Having found the eigenvalues of the example matrix $A=\begin{pmatrix}1&2\\9&4\\\end{pmatrix}$ in the last section to be $7$ and $-2$, 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> \end{equation} in the two cases $\lambda=7$ and $\lambda=-2$. In the first case, we have \begin{equation} \begin{pmatrix}1&2\\9&4\\\end{pmatrix} \begin{pmatrix}x\\ y\\\end{pmatrix} = 7\begin{pmatrix}x\\ y\\\end{pmatrix} \end{equation} or equivalently \begin{equation} \begin{pmatrix}x+2y\\ 9x+4y\\\end{pmatrix} = \begin{pmatrix}7x\\ 7y\\\end{pmatrix} . \end{equation} Thus, we must solve this system of two equations. But we quickly discover that these equations are reduntant, since the first implies $2y=6x$, while the second implies $3y=9x$, 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 $7$. To check an explicit example, choose any value for $x$, such as $x=1$, yielding the vector \begin{equation} v_7 = \begin{pmatrix}1\\ 3\\\end{pmatrix} \end{equation} and check by explicit computation that $Av_7=7v_7$, as expected.

Turning to the case $\lambda=-2$, a similar construction yields \begin{equation} \begin{pmatrix}1&2\\9&4\\\end{pmatrix} \begin{pmatrix}x\\ y\\\end{pmatrix} = -2\begin{pmatrix}x\\ y\\\end{pmatrix} \end{equation} or equivalently \begin{equation} \begin{pmatrix}x+2y\\ 9x+4y\\\end{pmatrix} = \begin{pmatrix}-2x\\ -2y\\\end{pmatrix} , \end{equation} and the first equation now yields $2y=-3x$, while the second yields $6y=-9x$. Again, these two equations are redundant; the eigenvectors with eigenvalue $-2$ satisfy $y=-\frac32$. An explicit example is \begin{equation} v_{-2} = \begin{pmatrix}2\\ -3\\\end{pmatrix} \end{equation} and you should again check by explicit computation that $Av_{-2}=-2v_{-2}$.