## Section5.3Finding Eigenvectors

Having found the eigenvalues of the example matrix $A=\begin{pmatrix}1\amp 2\\9\amp 4 \end{pmatrix}$ in the last section to be $7$ and $-2\text{,}$ we can now ask what the corresponding eigenvectors are. We must therefore solve the equation

\begin{equation} A \left|v\right> = \lambda \left|v\right>\tag{5.3.1} \end{equation}

in the two cases $\lambda=7$ and $\lambda=-2\text{.}$ In the first case, we have

\begin{equation} \begin{pmatrix}1\amp 2\\9\amp 4 \end{pmatrix} \begin{pmatrix}x\\ y \end{pmatrix} = 7\begin{pmatrix}x\\ y \end{pmatrix}\tag{5.3.2} \end{equation}

or equivalently

\begin{equation} \begin{pmatrix}x+2y\\ 9x+4y \end{pmatrix} = \begin{pmatrix}7x\\ 7y \end{pmatrix}\text{.}\tag{5.3.3} \end{equation}

Thus, we must solve this system of two equations. But we quickly discover that these equations are reduntant, since the first implies $2y=6x\text{,}$ while the second implies $3y=9x\text{,}$ which is the same condition. This is a good thing! (Why?)

We conclude that any vector $\begin{pmatrix}x\\ y \end{pmatrix}$ with $y=3x$ is an eigenvector of $A$ with eigenvalue $7\text{.}$ To check an explicit example, choose any value for $x\text{,}$ such as $x=1\text{,}$ yielding the vector

\begin{equation} v_7 = \begin{pmatrix}1\\ 3 \end{pmatrix}\tag{5.3.4} \end{equation}

and check by explicit computation that $Av_7=7v_7\text{,}$ as expected.

Turning to the case $\lambda=-2\text{,}$ a similar construction yields

\begin{equation} \begin{pmatrix}1\amp 2\\9\amp 4 \end{pmatrix} \begin{pmatrix}x\\ y \end{pmatrix} = -2\begin{pmatrix}x\\ y \end{pmatrix}\tag{5.3.5} \end{equation}

or equivalently

\begin{equation} \begin{pmatrix}x+2y\\ 9x+4y \end{pmatrix} = \begin{pmatrix}-2x\\ -2y \end{pmatrix}\text{,}\tag{5.3.6} \end{equation}

and the first equation now yields $2y=-3x\text{,}$ while the second yields $6y=-9x\text{.}$ Again, these two equations are redundant; the eigenvectors with eigenvalue $-2$ satisfy $y=-\frac32\text{.}$ An explicit example is

\begin{equation} v_{-2} = \begin{pmatrix}2\\ -3 \end{pmatrix}\tag{5.3.7} \end{equation}

and you should again check by explicit computation that $Av_{-2}=-2v_{-2}\text{.}$