How to confirm the $k$-th power of this matrix? For any $k \in \mathbb N$,
$$\begin{bmatrix}2&-2\\0.5&0\end{bmatrix}^k = \begin{bmatrix}k+1&-2k\\k/2&-(k-1)\end{bmatrix}$$  
I found this empirically using numerical computation, but I cannot prove it. I have tried computing the eigendecomposition, but this matrix is not diagonalizable.

Using the Cayley-Hamilton Theorem, the characteristic polynomial is:
$$p(\lambda)=\det(\lambda I_2-A)=\begin{vmatrix}\lambda-2&2\\ -0.5&\lambda\end{vmatrix}=\lambda^2-2\lambda+1=0.$$
Hence:
$$p(A)=A^2-2A+I=0 \Rightarrow A^2=2A-I.$$
Thus:
$$\begin{align}A^2&=2A-I;\\
A^3&=A^2A=(2A-I)A=2A^2-A=2(2A-I)-A=3A-2I;\\
A^4&=A^3A=(3A-2I)A=3A^2-2A=3(2A-I)-2A=4A-3I;\\
\vdots \\
A^k&=kA-(k-1)I=k\begin{pmatrix}2&-2\\0.5&0\end{pmatrix}-(k-1)\begin{pmatrix}1&0\\0&1\end{pmatrix}=\begin{pmatrix}k+1&-2k\\0.5k&-(k-1)\end{pmatrix}.\\
\end{align}$$
