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Given a general inner product on $\mathbb{C}^n$ induced by the hermitian matrix $S$, i.e. $\langle v,w\rangle_S=v^\dagger S w$, a matrix $A$ is self-adjoined with respect to $\langle\cdot,\cdot\rangle_S$ if $\langle A v, w\rangle_S=\langle v, A w\rangle_S$ for all $v,w$. Because $A$ is self-adjoined, is has eigenvectors that are orthonormal also with respect to $\langle\cdot,\cdot\rangle_S$: $$AC=C\Lambda$$ with $$C^\dagger S C=\mathbb{1}$$ with the matrix $C$ containing the eigenvectors as columns, $\Lambda$ the diagonal matrix containing the eigenvalues and $\mathbb{1}$ the identity matrix.

The question is what can be said about the rows of $C$? I found numerically that $$SCC^\dagger=\mathbb{1}$$ and $$CC^\dagger S=\mathbb{1},$$ which also complies with the canonical case $S=\mathbb{1}$, but I do not see the reason why?

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  • $\begingroup$ When $X,Y$ are square matrices, $XY=I$ if and only if $YX=I$. $\endgroup$
    – user1551
    Commented Jan 9, 2015 at 18:42

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From $C^\dagger S C = 1$, multiply both sides from the left by $C$: $$ C C^\dagger S C = C $$ $$ (C C^\dagger S) C = (1) C $$ A similar argument with $C^\dagger$ from the right shows the other relation.

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