# Thread: Help with final step for eigenvector problem

1. ## Help with final step for eigenvector problem

B = $\left(\begin{array}{ccc}4&0&-1\\0&4&1\\-1&1&3\end{array}\right)$

Find all the other eigenvalues and eigenvectors.

I have found the eigenvalues for this matrix being $\lambda = 2$, $\lambda = 4$ & $\lambda = 5$

I have subtracted your eigenvalues from the diagonal to get the following matricies

for $\lambda = 2$ $\left(\begin{array}{ccc}2&0&-1\\0&2&1\\-1&-1&1\end{array}\right)$
for $\lambda = 5$ $\left(\begin{array}{ccc}-1&0&-1\\0&-1&1\\-1&-1&-2\end{array}\right)$

My problem is that I am stuck on where to go from here. I have used the elimination method but I always end up with a row of zeros and am unsure on where to go from there.

Any help would be appreciated.

2. You should get a row of zeroes. For every eigenvalue there are infinitely-many associated eigenvectors.

For example, the eigenvectors associated with $\lambda = 2$ are vectors of the form $\alpha \left(\begin{array}{c} 1 \over 2 \\ -\frac{1}{2} \\ 1\end{array}\right)$

3. Ok. Im still confused as to how you got that answer. Are you able to show me the working out that you did?

4. The reduced row echelon form of $\left(\begin{array}{ccc}2&0&-1\\0&2&1\\-1&1&1\end{array}\right)$ is $\left(\begin{array}{ccc}1&0&-\frac{1}{2}\\0&1&1 \over 2\\0&0&0\end{array}\right)$

Since $x_{3}$ can be anything, just call it $\alpha$

Then $x_{2} + \frac {1}{2} x_{3} = 0$ which implies that $x_{2} = -\frac{\alpha}{2}$

And $x_{1} - \frac {1}{2} x_{3} =0$ which implies that $x_{1} = \frac {\alpha}{2}$

so solutions are of the form $\left(\begin{array}{c} \alpha \over 2 \\ -\frac{\alpha}{2} \\ \alpha\end{array}\right)$

and then you can factor out the $\alpha$