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 $A^5 v$.

Question

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*}

Answer 1

That's fine, to check note that for each eigenvectors

$$A\vec v=\lambda \vec v\implies A^n(c\vec v)=c\lambda^n \vec v$$

Answer 2

That's too compicated. You might have done just this: Since

• $A^5.b_1=b_1$;
• $A^5.(-4b_2)=-4A^5.b_2=-4\times2^5b_2=-128b_2$;
• $A^5.(3b_3)=3A^5.b_3=3\times3^5b_3=729b_3$,

then $A.(b_1-4b_2+3b_3)=b_1-128b_2+729b_3$.

Answer 3

$$ A^5v =A^5 b_1 - 4A^5b_2 + 3A^5b_3=$$

$$(1^5)b_1-4(2^5)b_2+3(3^5)b_3=$$

$$b_1-128b_2+729b_3$$