<s>[INST] Suppose $A \in M_{3,3}$ with eigenvalues $1,2,3$, eigenvectors $b_1, b_2, b_3$. Let $v = b_1 - 4b_2 + 3b_3$. Compute $A^5 v$. I think I did this right, but someone skim and double check?
Q: Suppose $A$ is a $3 \times 3$ matrix, with eigenvalues $1,2,3$ and corresponding eigenvectors $b_1, b_2, b_3$. Suppose that $v = b_1 - 4b_2 + 3b_3$. Compute $A^5 v$.
Let $\mathcal{B}$ be the eigenvector basis and $D$ be the diagonal eigenvalue matrix:
\begin{align*}
  D &= \begin{pmatrix}
    1 & 0 & 0 \\
    0 & 2 & 0 \\
    0 & 0 & 3 \\
  \end{pmatrix} \\  
  P_{\mathcal{BE}} &= \begin{pmatrix} \begin{pmatrix} b_1 \end{pmatrix} & \begin{pmatrix} b_2 \end{pmatrix} & \begin{pmatrix} b_3 \end{pmatrix} \end{pmatrix} \\
  [v]_{\mathcal{B}} &= (1,-4,3)^T \\
  A &= P_{\mathcal{BE}}^{-1} D P_{\mathcal{BE}} \\
  A^5 &= P_{\mathcal{BE}}^{-1} D^5 P_{\mathcal{BE}} \\
  A^5 v &= P_{\mathcal{BE}}^{-1} D^5 P_{\mathcal{BE}} [v]_{\mathcal{E}} \\
  [A^5 v]_{\mathcal{B}} &= D^5 [v]_{\mathcal{B}} \\
  [A^5 v]_{\mathcal{B}} &= (1^5 \cdot 1, 2^5 \cdot -4, 3^5 \cdot 3)^T \\
  [A^5 v]_{\mathcal{B}} &= (1, -128, 729)^T \\
  A^5 v &= b_1 - 128 b_2 + 729 b_3 \\
\end{align*}
 [/INST] That's fine, to check note that for each eigenvectors
$$A\vec v=\lambda \vec v\implies A^n(c\vec v)=c\lambda^n \vec v$$
</s>