Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 2 (a)/Solution 1

Since an eigenvector $\mathbf {x}$ of $A$ of eigenvalue $\lambda _{1}$ satisfies ${\textstyle A\mathbf {x} =\lambda _{1}\mathbf {x} }$, i.e., ${\textstyle (A-\lambda _{1}I)\mathbf {x} =\mathbf {0} }$, all we need to do is solve the following system.

$$(A-\lambda _{1}I)\mathbf {x} =\left({\begin{bmatrix}4&-1&7\\0&3&0\\1&2&-2\end{bmatrix}}-3{\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}}\right){\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}={\begin{bmatrix}1&-1&7\\0&0&0\\1&2&-5\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}=\mathbf {0} .$$

By applying Gaussian elimination,

$${\begin{bmatrix}1&-1&7\\0&0&0\\1&2&-5\end{bmatrix}}\xrightarrow {R_{2}\leftrightarrow R_{3},R_{2}-R_{1}} {\begin{bmatrix}1&-1&7\\0&3&-12\\0&0&0\end{bmatrix}}\xrightarrow {R_{2}/3} {\begin{bmatrix}1&-1&7\\0&1&-4\\0&0&0\end{bmatrix}}\xrightarrow {R_{1}-R_{2}} {\begin{bmatrix}1&0&3\\0&1&-4\\0&0&0\end{bmatrix}},$$

we obtain

$${\begin{bmatrix}1&0&3\\0&1&-4\\0&0&0\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}=\mathbf {0} .$$

Thus ${\textstyle x_{1}+3x_{3}=0}$ and ${\textstyle x_{2}-4x_{3}=0}$; in vector form we get

$${\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}={\begin{bmatrix}-3x_{3}\\4x_{3}\\x_{3}\end{bmatrix}}=x_{3}{\begin{bmatrix}-3\\4\\1\end{bmatrix}},$$

where $x_{3}\in \mathbb {R}$ is free. So an eigenvector corresponding to ${\textstyle \lambda _{1}=3}$ is

$$\color {blue}{\begin{bmatrix}-3\\4\\1\end{bmatrix}}.$$