# 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 $A^{-1}$. Add to solve later

## 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

### 4 Responses

1. alokin says:

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

• Yu says:

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.

2. Wizard says:

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.”

• Yu says:

Dear Wizard,

“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 $\lambda=2, \pm 1$.” Let $A$ be an $n\times n$ matrix. Assume that every vector $\mathbf{x}$ in $\R^n$ is an eigenvector for some eigenvalue...