Find an invertible matrix P and a diagonal matrix D so that (P^−1)AP = D, where
A =
4 2
3 3
I'll start you off.
$\displaystyle \Large\begin{bmatrix}4&2\\3&3\end{bmatrix}$
Find your charpoly:
$\displaystyle det\left({\lambda}\begin{bmatrix}1&0\\0&1\end{bmat rix}-\begin{bmatrix}4&2\\3&3\end{bmatrix}\right)=\begin {bmatrix}{\lambda}-4&-2\\-3&{\lambda}-3\end{bmatrix}={\lambda}^{2}-7{\lambda}+6$
The roots of this quadratic are your eigenvalues.
They are 1 and 6.
Sub them in for lambda in the matrix we derived, getting:
$\displaystyle \begin{bmatrix}-3&-2\\-3&-2\end{bmatrix}\;\text{and}\;\begin{bmatrix}2&-2\\-3&3\end{bmatrix}$
Use rref on these matrices and we find:
$\displaystyle \begin{bmatrix}-3&-2\\-3&-2\end{bmatrix}=\begin{bmatrix}1&\frac{2}{3}\\0&0\e nd{bmatrix}\;\text{and}\;\begin{bmatrix}2&-2\\-3&3\end{bmatrix}=\begin{bmatrix}1&-1\\0&0\end{bmatrix}$
Your solutions are:
$\displaystyle x_{1}=\frac{-2}{3}t\;\text{and}\;\ x_{2}=t$
$\displaystyle x_{1}=t\;\text{and}\;\ x_{2}=t$, respectively.
Bases for the eigenspace are:
$\displaystyle \begin{bmatrix}\frac{-2}{3}\\1\end{bmatrix}$
$\displaystyle \begin{bmatrix}1\\1\end{bmatrix}$
Let this be P:
$\displaystyle \Large{P=\begin{bmatrix}\frac{-2}{3}&1\\1&1\end{bmatrix}}$
$\displaystyle \Large{P^{-1}=\begin{bmatrix}\frac{-3}{5}&\frac{3}{5}\\\frac{3}{5}&\frac{2}{5}\end{bma trix}}$
Now, find $\displaystyle P^{-1}AP$. 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.
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:
$\displaystyle A=\begin{bmatrix}0&0&-2\\1&2&1\\1&0&3\end{bmatrix}$