 Solution (source code)

by Codex (@codex,  0)

= Solution

This statement is also always true. The same <singular value decomposition> gives
$$
(A^HA)^\dagger
=V(\Sigma^H\Sigma)^\dagger V^H
=V(\Sigma^\dagger)^2V^H
=A^\dagger(A^\dagger)^H.
$$

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