Let \(A = \begin{bmatrix} 0 & 1 \\ 6 & -1 \end{bmatrix}\). Prove that \(A\) is diagonalizable and find a closed-form expression for \(A^k \begin{bmatrix} 1 \\ 7 \end{bmatrix}\), \(k \in \mathbb{N}\).
Solution
The characteristic polynomial is \[ \det(\lambda I-A)=\lambda^2+\lambda-6=(\lambda-2)(\lambda+3). \] Thus \(A\) has two distinct eigenvalues and is diagonalizable. Corresponding eigenvectors are \[ v_2=\binom12,\qquad v_{-3}=\binom1{-3}. \] Since \[ \binom17=2v_2-v_{-3}, \] we obtain for every \(k\in\mathbb N\) \[ A^k\binom17 =2\,2^k v_2-(-3)^k v_{-3}. \] Equivalently, \[ \boxed{A^k\binom17= \binom{2^{k+1}-(-3)^k}{2^{k+2}+3(-3)^k}}. \]