0
$\begingroup$

I have the following hermitian matrix before me: \begin{pmatrix}2 &1+i\\ 1-i& 3\end{pmatrix} I calculated its characteristic polynomial as $k^2-5k+4$ which has $1$ and $4$ as its roots. Eigenvectors corresponding to these two eigenvalues are \begin{pmatrix}-(1+i) &1 \end{pmatrix} and \begin{pmatrix}1 &1-i\end{pmatrix} But these are not orthogonal. Why is it so? Am I doing something wrong?

$\endgroup$
3
  • 1
    $\begingroup$ make sure you are considering in the complex inner product i.e. $<v,w>=v \bar{w}$ $\endgroup$
    – Brozovic
    Mar 3 at 14:23
  • $\begingroup$ @Moo The second eigenvector of yours can be obtained by multiplying mine one by $1+i$. Even then the vectors are not orthogonal. $\endgroup$ Mar 3 at 14:54
  • $\begingroup$ @Brozovic Could you please elaborate on complex inner product? I have no knowledge about it. $\endgroup$ Mar 3 at 15:32
1
$\begingroup$

Check Eigen values are $4,1$ and eigenvectors are $V_1=\begin{bmatrix} 1+i \\2 \end{bmatrix}$ and $V_2=\begin{bmatrix} -1-i \\ 1 \end{bmatrix}$, then $V_1^{\dagger} V_2=0$, $\dagger$ means conjugate and transpose.

$\endgroup$

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.