invertable matrix P
Then a matrix C with the form:

|a -b|
|b a |
this comes from complex eigenvalues, and the formula
A= PC(P^-1)

my problem is

|5 -2|
|1 3 |
is the matrix

for the eigenvalues i got 4 +- 2i.
now i have to find eigenvectors and the C matrix
Can you help?
• Nov 23rd 2009, 04:43 PM
galactus
By subbing the eigenvalues back into A for lambda, we get:

$\begin{bmatrix}5-(4+i)&-2\\1&3-(4+i)\end{bmatrix}=\begin{bmatrix}1-i&-2\\1&-1-i\end{bmatrix}$

This leads to $\begin{bmatrix}1-i&-2\\1&-1-i\end{bmatrix}\begin{bmatrix}x_{1}\\x_{2}\end{bmat rix}=\begin{bmatrix}0\\0\end{bmatrix}$

This leads to eigenvector: $t\begin{bmatrix}\frac{2}{1-i}\\1\end{bmatrix}$

For the other eigenvalue, the corresponding eigenvector is: $t\begin{bmatrix}\frac{2}{1+i}\\1\end{bmatrix}$

Normalizing and using Gram-Schmidt

$||u||=\sqrt{|\frac{2}{1-i}|^{2}+1^{2}}=\sqrt{2}$

$p_{1}=\begin{bmatrix}\frac{2}{\sqrt{2}(1-i)}\\ \frac{1}{\sqrt{2}}\end{bmatrix}$

$p_{2}=\begin{bmatrix}\frac{2}{\sqrt{2}(1+i)}\\ \frac{1}{\sqrt{2}}\end{bmatrix}$

This gives us:

$P=\begin{bmatrix}\frac{2}{\sqrt{2}(1-i)}&\frac{2}{\sqrt{2}(1+i)}\\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{bmatrix}$

$P^{-1}=\begin{bmatrix}\frac{1}{\sqrt{2}i}&\frac{1}{\sq rt{2}}+\frac{1}{\sqrt{2}}i\\ \frac{1}{\sqrt{2}}i&\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i\end{bmatrix}$

Now, $P^{-1}AP=\begin{bmatrix}4+i&0\\0&4-i\end{bmatrix}$

Therefore, P diagonalizes A