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

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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> $?


1 Answer 1


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.

  • $\begingroup$ 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
    Dec 26, 2021 at 12:12

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