Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 31 / 3

Use block matrix elimination with the invertible square block $A$:The left multiplier is block triangular with identity diagonal blocks, so its determinant is $1$. Taking the determinant of the block-triangular product provesThe matrix $D-C{A}^{-1}B$ is the Schur complement of $A$. The proof needs $A$ invertible, but does not assume the Schur complement or the full matrix is invertible.

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.

Work with $z$ in the upper half-plane. Write the Stieltjes matrix resolvents and their normalized matrix traces asThe minor trace is normalized by $N$, not by $N-1$. This sign convention is the negative of the convention $(zI-X_N{)}^{-1}$ used in the general resolvent of an operator article. Here $g$ is the Stieltjes transform of a measure of the empirical spectral measure, with kernel $(x-z{)}^{-1}$.

The diagonal entries of $X_N$ are zero, so the preceding Schur complement formula gives $G_{ii}=-1/(z+q_i)$. Taking the matrix trace, subtracting the comparison value $-1/(z+g)$, and combining fractions gives the resolvent self-consistency defectThe positive numerator sign is fixed by this subtraction. All denominators are nonzero in the upper half-plane, as the imaginary-part estimate in the next part shows. The identity is deterministic and does not use entry independence or moment assumptions.

Put $\eta =\mathrm{Im}z>0$. For a real eigenvalue $\lambda$,The spectral theorem for real symmetric matrices therefore gives $\mathrm{Im}g>0$ and $\mathrm{Im}{q}_i\ge 0$, since $q_i$ is a sum of $(\lambda -z{)}^{-1}$ with nonnegative coefficients given by squared eigenvector coordinates of $x_i$. This is the positive imaginary part of a Stieltjes matrix resolvent. HenceApply the triangle inequality to the identity above, and add and subtract the minor trace:The allowed principal minor resolvent trace bound, with the printed $N$ normalization, bounds the second average by $c/(\eta N)$. Thus the explicit conclusion isIn particular, the final term is $O((\mathrm{Im}z{)}^{-3}{N}^{-1})$ with a constant independent of $N$. The upper-half-plane restriction gives the displayed positive denominators; in the lower half-plane the analogous bound uses $|\mathrm{Im}z|$.

The chosen centering also has a probabilistic meaning. The column vector $x_i$ is independent of the minor, has mean zero and coordinate variance $1/N$, so its conditional expectation satisfiesThis explains why the bound isolates fluctuations of a quadratic form around the minor trace. Mean centering alone is not a concentration estimate; no unrequested limiting assertion is being assumed.