Does symmetry of $AB$ implies symmetry of $A^\dagger B$? Let $A$, $B$ and $AB$ symmetric. Is $A^\dagger B$ also symmetric i.e. $$A^\dagger B = B A^\dagger$$,
where $A^\dagger$ is the pseudo-inverse of $A$ ?
 A: In short, yes!
We can show it using the spectral decomposition. As $A,\, B, \, AB$ are all symmetric this implies that $A$ and $B$ commute. If $A$ and $B$ commute they can be simultaneously diagonalized, i.e., they have spectral decompositions
$$
A = U D_A U^T \qquad U D_B U^T
$$
where $D_A$ and $D_B$ are real diagonal matrices containing the eigenvalues of $A$ and $B$ respectively and $U$ is a symmetric matrix. Now for symmetric matrices the pseudoinverse takes a special form,
$$
A^+ = U D_A^+ U^T
$$
where $D_A^+$ is the diagonal matrix where we have inverted all of the non-zero diagonal elements and kept the $0$ diagonal elements $0$. But now
$$
A^+ B = U D_A^+ U^T U D_B U^T = U D_A^+ D_B U^T = U D_B D_A^+ U^T = UD_B U^T U D_A^+ U = B A^+ \,.
$$
A: Yes, if $A,B$ and $AB$ are real symmetric. Since $AB$ is symmetric, $A$ commutes with $B$. Hence every polynomial in $A$ commutes with $B$. However, $A^+$ is precisely a polynomial in $A$ when $A$ is real symmetric (think about the orthogonal diagonalisation of $A$ if you don't see this). Therefore $A^+$ commutes with $B$.
The statement in question is false if $A$ or $B$ are complex symmetric. For a counterexample, consider $A=B=vv^\top$ where $v$ is any vector with both real and non-real non-zero elements. Then $A,B$ and $AB=A^2$ are complex symmetric, but
$$
AB^+
=(vv^\top)(vv^\top)^+
=\frac{1}{\|v\|^2}(vv^\top)(vv^\top)^\ast
=\frac{v^\top\overline{v}}{\|v\|^2}vv^\ast
=\frac{1}{\|v\|}vv^\ast
$$
is not, as it is Hermitian but non-real.
