## 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

$$A \left|v\right> = \lambda \left|v\right>\tag{5.3.1}$$

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

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

or equivalently

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

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

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

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

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

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

or equivalently

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

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

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

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