column space of positive semidefinite matrix Let be $A$ a symmetric matrix in block form
$$A= \begin{bmatrix} B & C  \\ C^T &  E\end{bmatrix}$$
and let $\operatorname{cs} A$ be the column space for $A$. Because $A$ is positive semidefinite $$\operatorname{cs} \left[\matrix{C^T &  E}\right] = \operatorname{cs} E.$$
Why? Thanks for all explanations.
 A: A possible proof strategy: 


*

*Show that for a quadratic function there are only two possibilities:
(a) the function is bounded from below and the minimum exists, and (b) the function is unbounded from below (the "minimum" is $-\infty$).

*For any vector $a$ consider the function
$$
f(x)=\left[\matrix{-a\\x}\right]^T\left[\matrix{B & C\\C^T & E}\right]\left[\matrix{-a\\x}\right]=x^TEx-2x^TC^Ta+a^TBa.
$$
Since the matrix is positive semidefinite, we have $f(x)\ge 0$, so the minimum must exists according to $1$a.

*The necessary condition for the minimum
$$
\nabla f(x)=2Ex-2C^Ta=0\quad\Leftrightarrow\quad C^Ta=Ex.
$$

*Since the minimum exists, the condition 3 can be interpreted as: for any vector $a$ there exists a vector $x$ such that
$$
C^Ta=Ex.
$$
It means precisely that any linear combination of columns of $C^T$ can be represented as a linear combination of columns 0f $E$, i.e.
$$
\operatorname{cs} C^T\subset \operatorname{cs} E\qquad\Rightarrow\qquad cs\,\left[\matrix{C^T & E}\right]\subset cs\, E.
$$

*The opposite inclusion $\operatorname{cs}\left[\matrix{C^T & E}\right]\supset \operatorname{cs} E$ is obvious.

A: We need to show that $\mathrm{Im}\,C^*\subset\mathrm{Im}\,E$ or, equivalently, that $\ker E\subset\ker C$. Let $x\in\ker E$. With
$$
y:=\pmatrix{0\\x}, \quad \text{we have}\quad
Ay=\pmatrix{Cx\\Ex}=\pmatrix{Cx\\0}.
$$
But $y^*Ay=x^*Ex=0$ so $Ay=0$ since $A$ is positive semidefinite (note that $A=G^*G$ for some $G$ so $0=y^*Ay=y^*G^*Gy=\|Gy\|_2^2$ implies $Gy=0$ and hence $Ay=G^*Gy=G^*0=0$). Hence $Cx=0$, that is, $x\in\ker C$.
