Is matrix hermitian and its eigenvectors orthogonal?

[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?

 

[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.

 

[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.

 

[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.

 

[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...?

 

[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.

 

[PeroK]

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...?

$v_1$ 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.

 

[bugatti79]

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

Onto my next problem :-)