$A \in \mathbb{C} ^{n\times n}$ be positive semidefinite. $ A= \begin{bmatrix} A_{11} & A_{12}\\ A^* _{12} &A_{22} \end{bmatrix}$ where $A_{11} \in \mathbb{C} ^{k\times k}$. Show that $N(A^* _{11}) \subset N(A^* _{12})$, $N$ denotes the nullspace.
Schur complement may provide a solution, but I am looking for a more basic solution.
My attempt: Consider $z= \begin{bmatrix} x\\y\end{bmatrix} \in \mathbb{C} ^n\, , x \in N(A^* _{11}). \begin{equation} \begin{split} z^* Az &= x^* A_{11}x + y^* A^* _{12} x + x^* A_{12} y + y^* A_{22} y\\ &= x^* A_{11}x + y^* A^* _{12} x + \overline{y^* A^* _{12} x} + y^* A_{22} y \\ &= x^* A_{11}x + 2Re( y^* A^* _{12} x) + y^* A_{22} y \\ &\geq 0 \end{split} \end{equation}$
$A_{22}$ is positive semidefinite. So $y^* A_{22} y \geq 0$. I don't know how to proceed. Suppose $A^* _{12} x\neq 0$. Can we choose $y$ such that $2Re( y^* A^* _{12} x) + y^* A_{22} y < 0$, hence a contradiction?
Thanks.