Prof. Gilbert Strang mentions, in this lecture, that a matrix $A$ has orthogonal eigenvectors when $A^T A=AA^T$; i.e., when it is normal. Could someone provide me a (preferably intuitive) proof of this?
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1This subsection of the Wikipedia page on the Spectral Theorem has a proof sketch. – angryavian Nov 22 '20 at 03:48
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Actually that should be $A^* A = A A^$, where $A^$ is the Hermitian transpose (the conjugate transpose). – Robert Israel Nov 22 '20 at 03:52
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This result holds for every symmetric matrix S. The basic idea is that column space and row space of a symmetric matrix are equal and that row space and null space are orthogonal.
Given a pair of eigenvectors $x, y$ corresponding to non-zero eigenvalues $\alpha, \beta$, we have $Sx = \alpha x \Rightarrow (S-\alpha I)x = 0$ and $(S - \alpha I)y = (\beta - \alpha)y$. Therefore, $x$ is in the null space of $S-\alpha I$ and $y$ is in the column space of $S-\alpha I$. Since column space is equal to row space ($S-\alpha I$ is also symmetric), we have $x^Ty = 0$. The case of zero eigen values is trivial.
balaji
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