$A^{-}A$ and $AA^{-}$ are symmetric when $A^{-}$ is reflexive

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I'm currently trying to solve the following statement regarding generalized inverse matrix $A^{-}$;

If $A^{-}$ is reflexive, then $(A^{-}A)' = A^{-}A$ and $(AA^{-})' = AA^{-}$

To start with, $A^{-}$ is reflexive, so $A$ is a generalize inverse of $A^{-}$. That is, $$A^{-} =A^{-}AA^{-}.$$

Then, I have to show $(A^{-}A)' =A'(A^{-})' = A^{-}A$, which seems unlcear. Simply substituting $A^{-}$ with $A^{-}AA^{-}$ does not give any useful information.

Any hints or other different way with respect to this problem would be grateful. Thank you.

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As it is formulated, the statement is false.

Consider for instance: $$ A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix},\hspace{3mm} G = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}. $$ We then have $G = GAG$ (and, analogously, $A = AGA$). However: $$ GA = \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix} \neq (GA)'. $$