Can't understand this pseudo-inverse relation. In the Answer to a different Question, a curious matrix relation came up:
M is symmetric and non-singular, G is non-symmetric and singular.
Theorem:
When $M$ is positive/negative definite, or more generally, if $G$ and $(G^\dagger G) M (G^\dagger G)$ have the same ranks then
$$G^{T}\left(GMG^{T}\right)^{\dagger}G=\left((G^\dagger G)M(G^\dagger G)\right)^\dagger$$ 
I have no idea how to prove this or why the rank relationships asserted would be true.
An example:
$M=\left(
\begin{array}{ccc}
 1. & 2. & 3. \\
 2. & 4. & 5. \\
 3. & 5. & 6. \\
\end{array}
\right)$   $G=\left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 2 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right)$   $GMG^T=\left(
\begin{array}{ccc}
 1. & 4. & 0. \\
 4. & 16. & 0. \\
 0. & 0. & 0. \\
\end{array}
\right)$
$GMG^T$ has a rank of 1 and the relation does not hold. But if we add a small number times the Identity Matrix to M, $GMG^T$ will have rank 2 and the relation will hold.
What is going on there? Can anyone help with an explanation or proof?  And under exactly what circumstances is this true or not true?

Primary matrices
The matrix 
$$
  \mathbf{M} = 
\left[
\begin{array}{ccc}
 1 & 2 & 3 \\
 2 & 4 & 5 \\
 3 & 5 & 6 \\
\end{array}
\right]
$$
has rank $\rho_{M} = 3$. The matrix 
$$
  \mathbf{G} = 
\left[
\begin{array}{ccc}
 1 & 0 & 0 \\
 2 & 0 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right]
$$
has rank $\rho_{G} = 2$. 
To make manipulation easier, introduce the submatrices
$$
m =
\left[
\begin{array}{ccc}
 1 & 2 \\
 2 & 4 
\end{array}
\right], \quad
  g = 
\left[
\begin{array}{ccc}
 1 & 0 \\
 2 & 0 
\end{array}
\right].
$$
We see a stencil matrix
$$
%
  \mathbf{G} =
%
\left[
\begin{array}{ccc}
 g & \mathbf{0} \\
 \mathbf{0} & 0 
\end{array}
\right], \quad
%
  \mathbf{G}^{\dagger} =
%
\left[
\begin{array}{ccc}
 g^{-1} & \mathbf{0} \\
 \mathbf{0} & 0 
\end{array}
\right], \quad
%
%
$$
Product matrices
The product matrix 
$$
  \mathbf{G} \mathbf{M} \mathbf{G}^{\mathrm{T}} = 
%
\left[
\begin{array}{cc}
 gmg^{\mathrm{T}} & \mathbf{0} \\
 \mathbf{0} & 0 \\
\end{array}
\right]
%
  =
%
\left[
\begin{array}{rrr}
 1 & 2 & 0 \\
 2 & 4 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right]
$$
has rank $\rho_{GMG^{T}} = 1$.
The pseudoinverse is 
$$
  \left( \mathbf{G} \mathbf{M} \mathbf{G}^{\mathrm{T}} \right)^{\dagger} = 
%
\frac{1}{25}
\left[
\begin{array}{rrr}
 1 & 2 & 0 \\
 2 & 4 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right]
%
  =
%
\frac{1}{25}
\mathbf{G} \mathbf{M} \mathbf{G}^{\mathrm{T}}.
$$
The final product matrix is
$$
\mathbf{G}^{\dagger}\mathbf{G} \mathbf{M} \mathbf{G}^{\dagger}\mathbf{G} =
\left[
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 0 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right].
$$
LHS
The left hand side of the identity is
$$
  \mathbf{G}^{\mathrm{T}} \left( \mathbf{G} \mathbf{M} \mathbf{G}^{\mathrm{T}} \right)^{\dagger} \mathbf{G} =
%
\left[
\begin{array}{rrr}
 1 & 0 & 0 \\
 0 & 0 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right]
$$
RHS
The right hand side of the identity is
$$
  \left( \mathbf{G}^{\dagger}\mathbf{G} \mathbf{M} \mathbf{G}^{\dagger}\mathbf{G} \right)^{\dagger} =
\left[
\begin{array}{rrr}
 1 & 0 & 0 \\
 0 & 0 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right]
$$
The equality holds in this case:
$$
 \mathbf{G}^{\mathrm{T}} \left( \mathbf{G} \mathbf{M} \mathbf{G}^{\mathrm{T}} \right)^{\dagger} \mathbf{G}
=
\left( \mathbf{G}^{\dagger}\mathbf{G} \mathbf{M} \mathbf{G}^{\dagger}\mathbf{G} \right)^{\dagger}
$$
Postscript Asymmetric $\mathbf{G}$ used after comments by @Dohleman.
