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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?

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    $\begingroup$ The theorem can be further generalized to polynomial case (your guess is right and not hard to verify). $\endgroup$ – Zhanxiong Jul 13 '15 at 21:16
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    $\begingroup$ If $Ax = \lambda x$ then $(A^{10} + A^7 + 5A)(x) = A^{10}x + A^7x + 5Ax = \lambda^{10}x + \lambda^7 x + 5 \cdot \lambda x = (\lambda^{10} + \lambda^7 + 5 \cdot \lambda) \cdot x$. $\endgroup$ – Krijn Jul 13 '15 at 21:19
  • $\begingroup$ You should re-check your algebra, the eigenvalues are $2,1,1$ $\endgroup$ – John McGee Jul 13 '15 at 21:20
  • $\begingroup$ Ah right, thanks. I wrote down the wrong minor. Also, for the second question, should I first find an invertible matrix $P$ such that $P^{-1} A P$ is a diagonal matrix, and then compute $A^{10}$? $\endgroup$ – Kamil Jul 13 '15 at 21:23
  • $\begingroup$ It might be easier to find the eigenvectors and write $X$ as a linear combination of these eigenvectors. Then $A^{10}X$ should be easy to calculate. $\endgroup$ – Krijn Jul 13 '15 at 21:25
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For the first question, observe that $$(A^{10}+A^{7}+5A)x=(\lambda^{10}+\lambda^7+5\lambda)x$$ 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 $$A^{10}X = A^{10}(3\cdot v_1 + 4\cdot v_2 + 1\cdot v_3) = 3\cdot A^{10}v_1 + 4\cdot A^{10} v_2 + A^{10}v_3$$ $$ = 3\cdot \lambda_2^{10} \cdot v_1 + 4\cdot \lambda_1^{10}\cdot v_2 + \lambda_1^{10}\cdot v_3 = 3\cdot 2^{10} \cdot v_1 + 4\cdot 1^{10}\cdot v_2 + 1^{10}\cdot v_3$$

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  • $\begingroup$ I don't understand how $v_3$ can be a different eigenvector from $v_2$, when the eigenvalue is $1$ in both cases? How did you compute $v_3$? $\endgroup$ – Kamil Jul 13 '15 at 22:35
  • $\begingroup$ If an eigenvalue $\lambda$ has multiplicity $> 1$, such as $\lambda = 1$ in our case, it can happen that there are two linearly independent eigenvectors $v$, $w$ such that $Av = \lambda v$ and $Aw = \lambda w$. There is an easy test for the amount of linearly independent eigenvectors that belong to a certain eigenvalue, namely: The amount of linearly independent eigenvalues for a certain eigenvalue $\lambda$ is equal to the dimension of the kernel of $A - \lambda I$. In our case, the dimension of the kernel of $A - 1\cdot I$ is equal to $2$, therefore, there are two eigenvectors. $\endgroup$ – Krijn Jul 13 '15 at 22:41
  • $\begingroup$ Thanks, that was very clear. I will look for a proof of this theorem. Also, is it correct that two distinct eigenvalues can never have the same eigenvector? $\endgroup$ – Kamil Jul 13 '15 at 22:53
  • $\begingroup$ That is indeed correct and very easy to proof, because if $Av = \lambda_1 v$ and $Av = \lambda_2 v$ then of course $\lambda_1 v = \lambda_2 v \Rightarrow \lambda_1 = \lambda_2$. $\endgroup$ – Krijn Jul 13 '15 at 22:54
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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: $$P(A)(v)=\sum_{k=0}^na_kA^kv=\sum_{k=0}^na_k\lambda^kv=P(\lambda)v$$

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

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By definition of eigenvalue:

$$Av=\lambda v$$

for $v\neq 0$ 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:

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

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

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

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

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

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

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