TITLE: Find high powers of a matrix with the Cayley Hamilton theorem
QUESTION [6 upvotes]: Let A = 
\begin{bmatrix}
0 & 1 & 0\\
0 & 0 & 1\\
-1 & -1 &-1\\
\end{bmatrix}
Compute $A^{10000} + A^{9998}$
I know this should be done by the Cayley-Hamilton theorem. I get as characteristic polynomial $-A^3 - A^2 - A - I = 0$ but I don't see how to calculate $A^{10000} + A^{9998}$ from there. I hope someone can help me out!

REPLY [7 votes]: Via the Cayley Hamilton Theorem:
$$
A^4 = I + (A-I)(A^3 + A^2+A+I)=I;\\
A^{10000} + A^{9998} = I^{2500} + I^{2499}A^2 = I + A^2.
$$