Equal nullspace requirement when defining an ordering I am just reading about ordering of matrices, i.e. in particular the definitions $A \preceq B$ iff $A-B$ is positive semi-definite.
Now, for any pd matrices $A,B$ we have $A \preceq (1+\epsilon)B$ and $B^{+/2}AB^{+/2} \preceq (1+\epsilon) I$. Similarly for singular matrices $C,D$ the relation $C \preceq (1+\epsilon) D$ implies that $D^{+/2}CD^{+/2} \preceq (1+\epsilon)I$ if $C,D$ have the same nullspace.
My question is just about the requirement of having the same nullspace in this case and I want to understand whether my understanding here is correct. The nullspace requirement is just important since for any vector $x$ in the nullspace of $D$, $D^{+/2}$ would map this to $1/0$ which is not valid. However if this $x$ is also in the nullspace of $C$, then $D^{+/2}CD^{+/2}$ would be a valid operator on $x$. Is that correct?
 A: I assume that $D^{+/2}$ is the pseudoinverse of the PSD square root $D^{1/2}$ of $D$. Then if $x$ is in the nullspace of $D$ you have $D^{+/2} x = 0$, so your reasoning is not correct. Moreover $C \preceq (1+\epsilon)D$ implies $D^{+/2} C D^{+/2} \preceq (1+\epsilon) I$ even if the nullspaces of the two matrices do not coincide. Indeed, for any matrix $W$ and PSD matrices $X$ and $Y$, $X \preceq Y$ implies $B^T X B \preceq B^T Y B$. In your case, you just need to use this fact and $D^{+/2} D D^{+/2} \preceq I$.
What you may have been thinking of is the reverse implication: if $D^{+/2} C D^{+/2} \preceq (1+\epsilon) I$ and the nullspaces of $C$ and $D$ coincide, then $C \preceq (1+\epsilon)D$. Here the assumption is necessary, because if there is a vector $x$ in the nullspace of $D$ but not in the nullspace of $C$, then $C \preceq (1+\epsilon)D$ could never hold, but it's certainly possible that $D^{+/2} C D^{+/2} \preceq (1+\epsilon) I$ does.
Let's also prove the reverse direction. Let $E$ be the common nullspace of $C$ and $D$, and let $F$ be its orthogonal complement. We need to show that for all $x$, $$
x^T C x \le (1+\epsilon) x^T D x.
$$ 
Write any $x$ as $x = y+z$ where $y \in E$ and $z \in F$. Then, $x^T C x = y^T C y$ and $x^T D x = y^T Dy$, so we only need to show the inequality for all $y \in F$. Write $u = D^{1/2} y$: by assumption we have 
$$
y^TD^{1/2} D^{+/2} C D^{+/2}D^{1/2}y \le (1+\epsilon) y^T D y.
$$
Notice that $D^{+/2}D^{1/2} = \Pi_F$, the orthogonal projection matrix onto $F$. $\Pi_F$ acts as the identity on $F$, i.e. $D^{+/2}D^{1/2}y = \Pi_F y = y$. This gives $y^T C y \le (1+\epsilon) y^T D y$, as we want. 
