A necessary and sufficient condition for a symmetric matrix to be positive semidefinite in terms of its Schur complement

According to the Wikipedia article Schur complement, if $X$ is a symmetric matrix of real numbers given by $X = \begin{bmatrix} A &B\\ B^T &C \end{bmatrix}$, then $X \succeq0$ (i.e. $X$ is positive semidefinite) iff the following conditions hold: \begin{align*} A&\succeq 0\\ C-B^TA^gB&\succeq0\\ (I-AA^g)B&=0\tag{*} \end{align*} where $A^g$ denotes the generalized inverse of $A$.

I've read somewhere that there is an interpretation of the third condition (the one marked with an asterisk) in terms of $A$'s column space. (Perhaps that $(*)$ is equivalent to $\mathcal{R}(B) \subseteq \mathcal{R}(A)$, i.e. that $B$'s column space is contained in $A$'s column space?)

How can $(*)$ be stated in terms of $A$'s column space, and why are these two representations equivalent?

If possible, explain it as simply as possible, assuming minimal to non-existent familiarity with Schur complements, generalized inverses, and generalized eigenvalue decompositions.

It is wrong to call $A^g$ the generalised inverse of $A$, because generalised inverses in general are not unique.

However, since $AA^gA=A$, the matrix $AA^g$ is always idempotent. As such, it represents a (possibly non-orthogonal) projection. If $A$ is $n\times n$, then from $$ A\mathbb R^n=AA^gA\mathbb R^n\subseteq AA^g\mathbb R^n\subseteq A\mathbb R^n, $$ we obtain $AA^gR^n=A\mathbb R^n$. Hence $AA^g$ has the same rank as $A$. In turn, \begin{aligned} \operatorname{nullity}(I-AA^g) &=\operatorname{rank}(AA^g)\quad\text{(because $AA^g$ is a projection)}\\ &=\operatorname{rank}(A). \end{aligned} This, together with the equality $(I-AA^g)A=0$, imply that the null space of $I-AA^g$ is precisely the column space of $A$. Hence the condition $(I-AA^g)B=0$ means that the column space of $B$ lies inside the column space of $A$. In other words, $B=AR$ for some matrix $R$.

Therefore, what Wikipedia says can be rephrased as follows: a real symmetric matrix $X=\pmatrix{A&B\\ B^T&C}$ is positive semidefinite if and only if

1. $A\succeq0$,
2. $\operatorname{range}(B)\subseteq\operatorname{range}(A)$, i.e., $B=AR$ for some matrix $R$, and
3. $C-R^TAR\succeq0$.

For necessity, if $X\succeq0$, clearly $A\succeq0$. Suppose $\operatorname{range}(B)\not\subseteq\operatorname{range}(A)$. Then $By\not\in\operatorname{range}(A)=(\ker A)^\perp$ for some vector $y$. Therefore there exists some $x\in\ker A$ such that $x^TBy\ne0$. If we keep $y$ fixed and scale $x$ by a large factor of an appropriate sign, we can make $x^TBy$ a large negative number. Hence $$ \pmatrix{x^T&y^T}\pmatrix{A&B\\ B^T&C}\pmatrix{x\\ y}=2x^TBy+y^TCy<0, $$ which is a contradiction to the positive semidefiniteness of $X$. So, we must have $\operatorname{range}(B)\subseteq\operatorname{range}(A)$, i.e. $B=AR$ for some matrix $R$. Finally, $C-R^TAR$ must be positive semidefinite because $X$ is congruent to $$ \pmatrix{I&0\\ -R^T&I}\pmatrix{A&AR\\ R^TA&C}\pmatrix{I&-R\\ 0&I}=\pmatrix{A&0\\ 0&C-R^TAR}. $$ This congruence relation also proves the sufficiency of $(1)-(3)$.