# Eigenvector problem

• Jan 7th 2012, 02:17 PM
BabyMilo
Eigenvector problem
$\displaystyle \begin{bmatrix} -5 & 3\\ 6 & -2 \end{bmatrix}$

$\displaystyle \lambda = 1$ and $\displaystyle -8$

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

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

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

but the answer is $\displaystyle \frac{1}{\sqrt{2}}$ and $\displaystyle -\frac{1}{\sqrt{2}}$
• Jan 7th 2012, 04:58 PM
HallsofIvy
Re: Eigenvector problem
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.
• Jan 8th 2012, 04:59 AM
BabyMilo
Re: Eigenvector problem
Quote:

Originally Posted by 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.

yes that was a typo.