Jordan similar matrix I have matrix
$B = \begin{bmatrix}1 & 1 & -2 & 0\\2 & 1 & 0 & 2 \\ 1 & 0 & 1 & 1 \\ 0 & -1 & 2 & 1\end{bmatrix}$.
I found the characteristic polynomial $(1-x)^4$ and was able to get my Jordan Matrix $J = \begin{bmatrix}1 & 1 & 0 & 0\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1\end{bmatrix}$. I need to find $P$ and $P^{-1}$ such that $B=PJP^{-1}$, but I got completely stuck after finding the eigenvectors.
Would really appreciate the help.
Thanks.
 A: Note: It is important to learn to type problems using MathJax for many reasons, see this wonderful MathJax Basic Tutorial and Quick Reference.
We have the matrix:
$$B = \begin{bmatrix}1 & 1 & -2 & 0\\2 & 1 & 0 & 2 \\ 1 & 0 & 1 & 1 \\ 0 & -1 & 2 & 1\end{bmatrix}$$
The characteristic polynomial is:
$$(\lambda - 1)^4 = 0 \implies \lambda_{1,2,3,4} = 1$$
To find the four linearly independent eigenvectors, we try solving $[A - \lambda_i I]v_i = 0$, which yields only two independent eigenvectors as:
$$v_1 = \begin{bmatrix}-1\\ 0 \\ 0 \\1 \end{bmatrix}, v_3 = \begin{bmatrix}0\\ 2 \\ 1 \\0 \end{bmatrix}$$
Note, I could have called these any two eigenvectors I wanted, but I am matching your $J$, so chose them as shown above.
Now, we need to find two generalized eigenvectors, we set up and solve $[A-I]v_2 = v_1$. After RREF, we have:
$$v_2 = \begin{bmatrix}0\\-1\\ 0 \\ 0 \end{bmatrix}$$
We repeat this process using $[A-I]v_4 = v_3$, yielding:
$$v_4 = \begin{bmatrix}1\\0\\ 0 \\ 0 \end{bmatrix}$$
We can now write $P$ as a linear combination of those column eigenvectors, yielding:
$$P = [v_1 | v_2 | v_3 | v_4] = \begin{bmatrix}-1 &  0 &  0 &  1 \\ 0 & -1 & 2 & 0 \\ 0 &  0 & 1 & 0 \\ 1 &  0 & 0 &  0 \end{bmatrix}$$
It is a simple matter to find $P^{-1}$ and now we have:
$$J = P^{-1}AP = \begin{bmatrix}1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1\end{bmatrix}$$
