2
$\begingroup$

Suppose I have a hermitian operator $\Omega$. The proof of the existence of a orthonormal eigenbasis as given in Shankar is given.

enter image description here

What I don't understand is why the second eigenvector $\left| \omega_2 \right> $ of matrix $\Omega$ belongs to the subspace $V^{n-1}_{\perp 1}$ orthogonal to the first eigenvector $\left| \omega_1 \right> $?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
1
$\begingroup$

Hints: One should use the facts that:

  1. Eigenvectors for different eigenvalues of a Hermitian operator are orthogonal, cf. e.g. this webpage.

  2. If the eigenvalues are the same one can choose appropriate linear combinations of the eigenvectors to be orthogonal.

Improve this answer
$\endgroup$
1
  • $\begingroup$ I had my doubt regarding this proof from the book. Of course, if eigenvalues are different and eigenvectors are orthogonal as a consequence, the proof becomes easy. But my doubt is the one mentioned in the question above. $\endgroup$
    – Anu3082
    Commented Dec 26, 2021 at 12:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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