1. ## Eigenvector problem

$\begin{bmatrix}
-5 & 3\\
6 & -2
\end{bmatrix}$

$\lambda = 1$ and $-8$

I could find the eigenvector when $\lambda = 1$

but how do i find the eigenvector when $\lambda = 8$?

I came up with $\begin{bmatrix}
1 & -3/13 \\
0 & 56/39
\end{bmatrix}$

but the answer is $\frac{1}{\sqrt{2}}$ and $-\frac{1}{\sqrt{2}}$

2. ## Re: Eigenvector problem

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

By definition, if $\begin{bmatrix}x \\ y\end{bmatrix}$ is an eigenvector corresponding to eigenvalue -8, then we must have
$\begin{bmatrix}-5 & 3 \\ 6 & -2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}-5x+ 3y \\ 6x- 2y\end{bmatrix}= \begin{bmatrix}-8x \\ -8y\end{bmatrix}$

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 $\begin{bmatrix}x \\ -x \end{bmatrix}$, which, of course, includes $\begin{bmatrix}\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}}\end{bmatrix}$, is an eigenvector corresponding to eigenvalue -8.

3. ## Re: Eigenvector problem

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

By definition, if $\begin{bmatrix}x \\ y\end{bmatrix}$ is an eigenvector corresponding to eigenvalue -8, then we must have
$\begin{bmatrix}-5 & 3 \\ 6 & -2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}-5x+ 3y \\ 6x- 2y\end{bmatrix}= \begin{bmatrix}-8x \\ -8y\end{bmatrix}$

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 $\begin{bmatrix}x \\ -x \end{bmatrix}$, which, of course, includes $\begin{bmatrix}\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}}\end{bmatrix}$, is an eigenvector corresponding to eigenvalue -8.
yes that was a typo.