Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 31 3 iii Solution Created 2026-10-03 Updated 2026-10-07
Work with in the upper half-plane. Write the Stieltjes matrix resolvents and their normalized matrix traces asThe minor trace is normalized by , not by . This sign convention is the negative of the convention used in the general resolvent of an operator article. Here is the Stieltjes transform of a measure of the empirical spectral measure, with kernel .
The diagonal entries of are zero, so the preceding Schur complement formula gives . Taking the matrix trace, subtracting the comparison value , 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.