when $\mathbf{P}$ and $\mathbf{Q}$ are orthogonal projectors on two complementary subspace, is $\mathbf{P}\mathbf{X}\mathbf{Q}= \mathbf{0}$? Suppose


*

*$\mathbf{P}$ is an orthogonal projector onto some space $\mathcal{M}\in\mathbb{R}^n$, and

*$\mathbf{Q}$ is an orthogonal projector onto its complementary subspace $\mathcal{M}^\perp$.


Is $\mathbf{P}\mathbf{X}\mathbf{Q}=\mathbf{Q}\mathbf{X}\mathbf{P}=\mathbf{0}$ for any $n\times n$ matrix $\mathbf{X}$? If it is, how can I prove that?
I know $\mathbf{P}\mathbf{Q} = \mathbf{Q}\mathbf{P} = \mathbf{0}$, but I'm wondering whether the above equation is also true. If it is, my computation will be very simple in my research.
 A: Nope it's not generally true. Consider, e.g.,
$$
P=\begin{bmatrix}1&0\\0&0\end{bmatrix},\quad
Q=\begin{bmatrix}0&0\\0&1\end{bmatrix},\quad
X=\begin{bmatrix}1&1\\1&1\end{bmatrix}.
$$
Then
$$
PXQ=\begin{bmatrix}0&1\\0&0\end{bmatrix}\neq 0.
$$
A: No: Consider any pair of nonzero linear maps $\alpha:\mathcal M\to\mathcal M^\perp$ and $\beta:\mathcal M^\perp \to\mathcal M$, and let 
$$X(v):=\alpha(Pv)+\beta(Qv)$$
for $v\in \Bbb R^n$. Then we will have
$\ QXP(v)=\alpha(Pv)\ $ and $\ PXQ(v)=\beta(Qv)$.
(By the way, the conditions mean exactly that $P^2=P=P^*$ and that $Q=I-P$ where $I$ is the identity.)
A: We construct a counterexample: Notice that a matrix is entirely defined by giving its action on a basis. By the hypothesis we have
$$\operatorname{Im }P=\mathcal M=\ker Q\quad;\quad \operatorname{Im }Q=\mathcal M^\perp=\ker P$$
so
$$\Bbb R^n=\operatorname{Im }P\oplus\operatorname{Im }Q$$
so let $(e_1,\ldots,e_p,e_{p+1},\ldots,e_n)$ a basis relative to the above decomposition and define $X$ such that
$$Xe_1=e_{p+1}\;\text{and} \;Pe_k=0\;\text{otherwise}$$
then we see that
$$QXPe_1=QXe_1=Qe_{p+1}=e_{p+1}$$
hence $QXP\ne0$.
