Is it true that $A\geq B$ implies $B^{\dagger}\geq A^{\dagger}$ for singular positive semi-definite matrices? Here $A^{\dagger}$ is the Moore-Penrose pseudo-inverse and $\geq$ denotes the Loewner partial order for positive semidefinite matrices.
I know that the statement is true for non-singular matrices, but cannot find whether this extends to singular matrices, and continuity arguments seem to not apply here. 
Any help will be greatly appreciated.
 A: It isn't true. The example could be the following:
$$
A=\begin{pmatrix}1&0\\0&1\end{pmatrix},\quad B=\begin{pmatrix}1&0\\0&0\end{pmatrix}.
$$
Their pseudoinverse matrices are
$$
A^+=A,\quad B^+=B.
$$
Which implies
$$
A\ge B,\quad B^+\not\ge A^+.
$$
A: It is true if and only if the matrices share the same nullspace.

Assume $null(A)=null(B)$. Let $P\in O(n)$ be the matrix of eigenvectors
of $A$ so that the first eigenvectors correspond to the nullspace. Then
$$A = P \begin{pmatrix} 0 & 0 \\ 0 & D\end{pmatrix} P^T,
\qquad
B = P \begin{pmatrix} 0 & 0 \\ 0 & \bar B\end{pmatrix} P^T
$$
where $D$ is diagonal with nonzero elements and $\bar B$ is invertible, and they have the same size. Then
$$A^\dagger = P \begin{pmatrix} 0 & 0 \\ 0 & D^{-1}\end{pmatrix} P^T,
\qquad
B^\dagger = P \begin{pmatrix} 0 & 0 \\ 0 & \bar B^{-1}\end{pmatrix} P^T.
$$
Then, since $D$ and $\bar B$ are invertible, using the arguments of https://math.stackexchange.com/a/3018669/484640,
 \begin{align}
A\ge B & \text{ iff } D \ge \bar B
\\ & \text{ iff } I\ge D^{1/2} \bar B D^{1/2}
\\ & \text{ iff } I\le D^{-1/2} \bar B^{-1} D^{-1/2}
\\ & \text{ iff } D^{-1} \le \bar B^{-1}
\\ & \text{ iff } A^\dagger \le B^\dagger.
\end{align}

Now assume $A \ge B$ but that the nullspaces are different, say there exists a vector $x$ with $Ax \ne 0$ but $Bx = 0$.
Then $x^T B^\dagger x =0$ by definition of the pseudo inverse  of $B$, but $x^T A^\dagger x \ne 0$ by definition of the pseudo inverse of $A$. This implies $x^T A^\dagger x > x^T B^\dagger x = 0$ so $A^\dagger \le B^\dagger$ cannot hold.
