# 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*)

Add to solve later

Sponsored Links

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.

Add to solve later

Sponsored Links