If $A$ is positive semi-definite matrix, partitioned, $A=\begin{bmatrix} B & C\\
C^* & D\end{bmatrix}$ with $B$ square, then prove that $\mathcal{C}(C^*)\subseteq\mathcal{C}(D)$.
I think we may instead show that $\mathcal{N}(D)\subseteq\mathcal{N}(C)$. So I tried as follows:
If $Dy=0$, then let $x=[z\ \ y]$. Now $x^*Ax=z^*Bz+y^*C^*z+z^*Cy\geq 0$. I am stuck here. Can I somehow proceed to show that $Cy=0$ too? or is there any other way of attacking this problem? Help needed.
Here $\mathcal{C}(.)$ is the column space and $\mathcal{N}(.)$ is the null space.
Thanks in advance.
Best Answer
Your approach is correct. We can continue it as follows.
As you have shown, we have $$ x^*Ax = z^*Bz + y^*C^*z + z^*Cy = z^*Bz + 2\operatorname{Re}(\langle z,Cy \rangle). $$ Now, suppose for the purpose of contradiction that $Cy \neq 0$. Let $z = -t \cdot Cy$ with $t > 0$. We find that $$ x^*Ax = ((Cy)^*B(Cy)) \cdot t^2 - 2 \|Cy\|^2 \cdot t. $$ Show that there must be value of $t$ for which $x^*Ax$ is negative, contradicting the positive semidefiniteness of $A$. Thus, $\mathcal N(C) \supseteq \mathcal N(D)$, which was what we wanted.
Another approach:
Because $A$ is positive semidefinite, the principal submatrix $D$ must be positive semidefinite. Sylvester's law of inertia guarantees the existence of an invertible matrix $P$ such that $PDP^{*}$ has the block-form $$ PDP^* = \pmatrix{I_r & 0\\0 & 0}, $$ where $I_r$ is the identity matrix of size $r$ (and $r$ is the rank of $D$). Note that because $A$ is positive semidefinite, the conjugate matrix $$ \pmatrix{I & 0\\ 0 & P} \pmatrix{B & C\\ C^* & D} \pmatrix{I & 0\\0 & P^*} = \pmatrix{B & CP^*\\PC^* & PDP^*} $$ must be positive semidefinite. Partition the matrices $PC^*$ into blocks $$ PC^* = \pmatrix{G_1\\G_2}, $$ where $G_1$ has $r$ rows. We now see that the block-matrix $$ \pmatrix{B & G_1^* & G_2^*\\ G_1 & I_r & 0\\ G_2 & 0 & 0 } $$ is positive semidefinite. It follows that the principal submatrix $$ \pmatrix{B & G_2^*\\ G_2 & 0} $$ is positive semidefinite. It follows that $G_2$ must be zero: otherwise, this matrix has a principal submatrix of the form $$ \pmatrix{a & \bar b\\b & 0} $$ with $b \neq 0$, which has negative determinant and therefore fails to be positive semidefinite. Conclude that $\mathcal C(PC^*) \subseteq \mathcal C(PDP^*)$, from which it follows that $\mathcal C(C^*)\subseteq \mathcal C(D)$.