Consider the following linear system (in block form) with s.p.d. matrix: $$ \begin{pmatrix} A & B\\ B^\top & C \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} = \begin{pmatrix} f\\g \end{pmatrix} $$ I'm wondering if elimination of some variables does improve conditioning of the system matrix. If we eliminate $y$ first, by substituting $y = C^{-1}g - C^{-1}B^\top x$ the following system is obtained: $$ (A - BC^{-1}B^\top) x = f - BC^{-1}g. $$ The matrix of the new system is simply the Schur complement of the block $C$.
The question is whether the conditioning number of the resulting system is less than the conditioning number of the original one? The case $C = I$ is particularly interesting.
I've tried using formula $$ 0 = \det \begin{pmatrix} A - \lambda I& B\\ B^\top & C - \lambda I \end{pmatrix} = \det(C - \lambda I) \det(A - \lambda I - B (C - \lambda I)^{-1}B^\top), $$ but with no luck, though $A - B (C - \lambda I)^{-1}B^\top$ seems to be quite close to $A - BC^{-1}B^\top$.
Numerical experiments show that the Schur complment is always better conditioned than the original matrix, here's my code.
Experiments also show that not only s.p.d, but also diagonally dominant M-matrices share this property.