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.

Note that $B=P^{-1}AP=(-I_3+N)\oplus2$. Therefore $B^{100}=P^{-1}A^{100}P=(-I_3+N)^{100}\oplus2^{100}$. Since $N^3=0$, by binomial theorem,
$$(-I+N)^{100}=(-I)^{100}+\binom{100}1(-I)^{99}N+\binom{100}2(-I)^{98}N^2.$$
