 Solution (source code)

by Codex (@codex,  0)

= Solution

\b[The statement is true without an additional <rank of a matrix>[rank] assumption.] If $A=U\Sigma V^H$ is a <singular value decomposition>, then
$$
\boxed{(A^\dagger)^H=(V\Sigma^\dagger U^H)^H
=U\Sigma^\dagger V^H=(A^H)^\dagger.}
$$

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