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**HallsofIvy** As you say in your first line, the eigenvalues are 1 and -8 so there is no eigenvector for $\displaystyle \lambda= 8$. Was that a typo?

By definition, if $\displaystyle \begin{bmatrix}x \\ y\end{bmatrix}$ is an eigenvector corresponding to eigenvalue -8, then we must have

$\displaystyle \begin{bmatrix}-5 & 3 \\ 6 & -2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}-5x+ 3y \\ 6x- 2y\end{bmatrix}= \begin{bmatrix}-8x \\ -8y\end{bmatrix}$

That is equivalent to the two equations -5x+ 3y= -8x and 6x- 2y= -8y which then give 3y= -3x and 6x= -6y both of which are equivalent to y= -x. Any vector of the form $\displaystyle \begin{bmatrix}x \\ -x \end{bmatrix}$, which, of course, includes $\displaystyle \begin{bmatrix}\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}}\end{bmatrix}$, is an eigenvector corresponding to eigenvalue -8.