1. ## Eigenvectors

I know how to get the eigenvalues, but hiow do i get the eigenvectors?

$\displaystyle \left(\begin{array}{cc}3&1\\1&3\end{array}\right)$

(3-$\displaystyle \lambda$)(3-$\displaystyle \lambda$) - 1

lambda = 2 and 4

how do i do the eigenvectors of this?

2. Originally Posted by adam_leeds
I know how to get the eigenvalues, but hiow do i get the eigenvectors?

$\displaystyle \left(\begin{array}{cc}3&1\\1&3\end{array}\right)$

(3-$\displaystyle \lambda$)(3-$\displaystyle \lambda$) - 1

lambda = 2 and 4

how do i do the eigenvectors of this?
For $\displaystyle \lambda=2$, the matrix $\displaystyle \begin{bmatrix}3-\lambda&1\\1&3-\lambda\end{bmatrix}$ becomes $\displaystyle \begin{bmatrix}1&1\\1&1\end{bmatrix}$. You then have to solve the system of equations $\displaystyle \begin{bmatrix}1&1\\1&1\end{bmatrix}\begin{bmatrix }x\\y\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}$, which reduces to $\displaystyle x+y=0$. This has the solution x=1, y=–1. So $\displaystyle \begin{bmatrix}1\\-1\end{bmatrix}$ is an eigenvector for the eigenvalue 2.

Now do the same thing for $\displaystyle \lambda=4$.

3. Or, just apply the definition of "eigenvector".

If 2 is an eigenvalue, then there exist a non-zero eigenvector, $\displaystyle \begin{pmatrix}x \\ y\end{pmatrix}$ such that

$\displaystyle \begin{pmatrix}3 & 1 \\ 1 & 3\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}= \begin{pmatrix}3x+ y\\ x+ 3y\end{pmatrix}= 2\begin{pmatrix}x \\ y\end{pmatrix}$

That gives the two equations 3x+ y= 2x and x+ 3y= 2y. Both of those reduce to the equation y= -x. That is
$\displaystyle \begin{pmatrix}x \\ y\end{pmatrix}= \begin{pmatrix}x \\ -x\end{pmatrix}= x\begin{pmatrix}1 \\ -1\end{pmatrix}$.

An eigenvector corresponding to eigenvalue 2 is $\displaystyle \begin{pmatrix}1 \\ -1\end{pmatrix}.$

With eigenvalue 4 instead of 2, those equations become 3x+ y= 4x and x+ 3y= 4y, both of which reduce to y= x. An eigenvector corresponding to eigenvalue 4 is $\displaystyle \begin{pmatrix}1 \\ 1\end{pmatrix}$.