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

2 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.

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Linear Algebra If Every Vector is Eigenvector, then Matrix is a Multiple of Identity Matrix

Let $A$ be an $n\times n$ matrix. Assume that every vector $\mathbf{x}$ in $\R^n$ is an eigenvector for some eigenvalue...

Close