# Find the Eigenvalues and Eigenvectors of the Matrix $A^4-3A^3+3A^2-2A+8E$.

## Problem 191

Let
$A=\begin{bmatrix} 1 & -1\\ 2& 3 \end{bmatrix}.$

Find the eigenvalues and the eigenvectors of the matrix
$B=A^4-3A^3+3A^2-2A+8E.$

(Nagoya University Linear Algebra Exam Problem)

Contents

## Hint.

Apply the Cayley-Hamilton theorem.
That is if $p_A(t)$ is the characteristic polynomial of the matrix $A$, then the matrix $p_A(A)$ is the zero matrix.

## Solution.

Let us first find the characteristic polynomial $p_A(t)$ of the matrix $A$.
We have
\begin{align*}
p_A(t)&=\det(A-tI)=\begin{vmatrix}
1-t & -1\\
2& 3-t
\end{vmatrix}\\
&=(1-t)(3-t)-(-1)(2)=t^2-4t+5.
\end{align*}
Solving $t^2-4t+5=0$, we see that the matrix $A$ has the eigenvalues $2\pm i$ but it is not a good idea to use this directly to find the eigenvalues of the matrix $B$.
Instead, note that by the Cayley-Hamilton theorem, we know that
$p_t(A)=A^2-4A+5I=O,$ where $I$ is the $2\times 2$ identity matrix and $O$ is the $2\times 2$ zero matrix.

Since we have
$B=A^4-3A^3+3A^2-2A+8E=(A^2-4A+5I)(A^2+A+2I)+A-2I,$ we have
$B=A-2I=\begin{bmatrix} -1 & -1\\ 2& 1 \end{bmatrix}.$ Since the eigenvalues of $A$ is $2\pm i$, the eigenvalues of $B=A-2I$ are
$(2\pm i)-2=\pm i.$

Next, we find eigenvectors.
Let us first find eigenvectors corresponding to the eigenvalue $i$.
We have
\begin{align*}
A-iI&=\begin{bmatrix}
-1-i & -1\\
2& 1-i
\end{bmatrix}
\xrightarrow{(-1+i)R_1}
\begin{bmatrix}
2 & 1-i\\
2 & 1-i
\end{bmatrix}\\
&
\xrightarrow{R_2-R_1}
\begin{bmatrix}
2 & 1-i\\
0 & 0
\end{bmatrix}
\xrightarrow{\frac{1}{2}R_1}
\begin{bmatrix}
1 & (1-i)/2\\
0 & 0
\end{bmatrix}.
\end{align*}
Thus we have
$x_1=-\frac{1-i}{2}$ and the eigenvectors associated with the eigenvalue $i$ are
$\mathbf{x}=x_2\begin{bmatrix} -\frac{1-i}{2} \\ 1 \end{bmatrix},$ where $x_2$ is any nonzero complex number.
Or equivalently, scaling the vector by $-1+i$, the eigenvectors corresponding to the eigenvalue $i$ are
$a\begin{bmatrix} 1 \\ -1-i \end{bmatrix},$ where $a$ is any nonzero complex number.

Since $B$ is a real matrix and the eigenvalues $i$ and $-i$ are complex conjugate to each other, the eigenvectors of $-i$ are just the conjugates of eigenvectors of $i$. Thus the eigenvectors corresponding to the eigenvalue $-i$ are
$b\begin{bmatrix} 1 \\ -1+i \end{bmatrix},$ where $b$ is any nonzero complex number.

### More from my site

#### You may also like...

##### A Matrix Having One Positive Eigenvalue and One Negative Eigenvalue

Prove that the matrix \[A=\begin{bmatrix} 1 & 1.00001 & 1 \\ 1.00001 &1 &1.00001 \\ 1 & 1.00001 & 1...

Close