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.
 A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$$
\expo{At}=\alpha\pars{t} + \beta\pars{t}A + \gamma\pars{t}A^{2}
$$

$$\dot{\alpha}\pars{t} + \dot{\beta}\pars{t}A + \dot{\gamma}\pars{t}A^{2}
=A\expo{At}
=\alpha\pars{t}A + \beta\pars{t}A^{2} + \gamma\pars{t}\ \overbrace{A^{3}}^{2A^{2}}\,,
\quad
\left\lbrace%
\begin{array}{l}
\alpha\pars{0} = 1
\\[1mm]
\beta\pars{0} = \gamma\pars{0} = 0
\end{array}\right.
$$

$$
\dot{\alpha}\pars{t} =  0\,,\quad
\dot{\beta}\pars{t} =  \alpha\pars{t}\,,\quad
\dot{\gamma}\pars{t} =  \beta\pars{t} + 2\gamma\pars{t}
\quad\imp\quad
\left\lbrace%
\begin{array}{rcl}
\alpha\pars{t} & = & 1
\\[1mm]
\beta\pars{t} & = & t
\\[1mm]
\gamma\pars{t} & = & {\expo{2t} - 2t - 1 \over 4}
\end{array}\right.
$$

$$
\expo{At} = 1 + tA + {\expo{2t} - 2t - 1 \over 4}\,A^{2}
$$

\begin{align}
A^{1000} &= \left.\totald[1000]{\pars{\expo{At}}}{t}\right\vert_{t = 0}
=\left. {A^{2} \over 4}\,
\totald[1000]{\bracks{\expo{2t} - 2t - 1}}{t}\right\vert_{t = 0}
={A^{2} \over 4}\,2^{1000}
\end{align}
$$
\boxed{\vphantom{\Huge {A \over B}}\quad\color{#00f}{\large A^{1000} = 2^{998}\ A^{2}}\quad}
$$
A: Your formula tells you, after you multiply through by $A^{997}$, that
$$A^{1000}=2A^{999}.$$
Similarly,
$$2A^{999}=4A^{998}.$$
This process can be repeated to find $A^{1000}$ in terms of $A^2$, which you can then compute. 
A: $$A^{1000}= A(A^3)^{333}=A (-2A^2)^{333}=(-2)^{333}A^{667}=\cdots$$
A: 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:
$A^{1000} = 2^{998}A^2$
A: Set $X_n=A^n$ since, $A^3 =2A^2$
then, we have $$X_{n+3} = A^{n+3} = 2A^2A^n = 2X_{n+2}$$
Hence, $(X_n)_n$ is geometric and 
$$X_n =2^{n-2}X_2\Longleftrightarrow A^n =2^{n-2} A^2$$
that is for $n=1000$ we get
$$A^{1000} = 2^{998}A^2$$
