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On Susskind book, exercise 3.1 say to prove the following:

"If a vector space is $N$-dimensional, an orthonormal basis of $N$ vectors can be constructed from the eigenvectors of a Hermitian operator."

Susskind wrote that the proof is easy.

From the book I understand that with two unequal eigenvalues of a Hermitian operator, then the corresponding eigenvectors are orthogonal. Even if the two eigenvalues are equal, the corresponding eigenvectors can be chosen to be orthogonal.

But I do not undesrtand how to prove if the space is N-dimensional, there will be N orthonormal eigenvectors.

Please Help.

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    https://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process If you have an N dimensional vector space, you can always find N linearly independent vectors. Through the Gram Schmidt process, iteratively subtract off the portions of the vectors which overlap with the other ones. So if N=2, you can always find two linearly independent vectors $v_1$ and $v_2$. Then subtract off the portion of $v_2$ which is parallel to $v_1$, i.e. define $v_2' = v_2 - \frac{\langle v_1, v_2 \rangle}{\langle v_1, v_1 \rangle} v_1$. Note that $\langle v_2', v_1 \rangle = 0$. – user1379857 Oct 13 '19 at 22:38
  • The previous comment does not answer the question - why can it be done with eigenvectors of a hermitian operator. You need to look up "spectral theorem" – lux Oct 14 '19 at 04:38
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    And here is the problem. The proof should be easy according with book. Spectral theorem with inverse matrix and diagonalization will be an hard proof, at least very hard for me. There are no other ways? –  Oct 14 '19 at 06:10
  • "Proof of c)" on page 3. – Keith McClary Oct 15 '19 at 21:05

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