<s>[INST] Find $A^{1000}$ by using Cayley-Hamilton Theorem I get stuck at the following question:
Consider the matrix
$$A=\begin{bmatrix}
0 & 2 & 0 \\
1 & 1 & -1 \\
-1 & 1 & 1\\
\end{bmatrix}$$
Find $A^{1000}$ by using the Cayley-Hamilton theorem.
I find the characteristic polynomial by $P(A) = -A^{3} + 2A^2 = 0$ (by Cayley-Hamilton) but I don't see how to find $A^{1000}$ by this characteristic polynomial.
 [/INST] There is another way of approaching this.
You could divide $x^{1000}$ by the characteristic polynomial:
$x^{1000} = (-x^3+2x^2)Q+R$ where $R$ is a polynomial of degree less than 3 with unknown coefficients.
write down $R=ax^2+bx+c$ and evaluate $R$ at the roots of the characteristic polynomial.
Meaning, write down $\lambda^{1000}=a\lambda ^2+b\lambda+c$
and 
$\xi^{1000} = a\xi ^2+b\xi+c$
and
$\rho^{1000} = a\rho ^2+b\rho+c$
where $\lambda$ and $\xi$ and $\rho$ are roots of the characteristic polynomial. as you can see, $Q$ wont matter because it is multiplied by zero.
Do this to find the coeffiecents of the remainder, $R$.
after you have done that, insert $x=A$ to get $A^{1000}=aA^2+bA+c$ with the coeffiecents $a,b,c$ that you found.
Edit: The problem here, is that you have a double root, so you need to use the derivative as well.
Full answer:
divide $x^{1000}$ by $(-x^3+2x^2)$ to get:
$x^{1000} = (-x^3+2x^2)Q+ax^2+bx+c$ where $Q$ is some polynomial unknown to us.
the roots of the char. polynomial are $0,2$. put $x=0$ to get:
$0^{1000}=0=0*Q+c=c$ so $c=0$.
now derive $x^{1000} = (-x^3+2x^2)Q+ax^2+bx$ to get:
$1000x^{999}=(-3x^2+4x)Q+Q'(-x^3+2x^2)+2ax+b$ and insert $x=0$ again t oget:
$1000*0^{999} = 0 =b$ meaning $b=0$.
Now back to our original formula with $b=c=0$:
$x^{1000} = (-x^3+2x^2)Q+ax^2$
Insert $x=2$ to get:
$2^{1000} = 4a$ meaning $a=2^{998}$.
Now our original formula looks like $x^{1000} = (-x^3+2x^2)Q+2^{998}x^2$
Inserts $x=A$ to get:
$A^{1000} = 2^{998}A^2$
</s>