Espero que me de la idea de derecho de lo que se consideraría un "geométrica" de la interpretación.
Usted puede pensar de Schur de complementación como la división de una transformación lineal de acuerdo a la forma en que las parejas de dos subespacios complementarios. Si usted representa a un vector como la suma de dos componentes en los subespacios, el resultado de actuar en él con el lineal de transformación tiene cuatro partes: dos correspondientes a la diagonal de bloques de la matriz, donde la transformación actúa sobre cada componente por separado, y dos correspondientes a la diagonal de bloques de la matriz, donde la transformación de la mezcla de los dos componentes. El complemento de Schur de los resultados de la división de la transformación en los pasos correspondientes a esta descomposición.
Imagina que estás de realizar estas operaciones en un equipo, y que te gustaría hacer de todo "en el lugar" -- desea permitir la multiplicación de un vector por una matriz (x *= A) y la adición de una matriz de múltiples de un vector a otro vector (x += B y), pero usted no desea agregar matriz de dos múltiplos de vectores (z = A x + B y). (Aquí a op= b es corto para a = a op b.) Aplicar directamente una transformación
$$Ax=\begin{pmatrix}P&K\\M&R\end{pmatrix}\begin{pmatrix}y\\z\end{pmatrix}$$
would require you to calculate $Py+Kz$ and $Mi+Rz$. To avoid that, you could first mix the lower component into the upper component (y += K z), then apply the diagonal transforms on the subspaces (y *= P and z *= R), and then mix the upper component into the lower component (z += M y), but the result wouldn't be quite right because you'd be mixing the lower component back into itself and multiplying the mixed components by the diagonal blocks. Correcting for this yields:
$$\begin{pmatrix}P&K\\M&R\end{pmatrix}
=
\begin{pmatrix}I&\\MP^{-1}&I\end{pmatrix}
\begin{pmatrix}P&\\&R-MP^{-1}K\end{pmatrix}
\begin{pmatrix}I&P^{-1}K\\&I\end{pmatrix}
=: LDU
\;,
$$
which decomposes the linear transform into a part $U$ that mixes the lower component into the upper component, a block-diagonal part $D$ that acts on the subspaces separately, and a part $L$ that mixes the upper component into the lower component; the block-diagonal part contains the Schur complement $S=R-MP^{-1}K$.
Now if you have
$$EA=AF$$
with $E$ lower triangular and $F$ upper triangular, that says that it doesn't matter whether you first mix the lower component into the upper component using $F$ and then apply $A$, or first apply $A$ and then mix the upper component into the lower component using $E$. With our decomposition of $A$, this becomes
$$ELDU=LDUF\;.$$
Now $E$ and $L$ are both lower diagonal and mix the upper component into the lower component, and $U$ and $F$ are both upper diagonal and mix the lower component into the upper component. Since the inverse of a lower/upper triangular matrix is again lower/upper triangular (the inverse of mixing one component into the other is simply subtracting what was added), multiplying by $L^{-1}$ on the left and $U^{-1}$ on the right collects all the lower diagonal parts and all the upper diagonal parts:
$$\left(L^{-1}EL\right)D=D\left(UFU^{-1}\right)\;.$$
The key here is that now there's only mixing in one direction on each side, upper into lower on the left and lower into upper on the right, and mixing in only one direction doesn't affect the diagonal blocks, which describe how the transform acts separately on the components. So we can simply equate the lower right diagonal blocks, and since $L$ and $U$ are pure mixing and their diagonal blocks are identities, this yields
$$E'S=SF'\;.$$
