Linear algebra, power of matrices $P^{-1}AP =
 \begin{pmatrix}
  -1 & 1 & 0 & 0 \\
  0 & -1 & 1 & 0 \\
  0 & 0 & -1 & 0 \\
  0 & 0 & 0 & 2 \\
 \end{pmatrix} 
$
with 
$P= 
 \begin{pmatrix}
 -1 & 1 & 0 & 0 \\
  0 & -1 & 1 & 0 \\
  0 & 0 & -1 & 0 \\
  0 & 0 & 0 & 1
\end{pmatrix}
$
and $P^{-1}$ is the inverse of $P$
Find $A$, $A^{100}$?
I found $A$, $P^{-1}AP=B$ and multiplied by $P$ and $P^{-1}$. How can I find $A^{100}$? I have to use eigenvalues/vectors.

Because of the way that $P$ and $A$ are constructed, it is not hard to see that $PA=AP$. Consequently, $P^{-1}AP=P^{-1}PA=A$. So that makes life easier. You can then write
$$
          A = 
     \left[\begin{array}{cccc}
          -1 & 0 & 0 & 0 \\
           0 & -1 & 0 & 0 \\
           0 & 0 & -1 & 0 \\
           0 & 0 & 0 & 2
          \end{array}\right]+
    \left[\begin{array}{cccc}
           0 & 1 & 0 & 0 \\
           0 & 0 & 1 & 0 \\
           0 & 0 & 0 & 0 \\
           0 & 0 & 0 & 0
          \end{array}\right] = D+N,
$$
where $D$ is the diagonal matrix and $N$ is nilpotent of order $3$ (meaning that $N^{3}=0$, but $N^{2} \ne 0$.) You have $DN=ND=-N$ and
$$
        N^{2} = \left[\begin{array}{cccc}
           0 & 0 & 1 & 0 \\
           0 & 0 & 0 & 0 \\
           0 & 0 & 0 & 0 \\
           0 & 0 & 0 & 0
          \end{array}\right],\;\;\; N^{3}=0.
$$
This is how the Jordan canonical form works. $N$ works on the same part of the space corresponding to the columns with $-1$ entries. So $DN=ND=-N$. And $DN^{2}=N^{2}D=-N^{2}$. So,
$$
(D+N)^{100} = D^{100}+(100)D^{99}N+(50)(99)D^{98}N^{2}=D^{100}-100N+50(99)N^{2}.
$$
You can write down the matrix $A^{100}$ from this. Nothing further is needed.
