Is matrix hermitian and its eigenvectors orthogonal?

1. Jun 22, 2015

bugatti79

I calculate

1) $\Omega= \begin{bmatrix} 1 & 3 &1 \\ 0 & 2 &0 \\ 0& 1 & 4 \end{bmatrix}$ as not Hermtian since $\Omega\ne\Omega^{\dagger}$ where$\Omega^{\dagger}=(\Omega^T)^*$

2) $\Omega\Omega^{T}\ne I$ implies eigenvectors are not orthogonal.

Is this correct?

2. Jun 22, 2015

Strilanc

The matrix $M = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ does not satisfy $M M^\dagger = I$, but its eigenvectors are orthogonal.

3. Jun 23, 2015

bugatti79

Ok then my 2)assumption is not correct thus I proceeded to calculate the eigenvectors for the following eigenvalues $\lambda=1,2,4$ and then check for orthogonality

$\lambda=1$

$\begin{bmatrix}1& 3 & 1\\ 0& 2&0 \\ 0& 1 &4 \end{bmatrix}\begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix}=\lambda \begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix}$

In component form I get

$(1-1)x_1+3x_2+x_3=0$
$0+(2-1)x_2+0=0$
$0+x_2+(4-1)x_3=0$
I am confused here for 2 reasons

1) How to deal with $(1-1)x_1$
2) the 2nd line suggest $x_2=0$ but on the 3rd line $x_2=-3x_3$

Not sure how to proceed from here...
Any help will be appreciated.

4. Jun 23, 2015

Strilanc

I'm not sure why $x_1$, $x_2$, and $x_3$ are coming into the process. To compute eigenvalues you solve for $\lambda$ by expanding $det(M - \lambda I) = 0$ then finding roots. You seem to be skipping ahead to the part where you know a specific lambda and want to find the associated vector.

The khan academy video on eigenvectors might help.

5. Jun 24, 2015

bugatti79

Thanks for the video, very useful

I have expanded $det(\Omega - \lambda I) = 0$ to get

$\begin{bmatrix}1-\lambda& 3 & 1\\ 0& 2-\lambda&0 \\ 0& 1 &4-\lambda \end{bmatrix}$

For $\lambda=4$ say we get the following

$\begin{bmatrix}1-4& 3 & 1\\ 0& 2-4&0 \\ 0& 1 &4-4 \end{bmatrix}\begin{bmatrix}v_1\\ v_2\\ v_3\end{bmatrix}=0$

$-3v_1+3v_2+v_3=0$
$-2v_2=0$
$v_2=0$
If we let $v_1=1$ then $v_3=3$. Our eigenvector for $\lambda=4$ is (1,0,3)

For $\lambda=1$ I have difficulty because

$\begin{bmatrix}1-1& 3 & 1\\ 0& 2-1&0 \\ 0& 1 &4-1 \end{bmatrix}\begin{bmatrix}v_1\\ v_2\\ v_3\end{bmatrix}=0$

$3v_2+v_3=0$
$v_2=0$
$v_2+3v_3=0$ but I cannot determine what $v_1$ is....?

6. Jun 24, 2015

JorisL

You can check that for $\lambda = 1$ every vector with $v_2=v_3=0$ and $v_1\in \mathbb{C}$ is an eigenvector.
Do that.

I would advise brushing up on solving systems of linear equations. Because that's where your main issue seems to be located.
If you have a solution, always check that the vectors you found actually give the correct eigenvalues.

7. Jun 24, 2015

PeroK

$v_1$ can be anything. You can choose it to be 1, for example. Remember that any multiple of an eigenvector is also an eigenvector with the same eigenvalue.

8. Jun 24, 2015

bugatti79

OK guys, thank you.
I can see how that holds for $\lambda=1$.

Onto my next problem :-)