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Problem: Let $A = begin{pmatrix} 1 & 2 & -1 \ 0 & 5 & -2 \ 0 & 6 & -2 end{pmatrix}$.

1) Compute the eigenvalues of $A^{10} + A^7 + 5A$.

2) Compute $A^{10} X$ for the vector $X = begin{pmatrix} 2 \ 4 \ 7 end{pmatrix}$.

Attempt at solution: I first computed the eigenvalues of $A$. The characteristic polynomial gives begin{align*} det(A - x mathbb{I}_3) = det begin{pmatrix} 1-x & 2 & -1 \ 0 & 5-x & -2 \ 0 & 6 & -2 -x end{pmatrix} end{align*} Laplace expansion along the first column gives begin{align*} det(A - x mathbb{I}_3) = (1-x) det begin{pmatrix} 5-x & -2 \ 6 & -2-x end{pmatrix} &= (1-x) [(5-x)(-2-x)+12] \ &= (1-x)(x^2-3x+2) \ &= (x-1)^2(2-x)end{align*} So the eigenvalues are $lambda_1 = 1, lambda_2 = 1$ and $lambda_3 = -2$.

Now, I'm aware that there is a theorem which says that if the matrix $A$ has an eigenvalue $lambda$, then $A^k$ has the eigenvalue $lambda^k$ corresponding to the same eigenvector. But I'm not sure what I should do about this sum here. Can I just add the corresponding eigenvalues? So for the first eigenvalue that would give me: $ 1^{10} + 1^7 + 5(1) = 7$?

Any help would be appreciated.

Edit: Figured out part 1), but I'm still wondering what to do with 2). Should I find an invertible matrix $P$ such that $P^{-1} A P$ is a diagonal matrix and then I can compute the powers easily?

For the first question, observe that
$$(A^{10}+A^{7}+5A)x=(lambda^{10}+lambda^7+5lambda)x$$ as I have shown in the comments.

For the second question, you observed correctly that we have three eigenvectors, for the eigenvalues $2$, $1$ and $1$, respectively
$$v_1 = begin{pmatrix} frac{1}{3}\ frac{2}{3} \ 1 \ end{pmatrix}, v_2 = begin{pmatrix} 0\ frac{1}{2} \ 1 \ end{pmatrix}, v_3= begin{pmatrix} 1\ 0 \ 0 \ end{pmatrix}$$

We see that $X = begin{pmatrix} 2\ 4 \ 7 \ end{pmatrix} = 3 cdot begin{pmatrix} frac{1}{3}\ frac{2}{3} \ 1 \ end{pmatrix} + 4 cdot begin{pmatrix} 0\ frac{1}{2} \ 1 \ end{pmatrix} + 1 cdot begin{pmatrix} 1\ 0 \ 0 \ end{pmatrix}$

With this we can calculate $A^{10}X$ quite easily, because $$A^{10}X = A^{10}(3cdot v_1 + 4cdot v_2 + 1cdot v_3) = 3cdot A^{10}v_1 + 4cdot A^{10} v_2 + A^{10}v_3$$ $$ = 3cdot lambda_2^{10} cdot v_1 + 4cdot lambda_1^{10}cdot v_2 + lambda_1^{10}cdot v_3 = 3cdot 2^{10} cdot v_1 + 4cdot 1^{10}cdot v_2 + 1^{10}cdot v_3$$

If $P(X)=sum_{k=0}^na_kX^k$ is a polynomial and a matrix $A$ has en eigenvalue $lambda$ with eigenvector $v$, we have:
$$P(A)(v)=sum_{k=0}^na_kA^kv=sum_{k=0}^na_klambda^kv=P(lambda)v$$

Thus, $P(lambda)$ is an eigenvalue of $P(A)$.

By definition of eigenvalue:

$$Av=lambda v$$

for $vneq 0$ and corresponding eigenvector. So:

$$A^{2}v=A (Av)=A(lambda v)=lambda Av=lambda lambda v=lambda^2 v$$

The same way we can show that:

$$A^k v=lambda^k v$$

So:

$$A^{10}v=lambda^{10}v$$

$$A^{7}v=lambda^{7}v$$

$$5A^{1}v=5lambda^{1}v$$

If we add these three equations side by side we have:

$$(A^{10}+A^{7}+5A)v=(lambda^{10}+lambda^7+5lambda)v$$

So if $lambda$ is eigenvalue of $A$, then $lambda^{10}+lambda^7+5lambda$ is eigenvalue of $A^{10}+A^{7}+5A$.