diagonal matrix

**luckyc1423** · July 12th 2006, 03:18 PM

Find an invertible matrix P and a diagonal matrix D so that (P^−1)AP = D, where
A =
4 2
3 3

**galactus** · July 12th 2006, 05:39 PM

I'll start you off.

$\Large\begin{bmatrix}4&2\\3&3\end{bmatrix}$

Find your charpoly:

$det\left({\lambda}\begin{bmatrix}1&0\\0&1\end{bmat rix}-\begin{bmatrix}4&2\\3&3\end{bmatrix}\right)=\begin {bmatrix}{\lambda}-4&-2\\-3&{\lambda}-3\end{bmatrix}={\lambda}^{2}-7{\lambda}+6$

The roots of this quadratic are your eigenvalues.

They are 1 and 6.

Sub them in for lambda in the matrix we derived, getting:

$\begin{bmatrix}-3&-2\\-3&-2\end{bmatrix}\;\text{and}\;\begin{bmatrix}2&-2\\-3&3\end{bmatrix}$

Use rref on these matrices and we find:

$\begin{bmatrix}-3&-2\\-3&-2\end{bmatrix}=\begin{bmatrix}1&\frac{2}{3}\\0&0\e nd{bmatrix}\;\text{and}\;\begin{bmatrix}2&-2\\-3&3\end{bmatrix}=\begin{bmatrix}1&-1\\0&0\end{bmatrix}$

Your solutions are:

$x_{1}=\frac{-2}{3}t\;\text{and}\;\ x_{2}=t$

$x_{1}=t\;\text{and}\;\ x_{2}=t$ , respectively.

Bases for the eigenspace are:

$\begin{bmatrix}\frac{-2}{3}\\1\end{bmatrix}$

$\begin{bmatrix}1\\1\end{bmatrix}$

Let this be P:

$\Large{P=\begin{bmatrix}\frac{-2}{3}&1\\1&1\end{bmatrix}}$

$\Large{P^{-1}=\begin{bmatrix}\frac{-3}{5}&\frac{3}{5}\\\frac{3}{5}&\frac{2}{5}\end{bma trix}}$

Now, find $P^{-1}AP$ . You have all the necessary info.

Check me out. Easy to flub up.

I hope this helps.

If you don't have one, get a good linear algebra text. Anton and Rorres publish some good ones.

**luckyc1423** · July 12th 2006, 06:39 PM

I am completely confused on this whole problem. If you could finish this off, then put an example for me to do and ill do it and post it to see if I got it right, thanks.

**galactus** · July 13th 2006, 04:09 AM

I went ahead and made some changes. I hope you can finish up. The worst is done.

Good luck on your distance learning class. I know they can be a booger.

Try this one:

Find a matrix P that diagonalizes:

$A=\begin{bmatrix}0&0&-2\\1&2&1\\1&0&3\end{bmatrix}$

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