Find All the Eigenvalues of Power of Matrix and Inverse Matrix

Linear algebra problems and solutions

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 $A^{-1}$.

 
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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.


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