Power a Matrix (Without calculating) So, I'm reading a linear algebra book (Fraleigh's - self learning) and one of the exercises about Matrix product is to power the matrix:
$$ 

\left( \begin{matrix} 

0 & 1\\
-1 & 1

\end{matrix} \right)

$$
to the 2001st power. But can't see how without calculating. Any tip?
I have the "feeling" that given that 1 only changes on its sig then its the same the power of 3 that the power of 2001 but I'm not sure. I'm correct? 
 A: Let $A$ be your matrix. It can be calculated that $A^3=-I$ where $I$ is the identity matrix. Then $A^{2001}=(A^3)^{667}=(-I)^{667}=-I$
A: If $A^n=I$ for a matrix $A$ and natural number $n$, then $A^{kn+r}=A^{kn}A^r=(A^n)^kA^r=I^kA^r=A^r$. Thus, $A^l\equiv A^m$ if $l=m\pmod n$. In the present case, $A^6=I$, and thus $A^{2001}=A^3$, since $2001=3\pmod6$.
A: Actually, it's not harder to compute the $n$th power of any $2$ by $2$ matrix:  
Let $a,b,c,d$ be complex numbers and consider the matrix 
$$A=\begin{pmatrix}a&b\\ c&d\end{pmatrix}.$$ 
Let $n\ge2$ be an integer. 
How to compute $A^n$? 
Here is a recipe: 
Assume first that the roots $u$ and $v$ of the polynomial 
$$f:=X^2-(a+d)\,X+ad-bc$$ 
are distinct. The equation of the secant line to the curve $y=x^n$ through the points $(u,u^n)$ and $(v,v^n)$ is 
$$y=\frac{u^n-v^n}{u-v}\ \ x-uv\ \ \frac{u^{n-1}-v^{n-1}}{u-v}\quad,$$ 
and we have 
$$A^n=\frac{u^n-v^n}{u-v}\ \ A-uv\ \ \frac{u^{n-1}-v^{n-1}}{u-v}\ \ I,$$ 
where $I$ is the identity matrix. 
For all nonnegative integer $k$ put 
$$s_k:=u^k+u^{k-1}v+u^{k-2}v^2+\cdots+v^k.$$ 
[In particular $s_0=1$.] As 
$$s_k=\frac{u^{k+1}-v^{k+1}}{u-v}\quad,$$ 
we have 
$$A^n=s_{n-1}\,A-u\,v\,s_{n-2}\,I.$$ 
This formula still makes sense, and is still true, if $u=v$, in which case it reads 
$$A^n=n\,u^{n-1}\,A-(n-1)\,u^n\,I.$$ 
Why does this recipe work?
The key point is the easily checked equality 
$$A^2-(a+d)\,A+(ad-bc)\,I,\quad(1)$$ 
which enables us to compute 
$$g(A):=a_n\,A^n+\cdots+a_2\,A^2+a_1\,A+a_0\,I$$ 
for all polynomial 
$$g=a_n\,X^n+\cdots+a_2\,X^2+a_1\,X+a_0\in\mathbb C[X]$$ 
as follows. 
Assume again $u\not=v$, and let $h\in\mathbb C[X]$ be the unique polynomial of degree $\le1$ which agrees with $g$ at $u$ and $v$: 
$$h=g(u)\ \frac{X-v}{u-v}+g(v)\ \frac{X-u}{v-u}\quad.$$ 
[In particular, the secant line to $y=g(x)$ through $(u,g(u))$ and $(u,g(u))$ is $y=h(x)$.] Then our polynomial $f$, which can be written as 
$$f=(X-u)(X-v),$$ 
divides $g-h$. That is, we have 
$$g(X)-h(X)=f(X)q(X)$$ 
for some polynomial $q$. On substituting $A$ for $X$, and remembering that $f(A)=0$ by (1), we get $g(A)=h(A)$, or 
$$g(A)=g(u)\ \frac{A-vI}{u-v}+g(v)\ \frac{A-uI}{v-u}\quad.$$ 
If $u=v$ we use the tangent line instead of the secant line: 
$$h:=g(u)+g'(u)\,(X-u),$$ 
and we get 
$$g(A)=g(u)\,I+g'(u)\,(A-u\,I).$$ 
This is the case of $2$ by $2$ matrices. For arbitrary square matrix, see for instance the last part of this answer. 
EDIT. The $n$th power of an $r$ by $r$ diagonalizable matrix $A$ is given by the beautiful Lagrange Interpolation Formula: 
$$A^n=\sum_{i=1}^k\ u_i^n\ \prod_{j\not=i}\ \frac{A-u_j\,I}{u_i-u_j}\quad,$$ 
where the $u_i$ are the distinct eigenvalues. 
Here the justification (which is essentially the same as above). The polynomial 
$$f:=(X-u_1)\cdots(X-u_k)$$ 
satisfies $f(A)=0$. Note that 
$$g:=\sum_{i=1}^k\ u_i^n\ \prod_{j\not=i}\ \frac{X-u_j}{u_i-u_j}$$ 
is the unique polynomial of degree less than $k$ which agrees with $X^n$ on all the $u_i$. Hence, $X^n-g$ is divisible by $f$, and this implies, as above, $A^n=g(A)$.
