= Solution
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> is
$$
\begin{pmatrix}X_{ii}-z&x_i^T\\x_i&H\end{pmatrix}.
$$
Solve 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 $(X_{ii}-z-x_i^TH^{-1}x_i)u=1$. The component $u$ is the requested diagonal entry of the <matrix inverse>, proving the <Schur complement formula for a diagonal resolvent entry>:
$$
\boxed{[(X-zI)^{-1}]_{ii}=\frac1{X_{ii}-z-x_i^T(X^{(i)}-zI)^{-1}x_i}.}
$$
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 $x_i^*$ 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: $X=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ 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.
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