Diagonalizing a 2×2 matrix

Eigenvalues and eigenvectors. From the earlier exercise: $\lambda_1=2$ with $v_1=(1,-2)$, and $\lambda_2=5$ with $v_2=(1,1)$.

Matrix $P$. The columns of $P$ are the eigenvectors:
$$P = \begin{pmatrix} 1 & 1 \\ -2 & 1 \end{pmatrix}.$$

Inverse of $P$. $\det P=1\cdot1-1\cdot(-2)=3$, so
$$P^{-1}=\frac13\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix}.$$

Diagonal matrix.
$$D = \begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix}.$$

Verification. $PDP^{-1}=A$ (direct computation).

Conclusion. $A=PDP^{-1}$ with the matrices above. Consequently $A^k=PD^kP^{-1}$.