How prove this matrix $rank (A_{11}A_{12})=rank(A_{12})$ let $A_{n\times n}$ is a positive semidefinite matrix,
and
$$A=\begin{bmatrix}
A_{11}&A_{12}\\
A_{21}&A_{22}
\end{bmatrix}$$
where $A_{11}$ is $m\times m,m<n $ matrix
prove or disprove 
$$rank(A_{11}A_{12})=rank(A_{12})$$
 A: Hint: the proof is comprised of two steps.


*

*Show that $\operatorname{Im}(A_{12})\subseteq \operatorname{Im}(A_{11})$. Suppose the contrary. Then there is a nonzero $v$ such that $A_{12}v \notin \operatorname{Im}(A_{11})$. Since $A_{11}$ is positive semidefinite, its kernel (i.e. nullspace) is precisely the orthogonal complement of its image. It follows that the orthogonal projection of $A_{12}v$ on $\ker(A_{11})$ is nonzero. Let $x$ be this projection and let $k\in\mathbb{R}$. Then
$$(kx^T,v^T)\,A\pmatrix{kx\\ v}=2kx^TA_{12}v + v^TA_{22}v.\tag{1}$$
Argue that when $|k|$ is sufficiently large and the sign of $k$ is chosen appropriately, $(1)$ will contradict the assumption that $A$ is positive semidefinite. Therefore $\operatorname{Im}(A_{12})\subseteq \operatorname{Im}(A_{11})$.

*Show that $A_{11}$ is invertible on $\operatorname{Im}(A_{11})$ (hint: $A_{11}$ is positive semidefinite). Hence the assertion follows from the result of step 1.


Alternatively, note that positive semidefinite matrices can be orthogonally diagonalised. So, WLOG you may assume that $A_{11}=D\oplus0$, where $D$ is an $r\times r$ positive diagonal matrix ($r=\operatorname{rank}(A_{11})$) and the zero block is $(m-r)\times(m-r)$. In this case, it is not hard to show that the last $m-r$ rows of $A_{12}$ must be zero and hence the assertion follows. This actually is the same proof as the above, but presented in a less linear algebraic but more matrix theoretic fashion. 
A: Hint: If $A$ is Hermitian, then $A_{11}$ must have full rank by Sylvester's criterion.
