What is the linear algebra of the Schur complement? I am interested in the combinatorics of electrical networks, and this leads naturally into the notion of the Schur complement:

Suppose we have a graph (possibly edge-weighted with 'resistances') with "boundary" and "interior" vertices— the idea is that we get to have control over the voltages on the boundary, but not on the interior. The graph, being a graph, has a Laplacian matrix $L=D-A$.
It turns out that the Schur complement of the Laplacian with respect to the internal vertices (thus, a square matrix of side length the number of boundary vertices) has electrical significance. Namely, this is the so-called "response matrix" which takes in a vector of voltages and spits out the corresponding vector of current: how much electricity will flow through the boundary vertices.

Regardless of application: I don't like matrices. I like linear maps (and I'm fine with bases). Is there any way for someone like me to think about the Schur complement without going all the way down to matrix-land?
 A: Understanding Schur complement through analogy
For a two-by-two matrix, you have
$$\left[\begin{matrix}a&b\\c&d\end{matrix}\right]\left[\begin{matrix}x\\y\end{matrix}\right] = \left[\begin{matrix}e\\f\end{matrix}\right]
    \quad\implies\quad
\left[\begin{matrix}a-bd^{-1}c&0\\0&d -ca^{-1}b\end{matrix}\right]\left[\begin{matrix}x\\y\end{matrix}\right] = \left[\begin{matrix}e-bd^{-1}f\\f-ca^{-1}e\end{matrix}\right].$$
This means that you can subtract a multiple of one row from the other row and obtain an equation of one variable, e.g.
$$(a-bd^{-1}c) x=e-bd^{-1}f,$$
from which point you can easily calculate the other variable by, e.g.
$y = d^{-1}(f - cx)$.
Schur complement is for two-by-two square block matrices, such as the block matrix $\mathbf M$ below.  You can subtract a linear combination of one row from another row, and obtain a linear system
$$\left[\begin{matrix}\mathbf A&\mathbf B\\\mathbf C&\mathbf D\end{matrix}\right]\left[\begin{matrix}\mathbf x\\\mathbf y\end{matrix}\right] = \left[\begin{matrix}\mathbf e\\\mathbf f\end{matrix}\right]
    \quad\implies\quad
\left[\begin{matrix}\mathbf A-\mathbf B\mathbf D^{-1}\mathbf C&\mathbf 0\\\mathbf 0&\mathbf D-\mathbf C\mathbf A^{-1}\mathbf B\end{matrix}\right]\left[\begin{matrix}\mathbf x\\\mathbf y\end{matrix}\right] = \left[\begin{matrix}\mathbf e-\mathbf B\mathbf D^{-1}\mathbf f\\\mathbf f-\mathbf C\mathbf A^{-1}\mathbf e\end{matrix}\right].$$
Instead of solving the linear system involving the two-by-two block matrix ($\mathbf x$ and $\mathbf y$ at the same time), you could solve $\mathbf x$ and then $\mathbf y$.
The schur complement of $\mathbf A$ is written as $\mathbf M/\mathbf A=\mathbf A-\mathbf B\mathbf D^{-1}\mathbf C$.

Subnet of an electric network
In terms of the equivalent electric network, one could write $\mathbf L\mathbf v = \mathbf i$.  Let $V$ be all vertices, $B$ a subnet of vertices (target of study) and $\bar B$ all vertices outside the subnet.  Then, one can rearrange the rows and columns to arrive at
$$\left[\begin{matrix}\mathbf L_{BB}&\mathbf L_{B\bar B}\\\mathbf L_{\bar BB}&\mathbf L_{\bar B\bar B}\end{matrix}\right]\left[\begin{matrix}\mathbf v_{B}\\\mathbf v_{\bar B}\end{matrix}\right] = \left[\begin{matrix}\mathbf i_{B}\\\mathbf i_{\bar B}\end{matrix}\right],$$
where the first block row gives $\mathbf L_{BB}\mathbf v_B + \mathbf L_{B\bar B}\mathbf v_{\bar B}=\mathbf i_B$.  We can restrict our attention to the voltage in the subnet $\mathbf v_B$ by eliminating $\mathbf v_{\bar B}$ using Schur complement:
$$(\mathbf L_{BB}-\mathbf L_{B\bar B}\mathbf L_{\bar B\bar B}^{-1}\mathbf L_{\bar BB})\mathbf v_B=\mathbf i_{\mathbf B}-\mathbf L_{B\bar B}\mathbf L_{\bar B\bar B}^{-1}\mathbf i_{\bar B}$$
The schur complement of $L$ over $\mathbf L_{BB}$ (resistance within subnet) is $\mathbf L/\mathbf L_{BB}$, which is $\mathbf L_{BB}-\mathbf L_{B\bar B}\mathbf L_{\bar B\bar B}^{-1}\mathbf L_{\bar BB}$, also called the response matrix.  By applying the Schur complement $\mathbf L/\mathbf L_{BB}$ to the voltage of the subnet $\mathbf v_{B}$ would mean
$(\mathbf L/\mathbf L_{BB})\mathbf v_{B}$, which should generate $\mathbf i_{\mathbf B}-\mathbf L_{B\bar B}\mathbf L_{\bar B\bar B}^{-1}\mathbf i_{\bar B}$.
Notation: $\mathbf L_{BB}=\mathbf L_\text{internal}$, $\mathbf L_{B\bar B}=\mathbf L_\text{mid}=\operatorname{Transpose}(\mathbf L_{\bar BB})$, $\mathbf L_{\bar B\bar B}=\mathbf L_\text{boundary}$.

Current through subnet boundary

... spits out the corresponding vector of current: how much electricity will flow through the boundary vertices.

This italic part must refer to $\mathbf i_{\mathbf B}-\mathbf L_{B\bar B}\mathbf L_{\bar B\bar B}^{-1}\mathbf i_{\bar B}$.  I'm not sure why this is the electricity flowing through the boundary vertices.
