Find the $n^{th}$ power of a $2$x$2$ matrix. Let $A=\begin{pmatrix}3&-2\\2&-2\end{pmatrix}$. Using Lagrange's interpolation compute $A^n$ for $n\in\mathbb{N} $
So far I've worked out the minimum polynomial of $A$ to be $(x-2)(x+1)$ but I'm unsure what to do next. Anyone got any ideas?
 A: Yes: Your minimal polynomial tells you that since A has two different eigenvalues (2 and -1), A is diagonalizable. Find P such as : 
$P^{-1}*AP = \begin{pmatrix}2&0\\0&-1\end{pmatrix}$
That can be done by calculating A-2*I, and find a combination of columns that gives 0. Such a combination will give an eigenvector of the eigenvalue 2. 
A-2*I = $\begin{pmatrix}1&-2\\2&-4\end{pmatrix}$ and 2*C1 + 1*C2 =0 $X_2 = \begin{pmatrix}2\\1\end{pmatrix}$
The same with A+I => $X_{-1} = \begin{pmatrix}1\\2\end{pmatrix}$ 
So it gives you : P = $\begin{pmatrix}2&1\\1&2\end{pmatrix}$ and : 
$P^{-1}*A*P = \begin{pmatrix}2&0\\0&-1\end{pmatrix}$ = D
Now $(P^{-1}*A*P)^n = P^{-1}*A^n*P = D^n = \begin{pmatrix}2^n&0\\0&(-1)^n\end{pmatrix}$
Now you get $A^n$ with a simple multiplication
More generally, let's say that you want to obtain a diagonal matrix with eigenvalues values ordered :$P^{-1}*A*P$= D =$diag(a_1,...,a_n )$, the good matrix P is the one that verifies : $C_i = X_i$,  $C_i$ being the i-th column of P starting from the left and $X_i$ the eigenvalu relatively to $a_i$
Edit: I'll do the interpolation way :) , in fact it's nice as well.
The idea is to write : $X^k = a_k*P_{-1}(X) + b_k*P_2(X) $
Conditions of interpolation : $P_{-1}(-1) = 1, P_{-1}(2) =0  ; P_2(2)=1, P_2(-1)=0$
-> $P_{-1}(X) = -\frac{1}{3}*(X-2)  ; P_2(X) = \frac{1}{3}*(X+1) $
 Evaluate the relation in -1 and 2 and you get : 
$a_k = -(-1)^k$ ; $b_k = 2^k$
Finally, you get : $A^n = (-1)^n*P_{-1}(A) +2^n*P_2(A)   $
I'll let you finish the calculations, but you can verify the relation by multiplying by $P^{-1}$ on the left and P on the right:
$P^{-1}*A^n*P = (P^{-1}*A*P)^n = -(-1)^n*(D-2*I) +2^n*(D+I)$
-> $P^{-1}*A^n*P = \begin{pmatrix}2^n&0\\0&(-1)^n\end{pmatrix}$
Same result as the one above :)
You can adapt this if you have more than two distincts eigenvalues, the result stays true and you write the relation with one polynomial for each value (like here for 2). 
A: I'll give another way. The Cayley-Hamilton theorem asserts that all polynomials satisfy their own characteristic polynomial. This means that
$$(A-2)(A+1)=A^2-A-2I=\mathbf{0}\implies A^2=A+2I$$$$\implies A^n=A^{n-1}+2A^{n-2}$$.
Using the substitution $a_n=A^n$, we now have the recurrence relation $$a_n=a_{n-1}+2a_{n-2},a_0=I,a_1=A$$
Solving this through standard means of solving recurrence relations, we obtain
$$A^n=a_n=\left(-\frac13A+\frac23I\right)\cdot(-1)^n+\left(\frac13A+\frac13I\right)\cdot2^n$$
$$=\pmatrix{-\frac13\cdot(-1)^n+\frac43\cdot2^n&&\frac23\cdot(-1)^n-\frac23\cdot2^n\\
-\frac23\cdot(-1)^n+\frac23\cdot2^n&&\frac43\cdot(-1)^n-\frac13\cdot2^n}$$
