Square matrix multiplication when raised to a power Is there any way to evaluate an expression that involves a matrix to power $n$, multiplied with some constant $D$:$$D\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{n} $$ I can obviously do it if this power wasn't there but with it in place I am somewhat stuck, are there any helpful linear algebra rules that can be used? I know the eigen values and vectors for this matrix as well if that helps
 A: By the Jordan Normal Form any $2\times 2$ matrix $M$ (over $\Bbb C$) is similar to a matrix $J$ that looks like one of the following:
$$J=\begin{bmatrix} \lambda _1 & 0\\ 0 & \lambda_2\end{bmatrix} \text{ or } J=\begin{bmatrix} \lambda_1 & 1\\ 0 & \lambda_1 \end{bmatrix}$$ where $\lambda_1 ,\lambda _2$ are the eigenvalues of $M$.
Therefore $M^n=(P^{-1}JP)^n=\color{grey}{P^{-1}J(PP^{-1})JP\ldots P^{-1}J(PP^{-1})JP}=P^{-1}J^nP$.
If $J=\begin{bmatrix} \lambda _1 & 0\\ 0 & \lambda_2\end{bmatrix}$, then $J^n=\begin{bmatrix} \lambda _1^n & 0\\ 0 & \lambda_2^n\end{bmatrix}$.
If $J=\begin{bmatrix} \lambda_1 & 1\\ 0 & \lambda_1 \end{bmatrix}$, then $J^n=\begin{bmatrix} \lambda_1^n & n\lambda_1^{n-1}\\ 0 & \lambda_1^n \end{bmatrix}$. (See this).
All that is left for you to do is find $P$.
A: The case when the matrix $A$ is diagonalzable is obvious. Now if the matrix is not diagonalizable then in $\mathbb{C}$ it's trigonalizable and $A$ can be reduced in the Jordan canonical form
$$A=P\left(\begin{array}{cc}\lambda&1\\0&\lambda\end{array}\right)P^{-1}=PBP^{-1}$$
and we have $B=D+N$ where $D=\mathrm{diag}(\lambda,\lambda)$ and $N$ is nilpotent matrix  and $N^2=0$ so we have 
$$B^n=(D+N)^n=D^n+nD^{n-1}N$$
hence we have $$A^n=PB^nP^{-1}=P(D^n+nD^{n-1}N)P^{-1}$$
A: Diagonalize (where S is matrix of eigenvectors and $\Lambda$ is the diagonal matrix with eigenvalues):
$$A=S \Lambda S^{-1}$$
$$A^n=S \Lambda S^{-1}S \Lambda S^{-1}... = S \Lambda^n S^{-1}$$
This works as long as A has 2 unique eigenvalues.
