Is Schur complement better conditioned than the original matrix? 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.
 A: Lemma. If $P$ and $Q$ are two $n\times n$ Hermitian matrices and $\operatorname{nullity}(Q)=k>0$, the minimum eigenvalue of $P+Q$ is bounded above by the $k$-th largest eigenvalue of $P$.
Proof of lemma. By Courant-Fischer minimax principle,
\begin{align}
\lambda_\min(P+Q)
&=\min_{\|x\|=1}x^\ast(P+Q)x\\
&\le\min\limits_{\substack{x\in\ker(Q)\\ \|x\|=1}}x^\ast(P+Q)x\\
&=\min\limits_{\substack{x\in\ker(Q)\\ \|x\|=1}}x^\ast Px\\
&\le\max\limits_{\substack{V\subseteq\mathbb C^n\\ \dim V=k}}
\min\limits_{\substack{x\in V\\ \|x\|=1}}x^\ast Px\\
&=\lambda_k^{\downarrow}(P).\ {}_{\large\square}
\end{align}
Now, return to your question. Suppose $A$ is $k\times k$ and $C$ is $(n-k)\times(n-k)$. Denote by $S$ the Schur complement $A-BC^{-1}B^\top$. Your block matrix is then $P+Q$, where
$$
P=\pmatrix{S\\ &0}\ \text{ and }\ Q=\pmatrix{BC^{-1/2}\\ C^{1/2}}\begin{matrix}\pmatrix{C^{-1/2}B^\top&C^{1/2}}\\ {}\end{matrix}.
$$
We have $\lambda_\min(P+Q)\le\lambda_k^{\downarrow}(P)=\lambda_\min(S)$ by our lemma and $\lambda_\max(S)=\lambda_\max(P)\le\lambda_\max(P+Q)$ because $Q\succeq0$. Since $\lambda_\min(P+Q)>0$, we conclude that
$$
\kappa(S)=\frac{\lambda_\max(S)}{\lambda_\min(S)}\le\frac{\lambda_\max(P+Q)}{\lambda_\min(P+Q)}=\kappa(P+Q).
$$
