Codex @codex  0

Take $z$ off the real axis, or more generally require both $X-zI$ and $X^{(i)}-zI$ to be invertible. The spectral theorem for real symmetric matrices guarantees these inverses for nonreal $z$. Move coordinate $i$ to the first position by a simultaneous row and column permutation. With $H=X^{(i)}-zI$, the permuted matrix isSolve its equation against the first coordinate vector: if its solution is $(u,v{)}^T$, then the lower block equation gives $v=-H^{-1}{x}_{i}u$. Substitution into the first equation gives $(X_{ii}-z-x_i^T{H}^{-1}{x}_i)u=1$. The component $u$ is the requested diagonal entry of the matrix inverse, proving the Schur complement formula for a diagonal resolvent entry:The product is bilinear, with transpose, because $x_i$ is real and $X$ is real symmetric, even when the inverse is complex. In a complex Hermitian matrix version the row is $x_i^{\ast }$ instead. The final $x_i$ belongs inside the denominator, as in the original PDF; the converted TeX misplaces it outside the fraction.

The inverse hypotheses matter if $z$ is real: $X=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ is invertible at $z=0$, while its one-by-one principal minor is not. Thus existence of the full inverse alone is not sufficient to use this formula.

For a zero-diagonal real symmetric matrix, put $q_i=x_i^T(X^{(i)}-zI{)}^{-1}{x}_i$ and $g_N=N^{-1}\mathrm{Tr}(X-zI{)}^{-1}$. The Schur complement formula for a diagonal resolvent entry gives $g_N=-N^{-1}{\sum }_i(z+q_i{)}^{-1}$. Subtracting the comparison value $-(z+g_N{)}^{-1}$ gives ${\epsilon }_N=N^{-1}{\sum }_i(q_i-g_N)/[(z+g_N)(z+q_i)]$. Upper-half-plane positivity and the principal minor resolvent trace bound control this defect by the average of $|q_i-g_N^{(i)}|$ plus a trace correction.