# Find All Eigenvalues and Corresponding Eigenvectors for the $3\times 3$ matrix

## Problem 720

Find all eigenvalues and corresponding eigenvectors for the matrix $A$ if
$A= \begin{bmatrix} 2 & -3 & 0 \\ 2 & -5 & 0 \\ 0 & 0 & 3 \end{bmatrix} .$

## Solution.

If $\lambda$ is an eigenvalue of $A$, then $\lambda$ satisfies
\begin{align*}
0
&=
\det(A-\lambda I)
=
\det\left(
\begin{bmatrix}
2 & -3 & 0 \\
2 & -5 & 0 \\
0 & 0 & 3
\end{bmatrix}

\begin{bmatrix}
\lambda & 0 & 0 \\
0 & \lambda & 0 \\
0 & 0 & \lambda
\end{bmatrix}
\right)
\\
&=
\begin{vmatrix}
2-\lambda & -3 & 0 \\
2 & -5-\lambda & 0 \\
0 & 0 & 3-\lambda
\end{vmatrix}
=
(3-\lambda)
\begin{vmatrix}
2-\lambda & -3 \\
2 & -5-\lambda
\end{vmatrix}
\\
&=
(3-\lambda)
\left[(2-\lambda)(-5-\lambda)+6\right] =
(3-\lambda)
(-10+5\lambda-2\lambda+\lambda^{2}+6)
\\
&=
(3-\lambda)(\lambda^{2}+3\lambda-4)
=
-(\lambda-3)(\lambda+4)(\lambda-1)
.
\end{align*}
In the above calculation, we calculated the determinant of the $3\times 3$ matrix by expanding along the third row. From the above equation, we can conclude that the eigenvalues of $A$ are $\lambda=3,1,-4$.

We will first find the eigenvectors corresponding to the eigenvalue $\lambda=3$. Any such eigenvector $\mathbf{x}$ must satisfy
\begin{align*}
\begin{bmatrix}
0 \\ 0 \\ 0
\end{bmatrix}
&=
(A-3I)\mathbf{x}
=
\left(
\begin{bmatrix}
2 & -3 & 0 \\
2 & -5 & 0 \\
0 & 0 & 3
\end{bmatrix}
-3
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\right)
\begin{bmatrix}
x_{1} \\ x_{2} \\ x_{3}
\end{bmatrix}
\\
&=
\begin{bmatrix}
-1 & -3 & 0 \\
2 & -8 & 0 \\
0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
x_{1} \\ x_{2} \\ x_{3}
\end{bmatrix}
.
\end{align*}
The augmented matrix for this system is given by
$\left[\begin{array}{ccc|c} -1 & -3 & 0 & 0 \\ 2 & -8 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \xrightarrow{R_{2}+2R_{1}} \left[\begin{array}{ccc|c} -1 & -3 & 0 & 0 \\ 0 & -14 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \xrightarrow[-\frac{1}{14}R_{2}]{-R_{1}}$ $\left[\begin{array}{ccc|c} 1 & 3 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \xrightarrow{R_{1}-3R_{2}} \left[\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] .$ Therefore, the solution is given by $x_{1}=0$ and $x_{2}=0$. Since no other restrictions are given on $x_{3}$, we can conclude that any eigenvector $\mathbf{x}$ of $A$ corresponding to $\lambda=3$ must be of the form
$\mathbf{x}=x_{3} \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$ where $x_{3}\neq 0$.

Next, any eigenvector $\mathbf{x}$ corresponding to the eigenvalue $\lambda=1$ must satisfy
\begin{align*}
\begin{bmatrix}
0 \\ 0 \\ 0
\end{bmatrix}
&=
(A-3I)\mathbf{x}
=
\left(
\begin{bmatrix}
2 & -3 & 0 \\
2 & -5 & 0 \\
0 & 0 & 3
\end{bmatrix}

\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\right)
\begin{bmatrix}
x_{1} \\ x_{2} \\ x_{3}
\end{bmatrix}
\\
&=
\begin{bmatrix}
1 & -3 & 0 \\
2 & -6 & 0 \\
0 & 0 & 2
\end{bmatrix}
\begin{bmatrix}
x_{1} \\ x_{2} \\ x_{3}
\end{bmatrix}
.
\end{align*}
The augmented matrix for this system is given by
$\left[\begin{array}{ccc|c} 1 & -3 & 0 & 0 \\ 2 & -6 & 0 & 0 \\ 0 & 0 & 2 & 0 \end{array}\right] \xrightarrow[\frac{1}{2}R_{3}]{R_{2}-2R_{1}} \left[\begin{array}{ccc|c} 1 & -3 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}\right] \xrightarrow{R_{2}\leftrightarrow R_{3}} \left[\begin{array}{ccc|c} 1 & -3 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] .$ Therefore, the solution to this system is given by $x_{1}-3x_{2}=0$, i.e. $x_{1}=3x_{2}$, and $x_{3}=0$. Therefore, any eigenvector $\mathbf{x}$ of $A$ corresponding to $\lambda=1$ must be of the form
$\mathbf{x} =x_{2} \begin{bmatrix} 3 \\ 1 \\ 0 \end{bmatrix}$ where $x_{2}\neq 0$.

Finally, any eigenvector $\mathbf{x}$ corresponding to the eigenvalue $\lambda=-4$ must satisfy
\begin{align*}
\begin{bmatrix}
0 \\ 0 \\ 0
\end{bmatrix}
&=
(A+4I)\mathbf{x}
=
\left(
\begin{bmatrix}
2 & -3 & 0 \\
2 & -5 & 0 \\
0 & 0 & 3
\end{bmatrix}
+4
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\right)
\begin{bmatrix}
x_{1} \\ x_{2} \\ x_{3}
\end{bmatrix}
\\
&=
\begin{bmatrix}
6 & -3 & 0 \\
2 & -1 & 0 \\
0 & 0 & 7
\end{bmatrix}
\begin{bmatrix}
x_{1} \\ x_{2} \\ x_{3}
\end{bmatrix}
.
\end{align*}
The augmented matrix for this system is given by
$\left[\begin{array}{ccc|c} 6 & -3 & 0 & 0 \\ 2 & -1 & 0 & 0 \\ 0 & 0 & 7 & 0 \end{array}\right] \xrightarrow[\frac{1}{7}R_{3}]{R_{2}-\frac{1}{6}R_{1}} \left[\begin{array}{ccc|c} 6 & -3 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}\right]$ $\xrightarrow[R_{2}\leftrightarrow R_{3}]{\frac{1}{6}R_{1}} \left[\begin{array}{ccc|c} 1 & -1/2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] .$ Therefore, the solution to this system satifies $x_{3}=0$ and $x_{1}-(1/2)x_{2}=0$, the latter of which reduces to $x_{1}=(1/2)x_{2}$. Thus any eigenvalue $\mathbf{x}$ of $A$ corresponding to $\lambda=-4$ must be of the form
$\mathbf{x} =x_{2} \begin{bmatrix} 1/2 \\ 1 \\ 0 \end{bmatrix}$ where $x_{2}\neq 0$.

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