0
$\begingroup$
Find eigen vectors for this:

enter image description here

I found that eigenvalues are $0,2,2$

And the eigen vector for $0$ is {$1,0,1$}

But I'm not sure how to find the eigenvector for $2$ since the matrix reduces to one equation. I already know what the eigenvectors are (from wolfram) however and there are two eigen vectors for $\lambda = 2$.

Any help would be great, Thanks

$\endgroup$
0
$\begingroup$

$$\ker{\left(\begin{bmatrix}1&1&-1\\0&2&0\\-1&1&1\end{bmatrix}-2I\right)}=\ker{\begin{bmatrix}-1&1&-1\\0&0&0\\-1&1&-1\end{bmatrix}}=\text{span}\left(\begin{bmatrix}1\\1\\0\end{bmatrix},\begin{bmatrix}1\\0\\-1\end{bmatrix}\right)$$

$\endgroup$
  • $\begingroup$ Yes but it reduces to {{-1,1,-1},{0,0,0},{0,0,0}} how do you get eigen vectors from that? $\endgroup$ – A A Apr 7 '14 at 18:56
  • $\begingroup$ You just find the null space / kernel of that matrix. In this case, adding the first two columns or taking the first column minus the third will give you the zero vector, so those two vectors span the kernel. $\endgroup$ – user140943 Apr 7 '14 at 18:58
  • $\begingroup$ That made no sense to me dude $\endgroup$ – A A Apr 7 '14 at 19:02
  • $\begingroup$ Well what happens when you multiply the matrix by the two vectors I put in the span? You should get the zero vector... $\endgroup$ – user140943 Apr 7 '14 at 19:03
  • $\begingroup$ I'm at this step: {{-1,1,-1},{0,0,0},{0,0,0}} Can you explain where to go next? $\endgroup$ – A A Apr 7 '14 at 19:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.