Jordan Form of a 3x3 Matrix

Question

I'm having some troubles figuring out Jordan Canonical Form. Can someone explain how to get there with this example matrix?

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

Also how i can get the tranformations matrix to write $A$=$PJP^{(-1)}$

Answer 1

Since the matrix have $\lambda_1=-1$ and $\lambda_{2,3}=2$ the Jordan form is given by

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

and to find the Jordan basis we can use that

$$A=PJP^{-1} \implies AP=PJ$$

and by $P=[u,v,w]$ we obtain

$$Au=2u$$ $$Av=u+2v$$ $$Aw=-w$$

once you find $u$ from the first equation you can find also $v$ from the second, the third one gives $w$.