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 $A^{-1}$.

 
LoadingAdd 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.


LoadingAdd to solve later

Sponsored Links

More from my site

  • Find the Inverse Matrix Using the Cayley-Hamilton TheoremFind the Inverse Matrix Using the Cayley-Hamilton Theorem Find the inverse matrix of the matrix \[A=\begin{bmatrix} 1 & 1 & 2 \\ 9 &2 &0 \\ 5 & 0 & 3 \end{bmatrix}\] using the Cayley–Hamilton theorem.   Solution. To use the Cayley-Hamilton theorem, we first compute the characteristic polynomial $p(t)$ of […]
  • How to Use the Cayley-Hamilton Theorem to Find the Inverse MatrixHow to Use the Cayley-Hamilton Theorem to Find the Inverse Matrix Find the inverse matrix of the $3\times 3$ matrix \[A=\begin{bmatrix} 7 & 2 & -2 \\ -6 &-1 &2 \\ 6 & 2 & -1 \end{bmatrix}\] using the Cayley-Hamilton theorem.   Solution. To apply the Cayley-Hamilton theorem, we first determine the characteristic […]
  • Eigenvalues and their Algebraic Multiplicities of a Matrix with a VariableEigenvalues and their Algebraic Multiplicities of a Matrix with a Variable Determine all eigenvalues and their algebraic multiplicities of the matrix \[A=\begin{bmatrix} 1 & a & 1 \\ a &1 &a \\ 1 & a & 1 \end{bmatrix},\] where $a$ is a real number.   Proof. To find eigenvalues we first compute the characteristic polynomial of the […]
  • Characteristic Polynomial, Eigenvalues, Diagonalization Problem (Princeton University Exam)Characteristic Polynomial, Eigenvalues, Diagonalization Problem (Princeton University Exam) Let \[\begin{bmatrix} 0 & 0 & 1 \\ 1 &0 &0 \\ 0 & 1 & 0 \end{bmatrix}.\] (a) Find the characteristic polynomial and all the eigenvalues (real and complex) of $A$. Is $A$ diagonalizable over the complex numbers? (b) Calculate $A^{2009}$. (Princeton University, […]
  • Rotation Matrix in Space and its Determinant and EigenvaluesRotation Matrix in Space and its Determinant and Eigenvalues For a real number $0\leq \theta \leq \pi$, we define the real $3\times 3$ matrix $A$ by \[A=\begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta &\cos\theta &0 \\ 0 & 0 & 1 \end{bmatrix}.\] (a) Find the determinant of the matrix $A$. (b) Show that $A$ is an […]
  • Find Inverse Matrices Using Adjoint MatricesFind Inverse Matrices Using Adjoint Matrices Let $A$ be an $n\times n$ matrix. The $(i, j)$ cofactor $C_{ij}$ of $A$ is defined to be \[C_{ij}=(-1)^{ij}\det(M_{ij}),\] where $M_{ij}$ is the $(i,j)$ minor matrix obtained from $A$ removing the $i$-th row and $j$-th column. Then consider the $n\times n$ matrix […]
  • Maximize the Dimension of the Null Space of $A-aI$Maximize the Dimension of the Null Space of $A-aI$ Let \[ A=\begin{bmatrix} 5 & 2 & -1 \\ 2 &2 &2 \\ -1 & 2 & 5 \end{bmatrix}.\] Pick your favorite number $a$. Find the dimension of the null space of the matrix $A-aI$, where $I$ is the $3\times 3$ identity matrix. Your score of this problem is equal to that […]
  • True of False Problems on Determinants and Invertible MatricesTrue of False Problems on Determinants and Invertible Matrices Determine whether each of the following statements is True or False. (a) If $A$ and $B$ are $n \times n$ matrices, and $P$ is an invertible $n \times n$ matrix such that $A=PBP^{-1}$, then $\det(A)=\det(B)$. (b) If the characteristic polynomial of an $n \times n$ matrix $A$ […]

You may also like...

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.

Leave a Reply to alokin Cancel reply

Your email address will not be published. Required fields are marked *

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

More in Linear Algebra
Problems and solutions 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