# Find All the Eigenvalues of Power of Matrix and Inverse Matrix

## Problem 361

Let

\[A=\begin{bmatrix}

3 & -12 & 4 \\

-1 &0 &-2 \\

-1 & 5 & -1

\end{bmatrix}.\]
Then find all eigenvalues of $A^5$. If $A$ is invertible, then find all the eigenvalues of $A^{-1}$.

## Proof.

We first determine all the eigenvalues of the matrix $A$.

The characteristic polynomial $p(t)$ of $A$ is given by

\begin{align*}

p(t)&=\det(A-tI)\\[6pt]
&=\begin{vmatrix}

3-t & -12 & 4 \\

-1 & -t &-2 \\

-1 & 5 & -1-t

\end{vmatrix}.

\end{align*}

Using the first row cofactor expansion, we compute

\begin{align*}

p(t)&=(3-t)\begin{vmatrix}

-t & -2\\

5& -1-t

\end{vmatrix}

-(-12)\begin{vmatrix}

-1 & -2\\

-1& -1-t

\end{vmatrix}+4\begin{vmatrix}

-1 & -t\\

-1& 5

\end{vmatrix}\\[6pt]
&=(3-t)(t^2+t+10)+12(t-1)+4(-5-t)\\

&=-t^3+2t^2+8t-2.

\end{align*}

Therefore the characteristic polynomial of $A$ is

\[p(t)=-t^3+2t^2+8t-2\]
and it can be factored as

\[p(t)=-(t-2)(t-1)(t+1).\]
The roots of the characteristic polynomials are all the eigenvalues of $A$.

Thus, $2, \pm 1$ are the eigenvalues of $A$.

To find the eigenvalues of $A^5$, recall that if $\lambda$ is an eigenvalue of $A$, then $\lambda^5$ is an eigenvalue of $A^5$.

It follows from this fact that $2^5, (-1)^5, 1^5$ are eigenvalues of $A^5$.

Since $A^5$ is a $3\times 3$ matrix, its characteristic polynomial has degree $3$, hence there are at most $3$ distinct eigenvalues of $A^5$.

Because we have found three eigenvalues, $32, -1, 1$, of $A^5$, these are all the eigenvalues of $A^5$.

Recall that a matrix is singular if and only if $\lambda=0$ is an eigenvalue of the matrix.

Since $0$ is not an eigenvalue of $A$, it follows that $A$ is nonsingular, and hence invertible. If $\lambda$ is an eigenvalue of $A$, then $\frac{1}{\lambda}$ is an eigenvalue of the inverse $A^{-1}$.

So $\frac{1}{\lambda}$ are eigenvalues of $A^{-1}$ for $\lambda=2, \pm 1$.

As above, the matrix $A^{-1}$ is $3\times 3$, hence it has at most three distinct eigenvalues. We have found $1/2, \pm 1$ are eigenvalues of $A^{-1}$, hence these are all the eigenvalues of $A^{-1}$.

In summary, all the eigenvalues of $A^5$ are $\pm 1, 32$. The matrix $A$ is invertible and all the eigenvalues of $A^{-1}$ are $\pm 1, 1/2$.

## Comment.

Do not try to compute $A^5$ and $A^{-1}$ and then find their eigenvalues.

It will be tedious for hand computation.

Add to solve later

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There is an error in the characteristical polynomial, it is not p(t)=−t^3+2t^2+8t−2, but p(t)=−t^3+2t^2+t−2

Dear alokin,

Thank you for catching the typo. I fixed the problem.

There was no change in the factorization and the rest of the argument.

I believe there is a typo in “So 1λ, λ=2,±1 are eigenvalues of A inverse.” towards the very end of your answer.

It should be “So 1λ, λ=2,±1 are eigenvalues of A.”

Dear Wizard,

Thank you for your comment.

“So $\frac{1}{\lambda}$, $\lambda=2, \pm 1$ are eigenvalues of $A^{-1}$” was not clear. I meant “So $\frac{1}{\lambda}$ are eigenvalues of $A^{-1}$ for $\lambda=2, \pm 1$.”