Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 31 (Random Matrices)

Let $V_m=\det [x_i^{j-1}{]}_{i,j=1}^m$. It is an alternating polynomial: exchanging two variables exchanges two rows of the Vandermonde matrix and reverses its determinant. Hence $V_m$ vanishes whenever $x_k=x_l$, so each factor $x_l-x_k$ divides it. The distinct linear factors are pairwise coprime polynomials, and their product therefore divides $V_m$. Both have total degree $0+1+\cdots +(m-1)=m(m-1)/2$, soThe coefficient of $x_m^{m-1}$ in the determinant is $V_{m-1}$, by expansion along its last row. The same coefficient in the product is ${\prod }_{k<l<m}(x_l-x_k)$. Thus $C_m=C_{m-1}$, and $C_1=1$. Consequently the Vandermonde determinant identity isThis is a polynomial identity, including repeated coordinates; no division by a possibly zero numerical Vandermonde product is needed.

Write ${\tilde{P}}_j(x)=x^{j-1}+{\sum }_{r=0}^{j-2}{b}_{rj}{x}^r$. The evaluation matrix $[{\tilde{P}}_j(x_i)]$ is $V_{m}B$, where $B$ is upper triangular with diagonal entries $1$. Its determinant is therefore unchanged. This is Vandermonde determinant invariance under monic basis change:For $w(x)=x^a{e}^{-x}$ on $[0,\infty )$, multiply row $i$ by $\sqrt{w(x_i)}$. The requested weighted product becomes the weighted Vandermonde determinant squareThe sign difference between the two orientations of the Vandermonde product disappears on squaring. For $a=0$ interpret the weight at zero by continuity, so $w(0)=1$; for $a>0$, $w(0)=0$.

Choose real monic orthogonal polynomials $p_j$ of degree $j$, $j=0,1,\dots$, for $w(x)=x^a{e}^{-x}$ on $[0,\infty )$, and letExtend each ${\varphi }_j$ by zero to negative $x$. The resulting functions form an orthonormal set in $L^2(\mathbb{R},dx)$. Define the orthogonal polynomial projection kernelThe formula on the right is for nonnegative arguments; the kernel is zero when either argument is negative. With ${\Phi }_{ij}={\varphi }_{j-1}(x_i)$, the kernel evaluation matrix is $\Phi {\Phi }^T$. By the determinant multiplication identity and the preceding monic basis change,Thus initially ${\hat{c}}_m=c_m{\prod }_j{h}_j$.

Use the symmetric labelled joint probability density on the full orthant. Expanding the two copies of $\det \Phi$ into permutations and integrating all variables, orthonormality makes a term vanish unless the two permutations agree. Each of the $m!$ surviving terms integrates to $1$. HenceIf the eigenvalues are instead listed in increasing order on a single ordered chamber, that chamber's density is $m!$ times this symmetric density. We keep the full-orthant convention throughout.

An explicit choice is $p_j(x)=(-1{)}^{j}j!L_j^{(a)}(x)$ in terms of the Generalized Laguerre polynomials. To verify the normalization independently, their Rodrigues' formula givesIntegrating by parts $j$ times proves orthogonality to every lower-degree polynomial. The boundary terms vanish at zero and infinity for $a\ge 0$. Taking the other factor to be $p_j$ and using $p_j^{(j)}=j!$ gives $h_j=j!\Gamma (a+j+1)$, where $\Gamma$ is the Gamma function. The Laguerre projection kernel is thereforefor $x,y\ge 0$.

Write $D_n=\det [K(x_i,x_j){]}_{i,j=1}^n$ and expand it by permutations. The assumptions ensure that the indicated single-variable integrals exist; the finite-rank projection kernel constructed above satisfies them with $r=m$. No symmetry of a general kernel is needed for the following argument.

If a permutation fixes $n$, its product contains the separate factor $K(x_n,x_n)$. Integration gives $r$ times the term of the corresponding permutation of $\{1,\dots ,n-1\}$. These terms contribute $r{D}_{n-1}$.

Otherwise $n$ belongs to a longer permutation cycle. Let $i$ precede $n$ and let $j$ follow it. The only factors containing $x_n$ are $K(x_i,x_n)K(x_n,x_j)$, whose integral is $K(x_i,x_j)$ by the reproducing assumption. Delete $n$ from the cycle to obtain a permutation $\tau$ on $n-1$ letters. Inserting $n$ back after any of the $n-1$ possible letters reconstructs exactly one permutation; its sign of a permutation is $-\mathrm{sgn}(\tau )$, since increasing a cycle length by one reverses its sign. Thus the nonfixed terms contribute $-(n-1)D_{n-1}$.

Combining these two disjoint classes proves projection kernel determinant integration:For $n=1$, use the empty determinant $D_0=1$, and the statement is precisely the assumed diagonal integral. For our kernel, orthonormality directly yieldsso every step of the marginal integration is justified.

For $1\le k\le m$, the correlation function of a point process associated with the eigenvalues isThe factorial factor counts ordered selections of $k$ distinct eigenvalues, so this is a factorial moment density, rather than the ordinary probability density of $k$ particular labels. Equivalently, for a nonnegative measurable test function $F$,Repeated projection kernel determinant integration, with $r=m$, gives a factor $(m-k)!$ on integrating an $m$-by-$m$ kernel determinant down to size $k$. Together with the normalizing $1/m!$ in $f_m$, the factors cancel:In particular $R_1(x)=K_m(x,x)$ and $\int R_{k}d\mathbf{x}=m!/(m-k)!$. Set $R_0=1$ and $R_k=0$ for $k>m$. The eigenvalue configuration is thus a determinantal point process with a rank-$m$ finite-rank projection kernel.

A function $f:\mathbb{R}\to \mathbb{R}$ satisfies Lipschitz continuity with bound $L\ge 0$ ifHere $L$ is an admissible Lipschitz bound; the least admissible value is the Lipschitz constant. The later parts use the admissible bound $1$.

The empirical spectral measures of the two real symmetric matrices arewhere ${\delta }_t$ is the Dirac measure at $t$. Thus $⟨L_N,f⟩=N^{-1}{\sum }_{i}f({\lambda }_i)$. Pair the eigenvalues in the specified increasing order and apply the triangle inequality followed by Lipschitz continuity with bound $1$:This deterministic estimate uses only the Lipschitz bound; no entry independence or distributional hypothesis is needed here.

The Hoffman–Wielandt inequality says that for normal matrices $A,B$ with eigenvalues ${\alpha }_i,{\beta }_i$, there is a permutation $\pi$ such thatwhere $\Vert \cdot {\Vert }_F$ is the Frobenius norm and $\ast$ denotes the conjugate transpose. For Hermitian matrices, the increasingly ordered real eigenvalues may be paired directly: this ordering minimizes the sum of squared distances over all permutations, by removing crossed pairs. In particular, for the real symmetric matrices in this question,The square on the right is justified by symmetry. For arbitrary matrices, the appropriate expression is the conjugate-transpose product, not the ordinary square.

Put $D=|⟨L_N,f⟩-⟨{\hat{L}}_N,f⟩|$. Apply the Cauchy-Schwarz inequality to the average of the nonnegative eigenvalue differences, and then the Hoffman–Wielandt inequality:Taking nonnegative square roots gives the spectral Lipschitz bound from Frobenius distanceThe factor $N^{-1/2}$ is essential: the empirical spectral measure has total mass $1$, not $N$.

Use the fixed common entry law from the printed setup: $Y=\sqrt{N}{X}_{ij}$ has mean zero and second moment $1$ for each upper-triangular entry, including the diagonal. LetSince $\mathbb{E}Y=0$, the removed part after centered truncation of a Wigner matrix isConsequently $\mathbb{E}{\Delta }_{ij}^2=N^{-1}\mathrm{Var}(Z_C)\le N^{-1}t(C)$. Symmetry gives $\mathrm{Tr}{\Delta }^2={\sum }_{i,j}{\Delta }_{ij}^2$. There are $N$ diagonal terms and $N(N-1)$ off-diagonal terms in this sum, soMirrored entries are counted twice, as they must be; independence of those mirrored entries is neither true nor needed for this expectation calculation.

By the preceding bound and the Markov inequality,The dominated convergence theorem applies to $Y^2{1}_{\{|Y|\ge C\}}\le Y^2$, so $t(C)\to 0$. Choose $C$ with $t(C)<{\epsilon }^3$. ThenThis second moment bound for spectral truncation depends only on $\epsilon$ and the common law of $Y$, and is uniform over functions with Lipschitz bound $1$. No fourth-moment hypothesis is required. If the scaled entry law were allowed to vary arbitrarily with $N$, a uniform choice would instead require uniform decay of its second-moment tails; the intended fixed-law setting supplies exactly that condition.

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.