Put a matrix $A$ in Jordan Normal Form and find a $P$ such that $P^{-1}AP=J$ I have a linear algebra exam tomorrow and this is a frequent question.
$A= 
\begin{pmatrix} 4 & 0 & 1 & 0 \\ 2 & 2 & 3 & 0 \\ -1 & 0 & 2 & 0 \\ 4 & 0 & 1 & 2
\end{pmatrix}$
$C_T(x)=(x-2)^2(x-3)^2$ so eigenvalues are $2$ and $3$.
$a_2=2$ so the sum of the sizes of Jordan blocks for eigenvalue $2$ equals $2=1+1$
$a_3=2$ so the sum of the sizes of Jordan blocks for eigenvalue $3$ equals $2=1+1$
$\\$
$g_2=dimE_2=dim \ ker \ (A-2I)$
$A-2I = \begin{pmatrix} 2 & 0 & 1 & 0 \\ 2 & 0 & 3 & 0 \\ -1 & 0 & 0 & 0 \\ 4 & 0 & 1 & 0 \end{pmatrix}$
$ker(A-2I) = \left\{ \begin{pmatrix} 0 \\ y \\ 0 \\ t \end{pmatrix} |\ y,t \in \mathbb{R} \right\} = Span \left\{ \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0\\0\\0\\1 \end{pmatrix} \right\}$
Therefore $g_2=2$. So there are 2 Jordan blocks, and they must both be of size $1$.
I put this in Jordan Normal form and got:
$\\$ $\\$
$J = \begin{pmatrix} 3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2
\end{pmatrix}$
How do I find an invertible matrix $P$ such that $P^{-1}AP=J$?
I found the Jordan Normal Form using geometric and algebraic multiplicities along with the minimum polynomial, if that helps at all!
Thanks!

The left two columns are just a basis of 2 eigenvectors. For 3, we take the far right vector as some $w$ such that $(A-3I)^2 w = 0 $ but $(A-3I) w \neq 0. $ Then the third column is $v = (A - 3I)w.$
$$
P =
\left(
\begin{array}{rrrr}
0 & 0 & 1&1 \\
1&0&-1&3 \\
0& 0 & -1& 0 \\
0&1&3&1
\end{array}
\right)
$$
determinant is $-1$ and
$$
P^{-1} =
\left(
\begin{array}{rrrr}
-3 & 1 & -4&0 \\
-1&0&2&1 \\
0& 0 & -1& 0 \\
1&0&1&0
\end{array}
\right)
$$
