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Peter Denton, Stephen Parke, Xining Zhang, and I have just uploaded to the arXiv a completely rewritten version of our previous paper, now titled “Eigenvectors from Eigenvalues: a survey of a basic identity in linear algebra“. This paper is now a survey of the various literature surrounding the following basic identity in linear algebra, which we propose to call the eigenvector-eigenvalue identity:

Theorem 1 (Eigenvector-eigenvalue identity) Let ${A}$ be an ${n \times n}$ Hermitian matrix, with eigenvalues ${\lambda_1(A),\dots,\lambda_n(A)}$. Let ${v_i}$ be a unit eigenvector corresponding to the eigenvalue ${\lambda_i(A)}$, and let ${v_{i,j}}$ be the ${j^{th}}$ component of ${v_i}$. Then

$\displaystyle |v_{i,j}|^2 \prod_{k=1; k \neq i}^n (\lambda_i(A) - \lambda_k(A)) = \prod_{k=1}^{n-1} (\lambda_i(A) - \lambda_k(M_j))$

where ${M_j}$ is the ${n-1 \times n-1}$ Hermitian matrix formed by deleting the ${j^{th}}$ row and column from ${A}$.

When we posted the first version of this paper, we were unaware of previous appearances of this identity in the literature; a related identity had been used by Erdos-Schlein-Yau and by myself and Van Vu for applications to random matrix theory, but to our knowledge this specific identity appeared to be new. Even two months after our preprint first appeared on the arXiv in August, we had only learned of one other place in the literature where the identity showed up (by Forrester and Zhang, who also cite an earlier paper of Baryshnikov).

The situation changed rather dramatically with the publication of a popular science article in Quanta on this identity in November, which gave this result significantly more exposure. Within a few weeks we became informed (through private communication, online discussion, and exploration of the citation tree around the references we were alerted to) of over three dozen places where the identity, or some other closely related identity, had previously appeared in the literature, in such areas as numerical linear algebra, various aspects of graph theory (graph reconstruction, chemical graph theory, and walks on graphs), inverse eigenvalue problems, random matrix theory, and neutrino physics. As a consequence, we have decided to completely rewrite our article in order to collate this crowdsourced information, and survey the history of this identity, all the known proofs (we collect seven distinct ways to prove the identity (or generalisations thereof)), and all the applications of it that we are currently aware of. The citation graph of the literature that this ad hoc crowdsourcing effort produced is only very weakly connected, which we found surprising:

The earliest explicit appearance of the eigenvector-eigenvalue identity we are now aware of is in a 1966 paper of Thompson, although this paper is only cited (directly or indirectly) by a fraction of the known literature, and also there is a precursor identity of Löwner from 1934 that can be shown to imply the identity as a limiting case. At the end of the paper we speculate on some possible reasons why this identity only achieved a modest amount of recognition and dissemination prior to the November 2019 Quanta article.

Peter Denton, Stephen Parke, Xining Zhang, and I have just uploaded to the arXiv the short unpublished note “Eigenvectors from eigenvalues“. This note gives two proofs of a general eigenvector identity observed recently by Denton, Parke and Zhang in the course of some quantum mechanical calculations. The identity is as follows:

Theorem 1 Let ${A}$ be an ${n \times n}$ Hermitian matrix, with eigenvalues ${\lambda_1(A),\dots,\lambda_n(A)}$. Let ${v_i}$ be a unit eigenvector corresponding to the eigenvalue ${\lambda_i(A)}$, and let ${v_{i,j}}$ be the ${j^{th}}$ component of ${v_i}$. Then

$\displaystyle |v_{i,j}|^2 \prod_{k=1; k \neq i}^n (\lambda_i(A) - \lambda_k(A)) = \prod_{k=1}^{n-1} (\lambda_i(A) - \lambda_k(M_j))$

where ${M_j}$ is the ${n-1 \times n-1}$ Hermitian matrix formed by deleting the ${j^{th}}$ row and column from ${A}$.

For instance, if we have

$\displaystyle A = \begin{pmatrix} a & X^* \\ X & M \end{pmatrix}$

for some real number ${a}$, ${n-1}$-dimensional vector ${X}$, and ${n-1 \times n-1}$ Hermitian matrix ${M}$, then we have

$\displaystyle |v_{i,1}|^2 = \frac{\prod_{k=1}^{n-1} (\lambda_i(A) - \lambda_k(M))}{\prod_{k=1; k \neq i}^n (\lambda_i(A) - \lambda_k(A))} \ \ \ \ \ (1)$

assuming that the denominator is non-zero.

Once one is aware of the identity, it is not so difficult to prove it; we give two proofs, each about half a page long, one of which is based on a variant of the Cauchy-Binet formula, and the other based on properties of the adjugate matrix. But perhaps it is surprising that such a formula exists at all; one does not normally expect to learn much information about eigenvectors purely from knowledge of eigenvalues. In the random matrix theory literature, for instance in this paper of Erdos, Schlein, and Yau, or this later paper of Van Vu and myself, a related identity has been used, namely

$\displaystyle |v_{i,1}|^2 = \frac{1}{1 + \| (M-\lambda_i(A))^{-1} X \|^2}, \ \ \ \ \ (2)$

but it is not immediately obvious that one can derive the former identity from the latter. (I do so below the fold; we ended up not putting this proof in the note as it was longer than the two other proofs we found. I also give two other proofs below the fold, one from a more geometric perspective and one proceeding via Cramer’s rule.) It was certainly something of a surprise to me that there is no explicit appearance of the ${a,X}$ components of ${A}$ in the formula (1) (though they do indirectly appear through their effect on the eigenvalues ${\lambda_k(A)}$; for instance from taking traces one sees that ${a = \sum_{k=1}^n \lambda_k(A) - \sum_{k=1}^{n-1} \lambda_k(M)}$).

One can get some feeling of the identity (1) by considering some special cases. Suppose for instance that ${A}$ is a diagonal matrix with all distinct entries. The upper left entry ${a}$ of ${A}$ is one of the eigenvalues of ${A}$. If it is equal to ${\lambda_i(A)}$, then the eigenvalues of ${M}$ are the other ${n-1}$ eigenvalues of ${A}$, and now the left and right-hand sides of (1) are equal to ${1}$. At the other extreme, if ${a}$ is equal to a different eigenvalue of ${A}$, then ${\lambda_i(A)}$ now appears as an eigenvalue of ${M}$, and both sides of (1) now vanish. More generally, if we order the eigenvalues ${\lambda_1(A) \leq \dots \leq \lambda_n(A)}$ and ${\lambda_1(M) \leq \dots \leq \lambda_{n-1}(M)}$, then the Cauchy interlacing inequalities tell us that

$\displaystyle 0 \leq \lambda_i(A) - \lambda_k(M) \leq \lambda_i(A) - \lambda_k(A)$

for ${1 \leq k < i}$, and

$\displaystyle \lambda_i(A) - \lambda_{k+1}(A) \leq \lambda_i(A) - \lambda_k(M) < 0$

for ${i \leq k \leq n-1}$, so that the right-hand side of (1) lies between ${0}$ and ${1}$, which is of course consistent with (1) as ${v_i}$ is a unit vector. Thus the identity relates the coefficient sizes of an eigenvector with the extent to which the Cauchy interlacing inequalities are sharp.

(This post is mostly intended for my own reference, as I found myself repeatedly looking up several conversions between polynomial bases on various occasions.)

Let ${\mathrm{Poly}_{\leq n}}$ denote the vector space of polynomials ${P:{\bf R} \rightarrow {\bf R}}$ of one variable ${x}$ with real coefficients of degree at most ${n}$. This is a vector space of dimension ${n+1}$, and the sequence of these spaces form a filtration:

$\displaystyle \mathrm{Poly}_{\leq 0} \subset \mathrm{Poly}_{\leq 1} \subset \mathrm{Poly}_{\leq 2} \subset \dots$

A standard basis for these vector spaces are given by the monomials ${x^0, x^1, x^2, \dots}$: every polynomial ${P(x)}$ in ${\mathrm{Poly}_{\leq n}}$ can be expressed uniquely as a linear combination of the first ${n+1}$ monomials ${x^0, x^1, \dots, x^n}$. More generally, if one has any sequence ${Q_0(x), Q_1(x), Q_2(x)}$ of polynomials, with each ${Q_n}$ of degree exactly ${n}$, then an easy induction shows that ${Q_0(x),\dots,Q_n(x)}$ forms a basis for ${\mathrm{Poly}_{\leq n}}$.

In particular, if we have two such sequences ${Q_0(x), Q_1(x), Q_2(x),\dots}$ and ${R_0(x), R_1(x), R_2(x), \dots}$ of polynomials, with each ${Q_n}$ of degree ${n}$ and each ${R_k}$ of degree ${k}$, then ${Q_n}$ must be expressible uniquely as a linear combination of the polynomials ${R_0,R_1,\dots,R_n}$, thus we have an identity of the form

$\displaystyle Q_n(x) = \sum_{k=0}^n c_{QR}(n,k) R_k(x)$

for some change of basis coefficients ${c_{QR}(n,k) \in {\bf R}}$. These coefficients describe how to convert a polynomial expressed in the ${Q_n}$ basis into a polynomial expressed in the ${R_k}$ basis.

Many standard combinatorial quantities ${c(n,k)}$ involving two natural numbers ${0 \leq k \leq n}$ can be interpreted as such change of basis coefficients. The most familiar example are the binomial coefficients ${\binom{n}{k}}$, which measures the conversion from the shifted monomial basis ${(x+1)^n}$ to the monomial basis ${x^k}$, thanks to (a special case of) the binomial formula:

$\displaystyle (x+1)^n = \sum_{k=0}^n \binom{n}{k} x^k,$

thus for instance

$\displaystyle (x+1)^3 = \binom{3}{0} x^0 + \binom{3}{1} x^1 + \binom{3}{2} x^2 + \binom{3}{3} x^3$

$\displaystyle = 1 + 3x + 3x^2 + x^3.$

More generally, for any shift ${h}$, the conversion from ${(x+h)^n}$ to ${x^k}$ is measured by the coefficients ${h^{n-k} \binom{n}{k}}$, thanks to the general case of the binomial formula.

But there are other bases of interest too. For instance if one uses the falling factorial basis

$\displaystyle (x)_n := x (x-1) \dots (x-n+1)$

then the conversion from falling factorials to monomials is given by the Stirling numbers of the first kind ${s(n,k)}$:

$\displaystyle (x)_n = \sum_{k=0}^n s(n,k) x^k,$

thus for instance

$\displaystyle (x)_3 = s(3,0) x^0 + s(3,1) x^1 + s(3,2) x^2 + s(3,3) x^3$

$\displaystyle = 0 + 2 x - 3x^2 + x^3$

and the conversion back is given by the Stirling numbers of the second kind ${S(n,k)}$:

$\displaystyle x^n = \sum_{k=0}^n S(n,k) (x)_k$

thus for instance

$\displaystyle x^3 = S(3,0) (x)_0 + S(3,1) (x)_1 + S(3,2) (x)_2 + S(3,3) (x)_3$

$\displaystyle = 0 + x + 3 x(x-1) + x(x-1)(x-2).$

If one uses the binomial functions ${\binom{x}{n} = \frac{1}{n!} (x)_n}$ as a basis instead of the falling factorials, one of course can rewrite these conversions as

$\displaystyle \binom{x}{n} = \sum_{k=0}^n \frac{1}{n!} s(n,k) x^k$

and

$\displaystyle x^n = \sum_{k=0}^n k! S(n,k) \binom{x}{k}$

thus for instance

$\displaystyle \binom{x}{3} = 0 + \frac{1}{3} x - \frac{1}{2} x^2 + \frac{1}{6} x^3$

and

$\displaystyle x^3 = 0 + \binom{x}{1} + 6 \binom{x}{2} + 6 \binom{x}{3}.$

As a slight variant, if one instead uses rising factorials

$\displaystyle (x)^n := x (x+1) \dots (x+n-1)$

then the conversion to monomials yields the unsigned Stirling numbers ${|s(n,k)|}$ of the first kind:

$\displaystyle (x)^n = \sum_{k=0}^n |s(n,k)| x^k$

thus for instance

$\displaystyle (x)^3 = 0 + 2x + 3x^2 + x^3.$

One final basis comes from the polylogarithm functions

$\displaystyle \mathrm{Li}_{-n}(x) := \sum_{j=1}^\infty j^n x^j.$

For instance one has

$\displaystyle \mathrm{Li}_1(x) = -\log(1-x)$

$\displaystyle \mathrm{Li}_0(x) = \frac{x}{1-x}$

$\displaystyle \mathrm{Li}_{-1}(x) = \frac{x}{(1-x)^2}$

$\displaystyle \mathrm{Li}_{-2}(x) = \frac{x}{(1-x)^3} (1+x)$

$\displaystyle \mathrm{Li}_{-3}(x) = \frac{x}{(1-x)^4} (1+4x+x^2)$

$\displaystyle \mathrm{Li}_{-4}(x) = \frac{x}{(1-x)^5} (1+11x+11x^2+x^3)$

and more generally one has

$\displaystyle \mathrm{Li}_{-n-1}(x) = \frac{x}{(1-x)^{n+2}} E_n(x)$

for all natural numbers ${n}$ and some polynomial ${E_n}$ of degree ${n}$ (the Eulerian polynomials), which when converted to the monomial basis yields the (shifted) Eulerian numbers

$\displaystyle E_n(x) = \sum_{k=0}^n A(n+1,k) x^k.$

For instance

$\displaystyle E_3(x) = A(4,0) x^0 + A(4,1) x^1 + A(4,2) x^2 + A(4,3) x^3$

$\displaystyle = 1 + 11x + 11x^2 + x^3.$

These particular coefficients also have useful combinatorial interpretations. For instance:

• The binomial coefficient ${\binom{n}{k}}$ is of course the number of ${k}$-element subsets of ${\{1,\dots,n\}}$.
• The unsigned Stirling numbers ${|s(n,k)|}$ of the first kind are the number of permutations of ${\{1,\dots,n\}}$ with exactly ${k}$ cycles. The signed Stirling numbers ${s(n,k)}$ are then given by the formula ${s(n,k) = (-1)^{n-k} |s(n,k)|}$.
• The Stirling numbers ${S(n,k)}$ of the second kind are the number of ways to partition ${\{1,\dots,n\}}$ into ${k}$ non-empty subsets.
• The Eulerian numbers ${A(n,k)}$ are the number of permutations of ${\{1,\dots,n\}}$ with exactly ${k}$ ascents.

These coefficients behave similarly to each other in several ways. For instance, the binomial coefficients ${\binom{n}{k}}$ obey the well known Pascal identity

$\displaystyle \binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}$

(with the convention that ${\binom{n}{k}}$ vanishes outside of the range ${0 \leq k \leq n}$). In a similar spirit, the unsigned Stirling numbers ${|s(n,k)|}$ of the first kind obey the identity

$\displaystyle |s(n+1,k)| = n |s(n,k)| + |s(n,k-1)|$

and the signed counterparts ${s(n,k)}$ obey the identity

$\displaystyle s(n+1,k) = -n s(n,k) + s(n,k-1).$

The Stirling numbers of the second kind ${S(n,k)}$ obey the identity

$\displaystyle S(n+1,k) = k S(n,k) + S(n,k-1)$

and the Eulerian numbers ${A(n,k)}$ obey the identity

$\displaystyle A(n+1,k) = (k+1) A(n,k) + (n-k+1) A(n,k-1).$

While talking mathematics with a postdoc here at UCLA (March Boedihardjo) we came across the following matrix problem which we managed to solve, but the proof was cute and the process of discovering it was fun, so I thought I would present the problem here as a puzzle without revealing the solution for now.

The problem involves word maps on a matrix group, which for sake of discussion we will take to be the special orthogonal group $SO(3)$ of real $3 \times 3$ matrices (one of the smallest matrix groups that contains a copy of the free group, which incidentally is the key observation powering the Banach-Tarski paradox).  Given any abstract word $w$ of two generators $x,y$ and their inverses (i.e., an element of the free group ${\bf F}_2$), one can define the word map $w: SO(3) \times SO(3) \to SO(3)$ simply by substituting a pair of matrices in $SO(3)$ into these generators.  For instance, if one has the word $w = x y x^{-2} y^2 x$, then the corresponding word map $w: SO(3) \times SO(3) \to SO(3)$ is given by

$\displaystyle w(A,B) := ABA^{-2} B^2 A$

for $A,B \in SO(3)$.  Because $SO(3)$ contains a copy of the free group, we see the word map is non-trivial (not equal to the identity) if and only if the word itself is nontrivial.

Anyway, here is the problem:

Problem. Does there exist a sequence $w_1, w_2, \dots$ of non-trivial word maps $w_n: SO(3) \times SO(3) \to SO(3)$ that converge uniformly to the identity map?

To put it another way, given any $\varepsilon > 0$, does there exist a non-trivial word $w$ such that $\|w(A,B) - 1 \| \leq \varepsilon$ for all $A,B \in SO(3)$, where $\| \|$ denotes (say) the operator norm, and $1$ denotes the identity matrix in $SO(3)$?

As I said, I don’t want to spoil the fun of working out this problem, so I will leave it as a challenge. Readers are welcome to share their thoughts, partial solutions, or full solutions in the comments below.

Apoorva Khare and I have updated our paper “On the sign patterns of entrywise positivity preservers in fixed dimension“, announced at this post from last month. The quantitative results are now sharpened using a new monotonicity property of ratios ${s_{\lambda}(u)/s_{\mu}(u)}$ of Schur polynomials, namely that such ratios are monotone non-decreasing in each coordinate of ${u}$ if ${u}$ is in the positive orthant, and the partition ${\lambda}$ is larger than that of ${\mu}$. (This monotonicity was also independently observed by Rachid Ait-Haddou, using the theory of blossoms.) In the revised version of the paper we give two proofs of this monotonicity. The first relies on a deep positivity result of Lam, Postnikov, and Pylyavskyy, which uses a representation-theoretic positivity result of Haiman to show that the polynomial combination

$\displaystyle s_{(\lambda \wedge \nu) / (\mu \wedge \rho)} s_{(\lambda \vee \nu) / (\mu \vee \rho)} - s_{\lambda/\mu} s_{\nu/\rho} \ \ \ \ \ (1)$

of skew-Schur polynomials is Schur-positive for any partitions ${\lambda,\mu,\nu,\rho}$ (using the convention that the skew-Schur polynomial ${s_{\lambda/\mu}}$ vanishes if ${\mu}$ is not contained in ${\lambda}$, and where ${\lambda \wedge \nu}$ and ${\lambda \vee \nu}$ denotes the pointwise min and max of ${\lambda}$ and ${\nu}$ respectively). It is fairly easy to derive the monotonicity of ${s_\lambda(u)/s_\mu(u)}$ from this, by using the expansion

$\displaystyle s_\lambda(u_1,\dots, u_n) = \sum_k u_1^k s_{\lambda/(k)}(u_2,\dots,u_n)$

of Schur polynomials into skew-Schur polynomials (as was done in this previous post).

The second proof of monotonicity avoids representation theory by a more elementary argument establishing the weaker claim that the above expression (1) is non-negative on the positive orthant. In fact we prove a more general determinantal log-supermodularity claim which may be of independent interest:

Theorem 1 Let ${A}$ be any ${n \times n}$ totally positive matrix (thus, every minor has a non-negative determinant). Then for any ${k}$-tuples ${I_1,I_2,J_1,J_2}$ of increasing elements of ${\{1,\dots,n\}}$, one has

$\displaystyle \det( A_{I_1 \wedge I_2, J_1 \wedge J_2} ) \det( A_{I_1 \vee I_2, J_1 \vee J_2} ) - \det(A_{I_1,J_1}) \det(A_{I_2,J_2}) \geq 0$

where ${A_{I,J}}$ denotes the ${k \times k}$ minor formed from the rows in ${I}$ and columns in ${J}$.

For instance, if ${A}$ is the matrix

$\displaystyle A = \begin{pmatrix} a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p \end{pmatrix}$

for some real numbers ${a,\dots,p}$, one has

$\displaystyle a h - de\geq 0$

(corresponding to the case ${k=1}$, ${I_1 = (1), I_2 = (2), J_1 = (4), J_2 = (1)}$), or

$\displaystyle \det \begin{pmatrix} a & c \\ i & k \end{pmatrix} \det \begin{pmatrix} f & h \\ n & p \end{pmatrix} - \det \begin{pmatrix} e & h \\ i & l \end{pmatrix} \det \begin{pmatrix} b & c \\ n & o \end{pmatrix} \geq 0$

(corresponding to the case ${k=2}$, ${I_1 = (2,3)}$, ${I_2 = (1,4)}$, ${J_1 = (1,4)}$, ${J_2 = (2,3)}$). It turns out that this claim can be proven relatively easy by an induction argument, relying on the Dodgson and Karlin identities from this previous post; the difficulties are largely notational in nature. Combining this result with the Jacobi-Trudi identity for skew-Schur polynomials (discussed in this previous post) gives the non-negativity of (1); it can also be used to directly establish the monotonicity of ratios ${s_\lambda(u)/s_\mu(u)}$ by applying the theorem to a generalised Vandermonde matrix.

(Log-supermodularity also arises as the natural hypothesis for the FKG inequality, though I do not know of any interesting application of the FKG inequality in this current setting.)

Suppose we have an ${n \times n}$ matrix ${M}$ that is expressed in block-matrix form as

$\displaystyle M = \begin{pmatrix} A & B \\ C & D \end{pmatrix}$

where ${A}$ is an ${(n-k) \times (n-k)}$ matrix, ${B}$ is an ${(n-k) \times k}$ matrix, ${C}$ is an ${k \times (n-k)}$ matrix, and ${D}$ is a ${k \times k}$ matrix for some ${1 < k < n}$. If ${A}$ is invertible, we can use the technique of Schur complementation to express the inverse of ${M}$ (if it exists) in terms of the inverse of ${A}$, and the other components ${B,C,D}$ of course. Indeed, to solve the equation

$\displaystyle M \begin{pmatrix} x & y \end{pmatrix} = \begin{pmatrix} a & b \end{pmatrix},$

where ${x, a}$ are ${(n-k) \times 1}$ column vectors and ${y,b}$ are ${k \times 1}$ column vectors, we can expand this out as a system

$\displaystyle Ax + By = a$

$\displaystyle Cx + Dy = b.$

Using the invertibility of ${A}$, we can write the first equation as

$\displaystyle x = A^{-1} a - A^{-1} B y \ \ \ \ \ (1)$

and substituting this into the second equation yields

$\displaystyle (D - C A^{-1} B) y = b - C A^{-1} a$

and thus (assuming that ${D - CA^{-1} B}$ is invertible)

$\displaystyle y = - (D - CA^{-1} B)^{-1} CA^{-1} a + (D - CA^{-1} B)^{-1} b$

and then inserting this back into (1) gives

$\displaystyle x = (A^{-1} + A^{-1} B (D - CA^{-1} B)^{-1} C A^{-1}) a - A^{-1} B (D - CA^{-1} B)^{-1} b.$

Comparing this with

$\displaystyle \begin{pmatrix} x & y \end{pmatrix} = M^{-1} \begin{pmatrix} a & b \end{pmatrix},$

we have managed to express the inverse of ${M}$ as

$\displaystyle M^{-1} =$

$\displaystyle \begin{pmatrix} A^{-1} + A^{-1} B (D - CA^{-1} B)^{-1} C A^{-1} & - A^{-1} B (D - CA^{-1} B)^{-1} \\ - (D - CA^{-1} B)^{-1} CA^{-1} & (D - CA^{-1} B)^{-1} \end{pmatrix}. \ \ \ \ \ (2)$

One can consider the inverse problem: given the inverse ${M^{-1}}$ of ${M}$, does one have a nice formula for the inverse ${A^{-1}}$ of the minor ${A}$? Trying to recover this directly from (2) looks somewhat messy. However, one can proceed as follows. Let ${U}$ denote the ${n \times k}$ matrix

$\displaystyle U := \begin{pmatrix} 0 \\ I_k \end{pmatrix}$

(with ${I_k}$ the ${k \times k}$ identity matrix), and let ${V}$ be its transpose:

$\displaystyle V := \begin{pmatrix} 0 & I_k \end{pmatrix}.$

Then for any scalar ${t}$ (which we identify with ${t}$ times the identity matrix), one has

$\displaystyle M + UtV = \begin{pmatrix} A & B \\ C & D+t \end{pmatrix},$

and hence by (2)

$\displaystyle (M+UtV)^{-1} =$

$\displaystyle \begin{pmatrix} A^{-1} + A^{-1} B (D + t - CA^{-1} B)^{-1} C A^{-1} & - A^{-1} B (D + t- CA^{-1} B)^{-1} \\ - (D + t - CA^{-1} B)^{-1} CA^{-1} & (D + t - CA^{-1} B)^{-1} \end{pmatrix}.$

noting that the inverses here will exist for ${t}$ large enough. Taking limits as ${t \rightarrow \infty}$, we conclude that

$\displaystyle \lim_{t \rightarrow \infty} (M+UtV)^{-1} = \begin{pmatrix} A^{-1} & 0 \\ 0 & 0 \end{pmatrix}.$

On the other hand, by the Woodbury matrix identity (discussed in this previous blog post), we have

$\displaystyle (M+UtV)^{-1} = M^{-1} - M^{-1} U (t^{-1} + V M^{-1} U)^{-1} V M^{-1}$

and hence on taking limits and comparing with the preceding identity, one has

$\displaystyle \begin{pmatrix} A^{-1} & 0 \\ 0 & 0 \end{pmatrix} = M^{-1} - M^{-1} U (V M^{-1} U)^{-1} V M^{-1}.$

This achieves the aim of expressing the inverse ${A^{-1}}$ of the minor in terms of the inverse of the full matrix. Taking traces and rearranging, we conclude in particular that

$\displaystyle \mathrm{tr} A^{-1} = \mathrm{tr} M^{-1} - \mathrm{tr} (V M^{-2} U) (V M^{-1} U)^{-1}. \ \ \ \ \ (3)$

In the ${k=1}$ case, this can be simplified to

$\displaystyle \mathrm{tr} A^{-1} = \mathrm{tr} M^{-1} - \frac{e_n^T M^{-2} e_n}{e_n^T M^{-1} e_n} \ \ \ \ \ (4)$

where ${e_n}$ is the ${n^{th}}$ basis column vector.

We can apply this identity to understand how the spectrum of an ${n \times n}$ random matrix ${M}$ relates to that of its top left ${n-1 \times n-1}$ minor ${A}$. Subtracting any complex multiple ${z}$ of the identity from ${M}$ (and hence from ${A}$), we can relate the Stieltjes transform ${s_M(z) := \frac{1}{n} \mathrm{tr}(M-z)^{-1}}$ of ${M}$ with the Stieltjes transform ${s_A(z) := \frac{1}{n-1} \mathrm{tr}(A-z)^{-1}}$ of ${A}$:

$\displaystyle s_A(z) = \frac{n}{n-1} s_M(z) - \frac{1}{n-1} \frac{e_n^T (M-z)^{-2} e_n}{e_n^T (M-z)^{-1} e_n} \ \ \ \ \ (5)$

At this point we begin to proceed informally. Assume for sake of argument that the random matrix ${M}$ is Hermitian, with distribution that is invariant under conjugation by the unitary group ${U(n)}$; for instance, ${M}$ could be drawn from the Gaussian Unitary Ensemble (GUE), or alternatively ${M}$ could be of the form ${M = U D U^*}$ for some real diagonal matrix ${D}$ and ${U}$ a unitary matrix drawn randomly from ${U(n)}$ using Haar measure. To fix normalisations we will assume that the eigenvalues of ${M}$ are typically of size ${O(1)}$. Then ${A}$ is also Hermitian and ${U(n)}$-invariant. Furthermore, the law of ${e_n^T (M-z)^{-1} e_n}$ will be the same as the law of ${u^* (M-z)^{-1} u}$, where ${u}$ is now drawn uniformly from the unit sphere (independently of ${M}$). Diagonalising ${M}$ into eigenvalues ${\lambda_j}$ and eigenvectors ${v_j}$, we have

$\displaystyle u^* (M-z)^{-1} u = \sum_{j=1}^n \frac{|u^* v_j|^2}{\lambda_j - z}.$

One can think of ${u}$ as a random (complex) Gaussian vector, divided by the magnitude of that vector (which, by the Chernoff inequality, will concentrate to ${\sqrt{n}}$). Thus the coefficients ${u^* v_j}$ with respect to the orthonormal basis ${v_1,\dots,v_j}$ can be thought of as independent (complex) Gaussian vectors, divided by that magnitude. Using this and the Chernoff inequality again, we see (for ${z}$ distance ${\sim 1}$ away from the real axis at least) that one has the concentration of measure

$\displaystyle u^* (M-z)^{-1} u \approx \frac{1}{n} \sum_{j=1}^n \frac{1}{\lambda_j - z}$

and thus

$\displaystyle e_n^T (M-z)^{-1} e_n \approx \frac{1}{n} \mathrm{tr} (M-z)^{-1} = s_M(z)$

(that is to say, the diagonal entries of ${(M-z)^{-1}}$ are roughly constant). Similarly we have

$\displaystyle e_n^T (M-z)^{-2} e_n \approx \frac{1}{n} \mathrm{tr} (M-z)^{-2} = \frac{d}{dz} s_M(z).$

Inserting this into (5) and discarding terms of size ${O(1/n^2)}$, we thus conclude the approximate relationship

$\displaystyle s_A(z) \approx s_M(z) + \frac{1}{n} ( s_M(z) - s_M(z)^{-1} \frac{d}{dz} s_M(z) ).$

This can be viewed as a difference equation for the Stieltjes transform of top left minors of ${M}$. Iterating this equation, and formally replacing the difference equation by a differential equation in the large ${n}$ limit, we see that when ${n}$ is large and ${k \approx e^{-t} n}$ for some ${t \geq 0}$, one expects the top left ${k \times k}$ minor ${A_k}$ of ${M}$ to have Stieltjes transform

$\displaystyle s_{A_k}(z) \approx s( t, z ) \ \ \ \ \ (6)$

where ${s(t,z)}$ solves the Burgers-type equation

$\displaystyle \partial_t s(t,z) = s(t,z) - s(t,z)^{-1} \frac{d}{dz} s(t,z) \ \ \ \ \ (7)$

with initial data ${s(0,z) = s_M(z)}$.

Example 1 If ${M}$ is a constant multiple ${M = cI_n}$ of the identity, then ${s_M(z) = \frac{1}{c-z}}$. One checks that ${s(t,z) = \frac{1}{c-z}}$ is a steady state solution to (7), which is unsurprising given that all minors of ${M}$ are also ${c}$ times the identity.

Example 2 If ${M}$ is GUE normalised so that each entry has variance ${\sigma^2/n}$, then by the semi-circular law (see previous notes) one has ${s_M(z) \approx \frac{-z + \sqrt{z^2-4\sigma^2}}{2\sigma^2} = -\frac{2}{z + \sqrt{z^2-4\sigma^2}}}$ (using an appropriate branch of the square root). One can then verify the self-similar solution

$\displaystyle s(t,z) = \frac{-z + \sqrt{z^2 - 4\sigma^2 e^{-t}}}{2\sigma^2 e^{-t}} = -\frac{2}{z + \sqrt{z^2 - 4\sigma^2 e^{-t}}}$

to (7), which is consistent with the fact that a top ${k \times k}$ minor of ${M}$ also has the law of GUE, with each entry having variance ${\sigma^2 / n \approx \sigma^2 e^{-t} / k}$ when ${k \approx e^{-t} n}$.

One can justify the approximation (6) given a sufficiently good well-posedness theory for the equation (7). We will not do so here, but will note that (as with the classical inviscid Burgers equation) the equation can be solved exactly (formally, at least) by the method of characteristics. For any initial position ${z_0}$, we consider the characteristic flow ${t \mapsto z(t)}$ formed by solving the ODE

$\displaystyle \frac{d}{dt} z(t) = s(t,z(t))^{-1} \ \ \ \ \ (8)$

with initial data ${z(0) = z_0}$, ignoring for this discussion the problems of existence and uniqueness. Then from the chain rule, the equation (7) implies that

$\displaystyle \frac{d}{dt} s( t, z(t) ) = s(t,z(t))$

and thus ${s(t,z(t)) = e^t s(0,z_0)}$. Inserting this back into (8) we see that

$\displaystyle z(t) = z_0 + s(0,z_0)^{-1} (1-e^{-t})$

and thus (7) may be solved implicitly via the equation

$\displaystyle s(t, z_0 + s(0,z_0)^{-1} (1-e^{-t}) ) = e^t s(0, z_0) \ \ \ \ \ (9)$

for all ${t}$ and ${z_0}$.

Remark 3 In practice, the equation (9) may stop working when ${z_0 + s(0,z_0)^{-1} (1-e^{-t})}$ crosses the real axis, as (7) does not necessarily hold in this region. It is a cute exercise (ultimately coming from the Cauchy-Schwarz inequality) to show that this crossing always happens, for instance if ${z_0}$ has positive imaginary part then ${z_0 + s(0,z_0)^{-1}}$ necessarily has negative or zero imaginary part.

Example 4 Suppose we have ${s(0,z) = \frac{1}{c-z}}$ as in Example 1. Then (9) becomes

$\displaystyle s( t, z_0 + (c-z_0) (1-e^{-t}) ) = \frac{e^t}{c-z_0}$

for any ${t,z_0}$, which after making the change of variables ${z = z_0 + (c-z_0) (1-e^{-t}) = c - e^{-t} (c - z_0)}$ becomes

$\displaystyle s(t, z ) = \frac{1}{c-z}$

as in Example 1.

Example 5 Suppose we have

$\displaystyle s(0,z) = \frac{-z + \sqrt{z^2-4\sigma^2}}{2\sigma^2} = -\frac{2}{z + \sqrt{z^2-4\sigma^2}}.$

as in Example 2. Then (9) becomes

$\displaystyle s(t, z_0 - \frac{z_0 + \sqrt{z_0^2-4\sigma^2}}{2} (1-e^{-t}) ) = e^t \frac{-z_0 + \sqrt{z_0^2-4\sigma^2}}{2\sigma^2}.$

If we write

$\displaystyle z := z_0 - \frac{z_0 + \sqrt{z_0^2-4\sigma^2}}{2} (1-e^{-t})$

$\displaystyle = \frac{(1+e^{-t}) z_0 - (1-e^{-t}) \sqrt{z_0^2-4\sigma^2}}{2}$

one can calculate that

$\displaystyle z^2 - 4 \sigma^2 e^{-t} = (\frac{(1-e^{-t}) z_0 - (1+e^{-t}) \sqrt{z_0^2-4\sigma^2}}{2})^2$

and hence

$\displaystyle \frac{-z + \sqrt{z^2 - 4\sigma^2 e^{-t}}}{2\sigma^2 e^{-t}} = e^t \frac{-z_0 + \sqrt{z_0^2-4\sigma^2}}{2\sigma^2}$

which gives

$\displaystyle s(t,z) = \frac{-z + \sqrt{z^2 - 4\sigma^2 e^{-t}}}{2\sigma^2 e^{-t}}. \ \ \ \ \ (10)$

One can recover the spectral measure ${\mu}$ from the Stieltjes transform ${s(z)}$ as the weak limit of ${x \mapsto \frac{1}{\pi} \mathrm{Im} s(x+i\varepsilon)}$ as ${\varepsilon \rightarrow 0}$; we write this informally as

$\displaystyle d\mu(x) = \frac{1}{\pi} \mathrm{Im} s(x+i0^+)\ dx.$

In this informal notation, we have for instance that

$\displaystyle \delta_c(x) = \frac{1}{\pi} \mathrm{Im} \frac{1}{c-x-i0^+}\ dx$

which can be interpreted as the fact that the Cauchy distributions ${\frac{1}{\pi} \frac{\varepsilon}{(c-x)^2+\varepsilon^2}}$ converge weakly to the Dirac mass at ${c}$ as ${\varepsilon \rightarrow 0}$. Similarly, the spectral measure associated to (10) is the semicircular measure ${\frac{1}{2\pi \sigma^2 e^{-t}} (4 \sigma^2 e^{-t}-x^2)_+^{1/2}}$.

If we let ${\mu_t}$ be the spectral measure associated to ${s(t,\cdot)}$, then the curve ${e^{-t} \mapsto \mu_t}$ from ${(0,1]}$ to the space of measures is the high-dimensional limit ${n \rightarrow \infty}$ of a Gelfand-Tsetlin pattern (discussed in this previous post), if the pattern is randomly generated amongst all matrices ${M}$ with spectrum asymptotic to ${\mu_0}$ as ${n \rightarrow \infty}$. For instance, if ${\mu_0 = \delta_c}$, then the curve is ${\alpha \mapsto \delta_c}$, corresponding to a pattern that is entirely filled with ${c}$‘s. If instead ${\mu_0 = \frac{1}{2\pi \sigma^2} (4\sigma^2-x^2)_+^{1/2}}$ is a semicircular distribution, then the pattern is

$\displaystyle \alpha \mapsto \frac{1}{2\pi \sigma^2 \alpha} (4\sigma^2 \alpha -x^2)_+^{1/2},$

thus at height ${\alpha}$ from the top, the pattern is semicircular on the interval ${[-2\sigma \sqrt{\alpha}, 2\sigma \sqrt{\alpha}]}$. The interlacing property of Gelfand-Tsetlin patterns translates to the claim that ${\alpha \mu_\alpha(-\infty,\lambda)}$ (resp. ${\alpha \mu_\alpha(\lambda,\infty)}$) is non-decreasing (resp. non-increasing) in ${\alpha}$ for any fixed ${\lambda}$. In principle one should be able to establish these monotonicity claims directly from the PDE (7) or from the implicit solution (9), but it was not clear to me how to do so.

An interesting example of such a limiting Gelfand-Tsetlin pattern occurs when ${\mu_0 = \frac{1}{2} \delta_{-1} + \frac{1}{2} \delta_1}$, which corresponds to ${M}$ being ${2P-I}$, where ${P}$ is an orthogonal projection to a random ${n/2}$-dimensional subspace of ${{\bf C}^n}$. Here we have

$\displaystyle s(0,z) = \frac{1}{2} \frac{1}{-1-z} + \frac{1}{2} \frac{1}{1-z} = \frac{z}{1-z^2}$

and so (9) in this case becomes

$\displaystyle s(t, z_0 + \frac{1-z_0^2}{z_0} (1-e^{-t}) ) = \frac{e^t z_0}{1-z_0^2}$

A tedious calculation then gives the solution

$\displaystyle s(t,z) = \frac{(2e^{-t}-1)z + \sqrt{z^2 - 4e^{-t}(1-e^{-t})}}{2e^{-t}(1-z^2)}. \ \ \ \ \ (11)$

For ${\alpha = e^{-t} > 1/2}$, there are simple poles at ${z=-1,+1}$, and the associated measure is

$\displaystyle \mu_\alpha = \frac{2\alpha-1}{2\alpha} \delta_{-1} + \frac{2\alpha-1}{2\alpha} \delta_1 + \frac{1}{2\pi \alpha(1-x^2)} (4\alpha(1-\alpha)-x^2)_+^{1/2}\ dx.$

This reflects the interlacing property, which forces ${\frac{2\alpha-1}{2\alpha} \alpha n}$ of the ${\alpha n}$ eigenvalues of the ${\alpha n \times \alpha n}$ minor to be equal to ${-1}$ (resp. ${+1}$). For ${\alpha = e^{-t} \leq 1/2}$, the poles disappear and one just has

$\displaystyle \mu_\alpha = \frac{1}{2\pi \alpha(1-x^2)} (4\alpha(1-\alpha)-x^2)_+^{1/2}\ dx.$

For ${\alpha=1/2}$, one has an inverse semicircle distribution

$\displaystyle \mu_{1/2} = \frac{1}{\pi} (1-x^2)_+^{-1/2}.$

There is presumably a direct geometric explanation of this fact (basically describing the singular values of the product of two random orthogonal projections to half-dimensional subspaces of ${{\bf C}^n}$), but I do not know of one off-hand.

The evolution of ${s(t,z)}$ can also be understood using the ${R}$-transform and ${S}$-transform from free probability. Formally, letlet ${z(t,s)}$ be the inverse of ${s(t,z)}$, thus

$\displaystyle s(t,z(t,s)) = s$

for all ${t,s}$, and then define the ${R}$-transform

$\displaystyle R(t,s) := z(t,-s) - \frac{1}{s}.$

The equation (9) may be rewritten as

$\displaystyle z( t, e^t s ) = z(0,s) + s^{-1} (1-e^{-t})$

and hence

$\displaystyle R(t, -e^t s) = R(0, -s)$

or equivalently

$\displaystyle R(t,s) = R(0, e^{-t} s). \ \ \ \ \ (12)$

See these previous notes for a discussion of free probability topics such as the ${R}$-transform.

Example 6 If ${s(t,z) = \frac{1}{c-z}}$ then the ${R}$ transform is ${R(t,s) = c}$.

Example 7 If ${s(t,z)}$ is given by (10), then the ${R}$ transform is

$\displaystyle R(t,s) = \sigma^2 e^{-t} s.$

Example 8 If ${s(t,z)}$ is given by (11), then the ${R}$ transform is

$\displaystyle R(t,s) = \frac{-1 + \sqrt{1 + 4 s^2 e^{-2t}}}{2 s e^{-t}}.$

This simple relationship (12) is essentially due to Nica and Speicher (thanks to Dima Shylakhtenko for this reference). It has the remarkable consequence that when ${\alpha = 1/m}$ is the reciprocal of a natural number ${m}$, then ${\mu_{1/m}}$ is the free arithmetic mean of ${m}$ copies of ${\mu}$, that is to say ${\mu_{1/m}}$ is the free convolution ${\mu \boxplus \dots \boxplus \mu}$ of ${m}$ copies of ${\mu}$, pushed forward by the map ${\lambda \rightarrow \lambda/m}$. In terms of random matrices, this is asserting that the top ${n/m \times n/m}$ minor of a random matrix ${M}$ has spectral measure approximately equal to that of an arithmetic mean ${\frac{1}{m} (M_1 + \dots + M_m)}$ of ${m}$ independent copies of ${M}$, so that the process of taking top left minors is in some sense a continuous analogue of the process of taking freely independent arithmetic means. There ought to be a geometric proof of this assertion, but I do not know of one. In the limit ${m \rightarrow \infty}$ (or ${\alpha \rightarrow 0}$), the ${R}$-transform becomes linear and the spectral measure becomes semicircular, which is of course consistent with the free central limit theorem.

In a similar vein, if one defines the function

$\displaystyle \omega(t,z) := \alpha \int_{\bf R} \frac{zx}{1-zx}\ d\mu_\alpha(x) = e^{-t} (- 1 - z^{-1} s(t, z^{-1}))$

and inverts it to obtain a function ${z(t,\omega)}$ with

$\displaystyle \omega(t, z(t,\omega)) = \omega$

for all ${t, \omega}$, then the ${S}$-transform ${S(t,\omega)}$ is defined by

$\displaystyle S(t,\omega) := \frac{1+\omega}{\omega} z(t,\omega).$

Writing

$\displaystyle s(t,z) = - z^{-1} ( 1 + e^t \omega(t, z^{-1}) )$

for any ${t}$, ${z}$, we have

$\displaystyle z_0 + s(0,z_0)^{-1} (1-e^{-t}) = z_0 \frac{\omega(0,z_0^{-1})+e^{-t}}{\omega(0,z_0^{-1})+1}$

and so (9) becomes

$\displaystyle - z_0^{-1} \frac{\omega(0,z_0^{-1})+1}{\omega(0,z_0^{-1})+e^{-t}} (1 + e^{t} \omega(t, z_0^{-1} \frac{\omega(0,z_0^{-1})+1}{\omega(0,z_0^{-1})+e^{-t}}))$

$\displaystyle = - e^t z_0^{-1} (1 + \omega(0, z_0^{-1}))$

which simplifies to

$\displaystyle \omega(t, z_0^{-1} \frac{\omega(0,z_0^{-1})+1}{\omega(0,z_0^{-1})+e^{-t}})) = \omega(0, z_0^{-1});$

replacing ${z_0}$ by ${z(0,\omega)^{-1}}$ we obtain

$\displaystyle \omega(t, z(0,\omega) \frac{\omega+1}{\omega+e^{-t}}) = \omega$

and thus

$\displaystyle z(0,\omega)\frac{\omega+1}{\omega+e^{-t}} = z(t, \omega)$

and hence

$\displaystyle S(0, \omega) = \frac{\omega+e^{-t}}{\omega+1} S(t, \omega).$

One can compute ${\frac{\omega+e^{-t}}{\omega+1}}$ to be the ${S}$-transform of the measure ${(1-\alpha) \delta_0 + \alpha \delta_1}$; from the link between ${S}$-transforms and free products (see e.g. these notes of Guionnet), we conclude that ${(1-\alpha)\delta_0 + \alpha \mu_\alpha}$ is the free product of ${\mu_1}$ and ${(1-\alpha) \delta_0 + \alpha \delta_1}$. This is consistent with the random matrix theory interpretation, since ${(1-\alpha)\delta_0 + \alpha \mu_\alpha}$ is also the spectral measure of ${PMP}$, where ${P}$ is the orthogonal projection to the span of the first ${\alpha n}$ basis elements, so in particular ${P}$ has spectral measure ${(1-\alpha) \delta_0 + \alpha \delta_1}$. If ${M}$ is unitarily invariant then (by a fundamental result of Voiculescu) it is asymptotically freely independent of ${P}$, so the spectral measure of ${PMP = P^{1/2} M P^{1/2}}$ is asymptotically the free product of that of ${M}$ and of ${P}$.

Fix a non-negative integer ${k}$. Define an (weak) integer partition of length ${k}$ to be a tuple ${\lambda = (\lambda_1,\dots,\lambda_k)}$ of non-increasing non-negative integers ${\lambda_1 \geq \dots \geq \lambda_k \geq 0}$. (Here our partitions are “weak” in the sense that we allow some parts of the partition to be zero. Henceforth we will omit the modifier “weak”, as we will not need to consider the more usual notion of “strong” partitions.) To each such partition ${\lambda}$, one can associate a Young diagram consisting of ${k}$ left-justified rows of boxes, with the ${i^{th}}$ row containing ${\lambda_i}$ boxes. A semi-standard Young tableau (or Young tableau for short) ${T}$ of shape ${\lambda}$ is a filling of these boxes by integers in ${\{1,\dots,k\}}$ that is weakly increasing along rows (moving rightwards) and strictly increasing along columns (moving downwards). The collection of such tableaux will be denoted ${{\mathcal T}_\lambda}$. The weight ${|T|}$ of a tableau ${T}$ is the tuple ${(n_1,\dots,n_k)}$, where ${n_i}$ is the number of occurrences of the integer ${i}$ in the tableau. For instance, if ${k=3}$ and ${\lambda = (6,4,2)}$, an example of a Young tableau of shape ${\lambda}$ would be

$\displaystyle \begin{tabular}{|c|c|c|c|c|c|} \hline 1 & 1 & 1 & 2 & 3 & 3 \\ \cline{1-6} 2 & 2 & 2 &3\\ \cline{1-4} 3 & 3\\ \cline{1-2} \end{tabular}$

The weight here would be ${|T| = (3,4,5)}$.

To each partition ${\lambda}$ one can associate the Schur polynomial ${s_\lambda(u_1,\dots,u_k)}$ on ${k}$ variables ${u = (u_1,\dots,u_k)}$, which we will define as

$\displaystyle s_\lambda(u) := \sum_{T \in {\mathcal T}_\lambda} u^{|T|}$

using the multinomial convention

$\displaystyle (u_1,\dots,u_k)^{(n_1,\dots,n_k)} := u_1^{n_1} \dots u_k^{n_k}.$

Thus for instance the Young tableau ${T}$ given above would contribute a term ${u_1^3 u_2^4 u_3^5}$ to the Schur polynomial ${s_{(6,4,2)}(u_1,u_2,u_3)}$. In the case of partitions of the form ${(n,0,\dots,0)}$, the Schur polynomial ${s_{(n,0,\dots,0)}}$ is just the complete homogeneous symmetric polynomial ${h_n}$ of degree ${n}$ on ${k}$ variables:

$\displaystyle s_{(n,0,\dots,0)}(u_1,\dots,u_k) := \sum_{n_1,\dots,n_k \geq 0: n_1+\dots+n_k = n} u_1^{n_1} \dots u_k^{n_k},$

thus for instance

$\displaystyle s_{(3,0)}(u_1,u_2) = u_1^3 + u_1^2 u_2 + u_1 u_2^2 + u_2^3.$

Schur polyomials are ubiquitous in the algebraic combinatorics of “type ${A}$ objects” such as the symmetric group ${S_k}$, the general linear group ${GL_k}$, or the unitary group ${U_k}$. For instance, one can view ${s_\lambda}$ as the character of an irreducible polynomial representation of ${GL_k({\bf C})}$ associated with the partition ${\lambda}$. However, we will not focus on these interpretations of Schur polynomials in this post.

This definition of Schur polynomials allows for a way to describe the polynomials recursively. If ${k > 1}$ and ${T}$ is a Young tableau of shape ${\lambda = (\lambda_1,\dots,\lambda_k)}$, taking values in ${\{1,\dots,k\}}$, one can form a sub-tableau ${T'}$ of some shape ${\lambda' = (\lambda'_1,\dots,\lambda'_{k-1})}$ by removing all the appearances of ${k}$ (which, among other things, necessarily deletes the ${k^{th}}$ row). For instance, with ${T}$ as in the previous example, the sub-tableau ${T'}$ would be

$\displaystyle \begin{tabular}{|c|c|c|c|} \hline 1 & 1 & 1 & 2 \\ \cline{1-4} 2 & 2 & 2 \\ \cline{1-3} \end{tabular}$

and the reduced partition ${\lambda'}$ in this case is ${(4,3)}$. As Young tableaux are required to be strictly increasing down columns, we can see that the reduced partition ${\lambda'}$ must intersperse the original partition ${\lambda}$ in the sense that

$\displaystyle \lambda_{i+1} \leq \lambda'_i \leq \lambda_i \ \ \ \ \ (1)$

for all ${1 \leq i \leq k-1}$; we denote this interspersion relation as ${\lambda' \prec \lambda}$ (though we caution that this is not intended to be a partial ordering). In the converse direction, if ${\lambda' \prec \lambda}$ and ${T'}$ is a Young tableau with shape ${\lambda'}$ with entries in ${\{1,\dots,k-1\}}$, one can form a Young tableau ${T}$ with shape ${\lambda}$ and entries in ${\{1,\dots,k\}}$ by appending to ${T'}$ an entry of ${k}$ in all the boxes that appear in the ${\lambda}$ shape but not the ${\lambda'}$ shape. This one-to-one correspondence leads to the recursion

$\displaystyle s_\lambda(u) = \sum_{\lambda' \prec \lambda} s_{\lambda'}(u') u_k^{|\lambda| - |\lambda'|} \ \ \ \ \ (2)$

where ${u = (u_1,\dots,u_k)}$, ${u' = (u_1,\dots,u_{k-1})}$, and the size ${|\lambda|}$ of a partition ${\lambda = (\lambda_1,\dots,\lambda_k)}$ is defined as ${|\lambda| := \lambda_1 + \dots + \lambda_k}$.

One can use this recursion (2) to prove some further standard identities for Schur polynomials, such as the determinant identity

$\displaystyle s_\lambda(u) V(u) = \det( u_i^{\lambda_j+k-j} )_{1 \leq i,j \leq k} \ \ \ \ \ (3)$

for ${u=(u_1,\dots,u_k)}$, where ${V(u)}$ denotes the Vandermonde determinant

$\displaystyle V(u) := \prod_{1 \leq i < j \leq k} (u_i - u_j), \ \ \ \ \ (4)$

or the Jacobi-Trudi identity

$\displaystyle s_\lambda(u) = \det( h_{\lambda_j - j + i}(u) )_{1 \leq i,j \leq k}, \ \ \ \ \ (5)$

with the convention that ${h_d(u) = 0}$ if ${d}$ is negative. Thus for instance

$\displaystyle s_{(1,1,0,\dots,0)}(u) = h_1^2(u) - h_0(u) h_2(u) = \sum_{1 \leq i < j \leq k} u_i u_j.$

We review the (standard) derivation of these identities via (2) below the fold. Among other things, these identities show that the Schur polynomials are symmetric, which is not immediately obvious from their definition.

One can also iterate (2) to write

$\displaystyle s_\lambda(u) = \sum_{() = \lambda^0 \prec \lambda^1 \prec \dots \prec \lambda^k = \lambda} \prod_{j=1}^k u_j^{|\lambda^j| - |\lambda^{j-1}|} \ \ \ \ \ (6)$

where the sum is over all tuples ${\lambda^1,\dots,\lambda^k}$, where each ${\lambda^j}$ is a partition of length ${j}$ that intersperses the next partition ${\lambda^{j+1}}$, with ${\lambda^k}$ set equal to ${\lambda}$. We will call such a tuple an integral Gelfand-Tsetlin pattern based at ${\lambda}$.

One can generalise (6) by introducing the skew Schur functions

$\displaystyle s_{\lambda/\mu}(u) := \sum_{\mu = \lambda^i \prec \dots \prec \lambda^k = \lambda} \prod_{j=i+1}^k u_j^{|\lambda^j| - |\lambda^{j-1}|} \ \ \ \ \ (7)$

for ${u = (u_{i+1},\dots,u_k)}$, whenever ${\lambda}$ is a partition of length ${k}$ and ${\mu}$ a partition of length ${i}$ for some ${0 \leq i \leq k}$, thus the Schur polynomial ${s_\lambda}$ is also the skew Schur polynomial ${s_{\lambda /()}}$ with ${i=0}$. (One could relabel the variables here to be something like ${(u_1,\dots,u_{k-i})}$ instead, but this labeling seems slightly more natural, particularly in view of identities such as (8) below.)

By construction, we have the decomposition

$\displaystyle s_{\lambda/\nu}(u_{i+1},\dots,u_k) = \sum_\mu s_{\mu/\nu}(u_{i+1},\dots,u_j) s_{\lambda/\mu}(u_{j+1},\dots,u_k) \ \ \ \ \ (8)$

whenever ${0 \leq i \leq j \leq k}$, and ${\nu, \mu, \lambda}$ are partitions of lengths ${i,j,k}$ respectively. This gives another recursive way to understand Schur polynomials and skew Schur polynomials. For instance, one can use it to establish the generalised Jacobi-Trudi identity

$\displaystyle s_{\lambda/\mu}(u) = \det( h_{\lambda_j - j - \mu_i + i}(u) )_{1 \leq i,j \leq k}, \ \ \ \ \ (9)$

with the convention that ${\mu_i = 0}$ for ${i}$ larger than the length of ${\mu}$; we do this below the fold.

The Schur polynomials (and skew Schur polynomials) are “discretised” (or “quantised”) in the sense that their parameters ${\lambda, \mu}$ are required to be integer-valued, and their definition similarly involves summation over a discrete set. It turns out that there are “continuous” (or “classical”) analogues of these functions, in which the parameters ${\lambda,\mu}$ now take real values rather than integers, and are defined via integration rather than summation. One can view these continuous analogues as a “semiclassical limit” of their discrete counterparts, in a manner that can be made precise using the machinery of geometric quantisation, but we will not do so here.

The continuous analogues can be defined as follows. Define a real partition of length ${k}$ to be a tuple ${\lambda = (\lambda_1,\dots,\lambda_k)}$ where ${\lambda_1 \geq \dots \geq \lambda_k \geq 0}$ are now real numbers. We can define the relation ${\lambda' \prec \lambda}$ of interspersion between a length ${k-1}$ real partition ${\lambda' = (\lambda'_1,\dots,\lambda'_{k-1})}$ and a length ${k}$ real partition ${\lambda = (\lambda_1,\dots,\lambda_{k})}$ precisely as before, by requiring that the inequalities (1) hold for all ${1 \leq i \leq k-1}$. We can then define the continuous Schur functions ${S_\lambda(x)}$ for ${x = (x_1,\dots,x_k) \in {\bf R}^k}$ recursively by defining

$\displaystyle S_{()}() = 1$

and

$\displaystyle S_\lambda(x) = \int_{\lambda' \prec \lambda} S_{\lambda'}(x') \exp( (|\lambda| - |\lambda'|) x_k ) \ \ \ \ \ (10)$

for ${k \geq 1}$ and ${\lambda}$ of length ${k}$, where ${x' := (x_1,\dots,x_{k-1})}$ and the integral is with respect to ${k-1}$-dimensional Lebesgue measure, and ${|\lambda| = \lambda_1 + \dots + \lambda_k}$ as before. Thus for instance

$\displaystyle S_{(\lambda_1)}(x_1) = \exp( \lambda_1 x_1 )$

and

$\displaystyle S_{(\lambda_1,\lambda_2)}(x_1,x_2) = \int_{\lambda_2}^{\lambda_1} \exp( \lambda'_1 x_1 + (\lambda_1+\lambda_2-\lambda'_1) x_2 )\ d\lambda'_1.$

More generally, we can define the continuous skew Schur functions ${S_{\lambda/\mu}(x)}$ for ${\lambda}$ of length ${k}$, ${\mu}$ of length ${j \leq k}$, and ${x = (x_{j+1},\dots,x_k) \in {\bf R}^{k-j}}$ recursively by defining

$\displaystyle S_{\mu/\mu}() = 1$

and

$\displaystyle S_{\lambda/\mu}(x) = \int_{\lambda' \prec \lambda} S_{\lambda'/\mu}(x') \exp( (|\lambda| - |\lambda'|) x_k )$

for ${k > j}$. Thus for instance

$\displaystyle S_{(\lambda_1,\lambda_2,\lambda_3)/(\mu_1,\mu_2)}(x_3) = 1_{\lambda_3 \leq \mu_2 \leq \lambda_2 \leq \mu_1 \leq \lambda_1} \exp( x_3 (\lambda_1+\lambda_2+\lambda_3 - \mu_1 - \mu_2 ))$

and

$\displaystyle S_{(\lambda_1,\lambda_2,\lambda_3)/(\mu_1)}(x_2, x_3) = \int_{\lambda_2 \leq \lambda'_2 \leq \lambda_2, \mu_1} \int_{\mu_1, \lambda_2 \leq \lambda'_1 \leq \lambda_1}$

$\displaystyle \exp( x_2 (\lambda'_1+\lambda'_2 - \mu_1) + x_3 (\lambda_1+\lambda_2+\lambda_3 - \lambda'_1 - \lambda'_2))\ d\lambda'_1 d\lambda'_2.$

By expanding out the recursion, one obtains the analogue

$\displaystyle S_\lambda(x) = \int_{\lambda^1 \prec \dots \prec \lambda^k = \lambda} \exp( \sum_{j=1}^k x_j (|\lambda^j| - |\lambda^{j-1}|))\ d\lambda^1 \dots d\lambda^{k-1},$

of (6), and more generally one has

$\displaystyle S_{\lambda/\mu}(x) = \int_{\mu = \lambda^i \prec \dots \prec \lambda^k = \lambda} \exp( \sum_{j=i+1}^k x_j (|\lambda^j| - |\lambda^{j-1}|))\ d\lambda^{i+1} \dots d\lambda^{k-1}.$

We will call the tuples ${(\lambda^1,\dots,\lambda^k)}$ in the first integral real Gelfand-Tsetlin patterns based at ${\lambda}$. The analogue of (8) is then

$\displaystyle S_{\lambda/\nu}(x_{i+1},\dots,x_k) = \int S_{\mu/\nu}(x_{i+1},\dots,x_j) S_{\lambda/\mu}(x_{j+1},\dots,x_k)\ d\mu$

where the integral is over all real partitions ${\mu}$ of length ${j}$, with Lebesgue measure.

By approximating various integrals by their Riemann sums, one can relate the continuous Schur functions to their discrete counterparts by the limiting formula

$\displaystyle N^{-k(k-1)/2} s_{\lfloor N \lambda \rfloor}( \exp[ x/N ] ) \rightarrow S_\lambda(x) \ \ \ \ \ (11)$

as ${N \rightarrow \infty}$ for any length ${k}$ real partition ${\lambda = (\lambda_1,\dots,\lambda_k)}$ and any ${x = (x_1,\dots,x_k) \in {\bf R}^k}$, where

$\displaystyle \lfloor N \lambda \rfloor := ( \lfloor N \lambda_1 \rfloor, \dots, \lfloor N \lambda_k \rfloor )$

and

$\displaystyle \exp[x/N] := (\exp(x_1/N), \dots, \exp(x_k/N)).$

More generally, one has

$\displaystyle N^{j(j-1)/2-k(k-1)/2} s_{\lfloor N \lambda \rfloor / \lfloor N \mu \rfloor}( \exp[ x/N ] ) \rightarrow S_{\lambda/\mu}(x)$

as ${N \rightarrow \infty}$ for any length ${k}$ real partition ${\lambda}$, any length ${j}$ real partition ${\mu}$ with ${0 \leq j \leq k}$, and any ${x = (x_{j+1},\dots,x_k) \in {\bf R}^{k-j}}$.

As a consequence of these limiting formulae, one expects all of the discrete identities above to have continuous counterparts. This is indeed the case; below the fold we shall prove the discrete and continuous identities in parallel. These are not new results by any means, but I was not able to locate a good place in the literature where they are explicitly written down, so I thought I would try to do so here (primarily for my own internal reference, but perhaps the calculations will be worthwhile to some others also).

The determinant ${\det_n(A)}$ of an ${n \times n}$ matrix (with coefficients in an arbitrary field) obey many useful identities, starting of course with the fundamental multiplicativity ${\det_n(AB) = \det_n(A) \det_n(B)}$ for ${n \times n}$ matrices ${A,B}$. This multiplicativity can in turn be used to establish many further identities; in particular, as shown in this previous post, it implies the Schur determinant identity

$\displaystyle \det_{n+k}\begin{pmatrix} A & B \\ C & D \end{pmatrix} = \det_n(A) \det_k( D - C A^{-1} B ) \ \ \ \ \ (1)$

whenever ${A}$ is an invertible ${n \times n}$ matrix, ${B}$ is an ${n \times k}$ matrix, ${C}$ is a ${k \times n}$ matrix, and ${D}$ is a ${k \times k}$ matrix. The matrix ${D - CA^{-1} B}$ is known as the Schur complement of the block ${A}$.

I only recently discovered that this identity in turn immediately implies what I always found to be a somewhat curious identity, namely the Dodgson condensation identity (also known as the Desnanot-Jacobi identity)

$\displaystyle \det_n(M) \det_{n-2}(M^{1,n}_{1,n}) = \det_{n-1}( M^1_1 ) \det_{n-1}(M^n_n)$

$\displaystyle - \det_{n-1}(M^1_n) \det_{n-1}(M^n_1)$

for any ${n \geq 3}$ and ${n \times n}$ matrix ${M}$, where ${M^i_j}$ denotes the ${n-1 \times n-1}$ matrix formed from ${M}$ by removing the ${i^{th}}$ row and ${j^{th}}$ column, and similarly ${M^{i,i'}_{j,j'}}$ denotes the ${n-2 \times n-2}$ matrix formed from ${M}$ by removing the ${i^{th}}$ and ${(i')^{th}}$ rows and ${j^{th}}$ and ${(j')^{th}}$ columns. Thus for instance when ${n=3}$ we obtain

$\displaystyle \det_3 \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \cdot e$

$\displaystyle = \det_2 \begin{pmatrix} e & f \\ h & i \end{pmatrix} \cdot \det_2 \begin{pmatrix} a & b \\ d & e \end{pmatrix}$

$\displaystyle - \det_2 \begin{pmatrix} b & c \\ e & f \end{pmatrix} \cdot \det_2 \begin{pmatrix} d & e \\ g & h \end{pmatrix}$

for any scalars ${a,b,c,d,e,f,g,h,i}$. (Charles Dodgson, better known by his pen name Lewis Caroll, is of course also known for writing “Alice in Wonderland” and “Through the Looking Glass“.)

The derivation is not new; it is for instance noted explicitly in this paper of Brualdi and Schneider, though I do not know if this is the earliest place in the literature where it can be found. (EDIT: Apoorva Khare has pointed out to me that the original arguments of Dodgson can be interpreted as implicitly following this derivation.) I thought it is worth presenting the short derivation here, though.

Firstly, by swapping the first and ${(n-1)^{th}}$ rows, and similarly for the columns, it is easy to see that the Dodgson condensation identity is equivalent to the variant

$\displaystyle \det_n(M) \det_{n-2}(M^{n-1,n}_{n-1,n}) = \det_{n-1}( M^{n-1}_{n-1} ) \det_{n-1}(M^n_n) \ \ \ \ \ (2)$

$\displaystyle - \det_{n-1}(M^{n-1}_n) \det_{n-1}(M^n_{n-1}).$

Now write

$\displaystyle M = \begin{pmatrix} A & B_1 & B_2 \\ C_1 & d_{11} & d_{12} \\ C_2 & d_{21} & d_{22} \end{pmatrix}$

where ${A}$ is an ${n-2 \times n-2}$ matrix, ${B_1, B_2}$ are ${n-2 \times 1}$ column vectors, ${C_1, C_2}$ are ${1 \times n-2}$ row vectors, and ${d_{11}, d_{12}, d_{21}, d_{22}}$ are scalars. If ${A}$ is invertible, we may apply the Schur determinant identity repeatedly to conclude that

$\displaystyle \det_n(M) = \det_{n-2}(A) \det_2 \begin{pmatrix} d_{11} - C_1 A^{-1} B_1 & d_{12} - C_1 A^{-1} B_2 \\ d_{21} - C_2 A^{-1} B_1 & d_{22} - C_2 A^{-1} B_2 \end{pmatrix}$

$\displaystyle \det_{n-2} (M^{n-1,n}_{n-1,n}) = \det_{n-2}(A)$

$\displaystyle \det_{n-1}( M^{n-1}_{n-1} ) = \det_{n-2}(A) (d_{22} - C_2 A^{-1} B_2 )$

$\displaystyle \det_{n-1}( M^{n-1}_{n} ) = \det_{n-2}(A) (d_{21} - C_2 A^{-1} B_1 )$

$\displaystyle \det_{n-1}( M^{n}_{n-1} ) = \det_{n-2}(A) (d_{12} - C_1 A^{-1} B_2 )$

$\displaystyle \det_{n-1}( M^{n}_{n} ) = \det_{n-2}(A) (d_{11} - C_1 A^{-1} B_1 )$

and the claim (2) then follows by a brief calculation (and the explicit form ${\det_2 \begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad-bc}$ of the ${2 \times 2}$ determinant). To remove the requirement that ${A}$ be invertible, one can use a limiting argument, noting that one can work without loss of generality in an algebraically closed field, and in such a field, the set of invertible matrices is dense in the Zariski topology. (In the case when the scalars are reals or complexes, one can just use density in the ordinary topology instead if desired.)

The same argument gives the more general determinant identity of Sylvester

$\displaystyle \det_n(M) \det_{n-k}(M^S_S)^{k-1} = \det_k \left( \det_{n-k+1}(M^{S \backslash \{i\}}_{S \backslash \{j\}}) \right)_{i,j \in S}$

whenever ${n > k \geq 1}$, ${S}$ is a ${k}$-element subset of ${\{1,\dots,n\}}$, and ${M^S_{S'}}$ denotes the matrix formed from ${M}$ by removing the rows associated to ${S}$ and the columns associated to ${S'}$. (The Dodgson condensation identity is basically the ${k=2}$ case of this identity.)

A closely related proof of (2) proceeds by elementary row and column operations. Observe that if one adds some multiple of one of the first ${n-2}$ rows of ${M}$ to one of the last two rows of ${M}$, then the left and right sides of (2) do not change. If the minor ${A}$ is invertible, this allows one to reduce to the case where the components ${C_1,C_2}$ of the matrix vanish. Similarly, using elementary column operations instead of row operations we may assume that ${B_1,B_2}$ vanish. All matrices involved are now block-diagonal and the identity follows from a routine computation.

The latter approach can also prove the cute identity

$\displaystyle \det_2 \begin{pmatrix} \det_n( X_1, Y_1, A ) & \det_n( X_1, Y_2, A ) \\ \det_n(X_2, Y_1, A) & \det_n(X_2,Y_2, A) \end{pmatrix} = \det_n( X_1,X_2,A) \det_n(Y_1,Y_2,A)$

for any ${n \geq 2}$, any ${n \times 1}$ column vectors ${X_1,X_2,Y_1,Y_2}$, and any ${n \times n-2}$ matrix ${A}$, which can for instance be found in page 7 of this text of Karlin. Observe that both sides of this identity are unchanged if one adds some multiple of any column of ${A}$ to one of ${X_1,X_2,Y_1,Y_2}$; for generic ${A}$, this allows one to reduce ${X_1,X_2,Y_1,Y_2}$ to have only the first two entries allowed to be non-zero, at which point the determinants split into ${2 \times 2}$ and ${n -2 \times n-2}$ determinants and we can reduce to the ${n=2}$ case (eliminating the role of ${A}$). One can now either proceed by a direct computation, or by observing that the left-hand side is quartilinear in ${X_1,X_2,Y_1,Y_2}$ and antisymmetric in ${X_1,X_2}$ and ${Y_1,Y_2}$ which forces it to be a scalar multiple of ${\det_2(X_1,X_2) \det_2(Y_1,Y_2)}$, at which point one can test the identity at a single point (e.g. ${X_1=Y_1 = e_1}$ and ${X_2=Y_2=e_2}$ for the standard basis ${e_1,e_2}$) to conclude the argument. (One can also derive this identity from the Sylvester determinant identity but I think the calculations are a little messier if one goes by that route. Conversely, one can recover the Dodgson condensation identity from Karlin’s identity by setting ${X_1=e_1}$, ${X_2=e_2}$ (for instance) and then permuting some rows and columns.)

In July I will be spending a week at Park City, being one of the mini-course lecturers in the Graduate Summer School component of the Park City Summer Session on random matrices.  I have chosen to give some lectures on least singular values of random matrices, the circular law, and the Lindeberg exchange method in random matrix theory; this is a slightly different set of topics than I had initially advertised (which was instead about the Lindeberg exchange method and the local relaxation flow method), but after consulting with the other mini-course lecturers I felt that this would be a more complementary set of topics.  I have uploaded an draft of my lecture notes (some portion of which is derived from my monograph on the subject); as always, comments and corrections are welcome.

[Update, June 23: notes revised and reformatted to PCMI format. -T.]

[Update, Mar 19 2018: further revision. -T.]

Suppose ${F: X \rightarrow Y}$ is a continuous (but nonlinear) map from one normed vector space ${X}$ to another ${Y}$. The continuity means, roughly speaking, that if ${x_0, x \in X}$ are such that ${\|x-x_0\|_X}$ is small, then ${\|F(x)-F(x_0)\|_Y}$ is also small (though the precise notion of “smallness” may depend on ${x}$ or ${x_0}$, particularly if ${F}$ is not known to be uniformly continuous). If ${F}$ is known to be differentiable (in, say, the Fréchet sense), then we in fact have a linear bound of the form

$\displaystyle \|F(x)-F(x_0)\|_Y \leq C(x_0) \|x-x_0\|_X$

for some ${C(x_0)}$ depending on ${x_0}$, if ${\|x-x_0\|_X}$ is small enough; one can of course make ${C(x_0)}$ independent of ${x_0}$ (and drop the smallness condition) if ${F}$ is known instead to be Lipschitz continuous.

In many applications in analysis, one would like more explicit and quantitative bounds that estimate quantities like ${\|F(x)-F(x_0)\|_Y}$ in terms of quantities like ${\|x-x_0\|_X}$. There are a number of ways to do this. First of all, there is of course the trivial estimate arising from the triangle inequality:

$\displaystyle \|F(x)-F(x_0)\|_Y \leq \|F(x)\|_Y + \|F(x_0)\|_Y. \ \ \ \ \ (1)$

This estimate is usually not very good when ${x}$ and ${x_0}$ are close together. However, when ${x}$ and ${x_0}$ are far apart, this estimate can be more or less sharp. For instance, if the magnitude of ${F}$ varies so much from ${x_0}$ to ${x}$ that ${\|F(x)\|_Y}$ is more than (say) twice that of ${\|F(x_0)\|_Y}$, or vice versa, then (1) is sharp up to a multiplicative constant. Also, if ${F}$ is oscillatory in nature, and the distance between ${x}$ and ${x_0}$ exceeds the “wavelength” of the oscillation of ${F}$ at ${x_0}$ (or at ${x}$), then one also typically expects (1) to be close to sharp. Conversely, if ${F}$ does not vary much in magnitude from ${x_0}$ to ${x}$, and the distance between ${x}$ and ${x_0}$ is less than the wavelength of any oscillation present in ${F}$, one expects to be able to improve upon (1).

When ${F}$ is relatively simple in form, one can sometimes proceed simply by substituting ${x = x_0 + h}$. For instance, if ${F: R \rightarrow R}$ is the squaring function ${F(x) = x^2}$ in a commutative ring ${R}$, one has

$\displaystyle F(x_0+h) = (x_0+h)^2 = x_0^2 + 2x_0 h+ h^2$

and thus

$\displaystyle F(x_0+h) - F(x_0) = 2x_0 h + h^2$

or in terms of the original variables ${x, x_0}$ one has

$\displaystyle F(x) - F(x_0) = 2 x_0 (x-x_0) + (x-x_0)^2.$

If the ring ${R}$ is not commutative, one has to modify this to

$\displaystyle F(x) - F(x_0) = x_0 (x-x_0) + (x-x_0) x_0 + (x-x_0)^2.$

Thus, for instance, if ${A, B}$ are ${n \times n}$ matrices and ${\| \|_{op}}$ denotes the operator norm, one sees from the triangle inequality and the sub-multiplicativity ${\| AB\|_{op} \leq \| A \|_{op} \|B\|_{op}}$ of operator norm that

$\displaystyle \| A^2 - B^2 \|_{op} \leq \| A - B \|_{op} ( 2 \|B\|_{op} + \|A - B \|_{op} ). \ \ \ \ \ (2)$

If ${F(x)}$ involves ${x}$ (or various components of ${x}$) in several places, one can sometimes get a good estimate by “swapping” ${x}$ with ${x_0}$ at each of the places in turn, using a telescoping series. For instance, if we again use the squaring function ${F(x) = x^2 = x x}$ in a non-commutative ring, we have

$\displaystyle F(x) - F(x_0) = x x - x_0 x_0$

$\displaystyle = (x x - x_0 x) + (x_0 x - x_0 x_0)$

$\displaystyle = (x-x_0) x + x_0 (x-x_0)$

which for instance leads to a slight improvement of (2):

$\displaystyle \| A^2 - B^2 \|_{op} \leq \| A - B \|_{op} ( \| A\|_{op} + \|B\|_{op} ).$

More generally, for any natural number ${n}$, one has the identity

$\displaystyle x^n - x_0^n = (x-x_0) (x^{n-1} + x^{n-2} x_0 + \dots + x x_0^{n-2} + x_0^{n-1}) \ \ \ \ \ (3)$

in a commutative ring, while in a non-commutative ring one must modify this to

$\displaystyle x^n - x_0^n = \sum_{i=0}^{n-1} x_0^i (x-x_0) x^{n-1-i},$

and for matrices one has

$\displaystyle \| A^n - B^n \|_{op} \leq \| A-B\|_{op} ( \|A\|_{op}^{n-1} + \| A\|_{op}^{n-2} \| B\|_{op} + \dots + \|B\|_{op}^{n-1} ).$

Exercise 1 If ${U}$ and ${V}$ are unitary ${n \times n}$ matrices, show that the commutator ${[U,V] := U V U^{-1} V^{-1}}$ obeys the inequality

$\displaystyle \| [U,V] - I \|_{op} \leq 2 \| U - I \|_{op} \| V - I \|_{op}.$

(Hint: first control ${\| UV - VU \|_{op}}$.)

Now suppose (for simplicity) that ${F: {\bf R}^d \rightarrow {\bf R}^{d'}}$ is a map between Euclidean spaces. If ${F}$ is continuously differentiable, then one can use the fundamental theorem of calculus to write

$\displaystyle F(x) - F(x_0) = \int_0^1 \frac{d}{dt} F( \gamma(t) )\ dt$

where ${\gamma: [0,1] \rightarrow Y}$ is any continuously differentiable path from ${x_0}$ to ${x}$. For instance, if one uses the straight line path ${\gamma(t) := (1-t) x_0 + tx}$, one has

$\displaystyle F(x) - F(x_0) = \int_0^1 ((x-x_0) \cdot \nabla F)( (1-t) x_0 + t x )\ dt.$

In the one-dimensional case ${d=1}$, this simplifies to

$\displaystyle F(x) - F(x_0) = (x-x_0) \int_0^1 F'( (1-t) x_0 + t x )\ dt. \ \ \ \ \ (4)$

Among other things, this immediately implies the factor theorem for ${C^k}$ functions: if ${F}$ is a ${C^k({\bf R})}$ function for some ${k \geq 1}$ that vanishes at some point ${x_0}$, then ${F(x)}$ factors as the product of ${x-x_0}$ and some ${C^{k-1}}$ function ${G}$. Another basic consequence is that if ${\nabla F}$ is uniformly bounded in magnitude by some constant ${C}$, then ${F}$ is Lipschitz continuous with the same constant ${C}$.

Applying (4) to the power function ${x \mapsto x^n}$, we obtain the identity

$\displaystyle x^n - x_0^n = n (x-x_0) \int_0^1 ((1-t) x_0 + t x)^{n-1}\ dt \ \ \ \ \ (5)$

which can be compared with (3). Indeed, for ${x_0}$ and ${x}$ close to ${1}$, one can use logarithms and Taylor expansion to arrive at the approximation ${((1-t) x_0 + t x)^{n-1} \approx x_0^{(1-t) (n-1)} x^{t(n-1)}}$, so (3) behaves a little like a Riemann sum approximation to (5).

Exercise 2 For each ${i=1,\dots,n}$, let ${X^{(1)}_i}$ and ${X^{(0)}_i}$ be random variables taking values in a measurable space ${R_i}$, and let ${F: R_1 \times \dots \times R_n \rightarrow {\bf R}^m}$ be a bounded measurable function.

• (i) (Lindeberg exchange identity) Show that

$\displaystyle \mathop{\bf E} F(X^{(1)}_1,\dots,X^{(1)}_n) - \mathop{\bf E} F(X^{(0)}_1,\dots,X^{(0)}_n)$

$\displaystyle = \sum_{i=1}^n \mathop{\bf E} F( X^{(1)}_1,\dots, X^{(1)}_{i-1}, X^{(1)}_i, X^{(0)}_{i+1}, \dots, X^{(0)}_n)$

$\displaystyle - \mathop{\bf E} F( X^{(1)}_1,\dots, X^{(1)}_{i-1}, X^{(0)}_i, X^{(0)}_{i+1}, \dots, X^{(0)}_n).$

• (ii) (Knowles-Yin exchange identity) Show that

$\displaystyle \mathop{\bf E} F(X^{(1)}_1,\dots,X^{(1)}_n) - \mathop{\bf E} F(X^{(0)}_1,\dots,X^{(0)}_n)$

$\displaystyle = \int_0^1 \sum_{i=1}^n \mathop{\bf E} F( X^{(t)}_1,\dots, X^{(t)}_{i-1}, X^{(1)}_i, X^{(t)}_{i+1}, \dots, X^{(t)}_n)$

$\displaystyle - \mathop{\bf E} F( X^{(t)}_1,\dots, X^{(t)}_{i-1}, X^{(0)}_i, X^{(t)}_{i+1}, \dots, X^{(t)}_n)\ dt,$

where ${X^{(t)}_i = 1_{I_i \leq t} X^{(0)}_i + 1_{I_i > t} X^{(1)}_i}$ is a mixture of ${X^{(0)}_i}$ and ${X^{(1)}_i}$, with ${I_1,\dots,I_n}$ uniformly drawn from ${[0,1]}$ independently of each other and of the ${X^{(0)}_1,\dots,X^{(0)}_n, X^{(1)}_0,\dots,X^{(1)}_n}$.

• (iii) Discuss the relationship between the identities in parts (i), (ii) with the identities (3), (5).

(The identity in (i) is the starting point for the Lindeberg exchange method in probability theory, discussed for instance in this previous post. The identity in (ii) can also be used in the Lindeberg exchange method; the terms in the right-hand side are slightly more symmetric in the indices ${1,\dots,n}$, which can be a technical advantage in some applications; see this paper of Knowles and Yin for an instance of this.)

Exercise 3 If ${F: {\bf R}^d \rightarrow {\bf R}^{d'}}$ is continuously ${k}$ times differentiable, establish Taylor’s theorem with remainder

$\displaystyle F(x) = \sum_{j=0}^{k-1} \frac{1}{j!} (((x-x_0) \cdot \nabla)^j F)( x_0 )$

$\displaystyle + \int_0^1 \frac{(1-t)^{k-1}}{(k-1)!} (((x-x_0) \cdot \nabla)^k F)((1-t) x_0 + t x)\ dt.$

If ${\nabla^k F}$ is bounded, conclude that

$\displaystyle |F(x) - \sum_{j=0}^{k-1} \frac{1}{j!} (((x-x_0) \cdot \nabla)^j F)( x_0 )|$

$\displaystyle \leq \frac{|x-x_0|^k}{k!} \sup_{y \in {\bf R}^d} |\nabla^k F(y)|.$

For real scalar functions ${F: {\bf R}^d \rightarrow {\bf R}}$, the average value of the continuous real-valued function ${(x - x_0) \cdot \nabla F((1-t) x_0 + t x)}$ must be attained at some point ${t}$ in the interval ${[0,1]}$. We thus conclude the mean-value theorem

$\displaystyle F(x) - F(x_0) = ((x - x_0) \cdot \nabla F)((1-t) x_0 + t x)$

for some ${t \in [0,1]}$ (that can depend on ${x}$, ${x_0}$, and ${F}$). This can for instance give a second proof of fact that continuously differentiable functions ${F}$ with bounded derivative are Lipschitz continuous. However it is worth stressing that the mean-value theorem is only available for real scalar functions; it is false for instance for complex scalar functions. A basic counterexample is given by the function ${e(x) := e^{2\pi i x}}$; there is no ${t \in [0,1]}$ for which ${e(1) - e(0) = e'(t)}$. On the other hand, as ${e'}$ has magnitude ${2\pi}$, we still know from (4) that ${e}$ is Lipschitz of constant ${2\pi}$, and when combined with (1) we obtain the basic bounds

$\displaystyle |e(x) - e(y)| \leq \min( 2, 2\pi |x-y| )$

which are already very useful for many applications.

Exercise 4 Let ${H_0, V}$ be ${n \times n}$ matrices, and let ${t}$ be a non-negative real.

• (i) Establish the Duhamel formula

$\displaystyle e^{t(H_0+V)} = e^{tH_0} + \int_0^t e^{(t-s) H_0} V e^{s (H_0+V)}\ ds$

$\displaystyle = e^{tH_0} + \int_0^t e^{(t-s) (H_0+V)} V e^{s H_0}\ ds$

where ${e^A}$ denotes the matrix exponential of ${A}$. (Hint: Differentiate ${e^{(t-s) H_0} e^{s (H_0+V)}}$ or ${e^{(t-s) (H_0+V)} e^{s H_0}}$ in ${s}$.)

• (ii) Establish the iterated Duhamel formula

$\displaystyle e^{t(H_0+V)} = e^{tH_0} + \sum_{j=1}^k \int_{0 \leq t_1 \leq \dots \leq t_j \leq t}$

$\displaystyle e^{(t-t_j) H_0} V e^{(t_j-t_{j-1}) H_0} V \dots e^{(t_2-t_1) H_0} V e^{t_1 H_0}\ dt_1 \dots dt_j$

$\displaystyle + \int_{0 \leq t_1 \leq \dots \leq t_{k+1} \leq t}$

$\displaystyle e^{(t-t_{k+1}) (H_0+V)} V e^{(t_{k+1}-t_k) H_0} V \dots e^{(t_2-t_1) H_0} V e^{t_1 H_0}\ dt_1 \dots dt_{k+1}$

for any ${k \geq 0}$.

• (iii) Establish the infinitely iterated Duhamel formula

$\displaystyle e^{t(H_0+V)} = e^{tH_0} + \sum_{j=1}^\infty \int_{0 \leq t_1 \leq \dots \leq t_j \leq t}$

$\displaystyle e^{(t-t_j) H_0} V e^{(t_j-t_{j-1}) H_0} V \dots e^{(t_2-t_1) H_0} V e^{t_1 H_0}\ dt_1 \dots dt_j.$

• (iv) If ${H(t)}$ is an ${n \times n}$ matrix depending in a continuously differentiable fashion on ${t}$, establish the variation formula

$\displaystyle \frac{d}{dt} e^{H(t)} = (F(\mathrm{ad}(H(t))) H'(t)) e^{H(t)}$

where ${\mathrm{ad}(H)}$ is the adjoint representation ${\mathrm{ad}(H)(V) = HV - VH}$ applied to ${H}$, and ${F}$ is the function

$\displaystyle F(z) := \int_0^1 e^{sz}\ ds$

(thus ${F(z) = \frac{e^z-1}{z}}$ for non-zero ${z}$), with ${F(\mathrm{ad}(H(t)))}$ defined using functional calculus.

We remark that further manipulation of (iv) of the above exercise using the fundamental theorem of calculus eventually leads to the Baker-Campbell-Hausdorff-Dynkin formula, as discussed in this previous blog post.

Exercise 5 Let ${A, B}$ be positive definite ${n \times n}$ matrices, and let ${Y}$ be an ${n \times n}$ matrix. Show that there is a unique solution ${X}$ to the Sylvester equation

$\displaystyle AX + X B = Y$

which is given by the formula

$\displaystyle X = \int_0^\infty e^{-tA} Y e^{-tB}\ dt.$

In the above examples we had applied the fundamental theorem of calculus along linear curves ${\gamma(t) = (1-t) x_0 + t x}$. However, it is sometimes better to use other curves. For instance, the circular arc ${\gamma(t) = \cos(\pi t/2) x_0 + \sin(\pi t/2) x}$ can be useful, particularly if ${x_0}$ and ${x}$ are “orthogonal” or “independent” in some sense; a good example of this is the proof by Maurey and Pisier of the gaussian concentration inequality, given in Theorem 8 of this previous blog post. In a similar vein, if one wishes to compare a scalar random variable ${X}$ of mean zero and variance one with a Gaussian random variable ${G}$ of mean zero and variance one, it can be useful to introduce the intermediate random variables ${\gamma(t) := (1-t)^{1/2} X + t^{1/2} G}$ (where ${X}$ and ${G}$ are independent); note that these variables have mean zero and variance one, and after coupling them together appropriately they evolve by the Ornstein-Uhlenbeck process, which has many useful properties. For instance, one can use these ideas to establish monotonicity formulae for entropy; see e.g. this paper of Courtade for an example of this and further references. More generally, one can exploit curves ${\gamma}$ that flow according to some geometrically natural ODE or PDE; several examples of this occur famously in Perelman’s proof of the Poincaré conjecture via Ricci flow, discussed for instance in this previous set of lecture notes.

In some cases, it is difficult to compute ${F(x)-F(x_0)}$ or the derivative ${\nabla F}$ directly, but one can instead proceed by implicit differentiation, or some variant thereof. Consider for instance the matrix inversion map ${F(A) := A^{-1}}$ (defined on the open dense subset of ${n \times n}$ matrices consisting of invertible matrices). If one wants to compute ${F(B)-F(A)}$ for ${B}$ close to ${A}$, one can write temporarily write ${F(B) - F(A) = E}$, thus

$\displaystyle B^{-1} - A^{-1} = E.$

Multiplying both sides on the left by ${B}$ to eliminate the ${B^{-1}}$ term, and on the right by ${A}$ to eliminate the ${A^{-1}}$ term, one obtains

$\displaystyle A - B = B E A$

and thus on reversing these steps we arrive at the basic identity

$\displaystyle B^{-1} - A^{-1} = B^{-1} (A - B) A^{-1}. \ \ \ \ \ (6)$

For instance, if ${H_0, V}$ are ${n \times n}$ matrices, and we consider the resolvents

$\displaystyle R_0(z) := (H_0 - z I)^{-1}; \quad R_V(z) := (H_0 + V - zI)^{-1}$

then we have the resolvent identity

$\displaystyle R_V(z) - R_0(z) = - R_V(z) V R_0(z) \ \ \ \ \ (7)$

as long as ${z}$ does not lie in the spectrum of ${H_0}$ or ${H_0+V}$ (for instance, if ${H_0}$, ${V}$ are self-adjoint then one can take ${z}$ to be any strictly complex number). One can iterate this identity to obtain

$\displaystyle R_V(z) = \sum_{j=0}^k (-R_0(z) V)^j R_0(z) + (-R_V(z) V) (-R_0(z) V)^k R_0(z)$

for any natural number ${k}$; in particular, if ${R_0(z) V}$ has operator norm less than one, one has the Neumann series

$\displaystyle R_V(z) = \sum_{j=0}^\infty (-R_0(z) V)^j R_0(z).$

Similarly, if ${A(t)}$ is a family of invertible matrices that depends in a continuously differentiable fashion on a time variable ${t}$, then by implicitly differentiating the identity

$\displaystyle A(t) A(t)^{-1} = I$

in ${t}$ using the product rule, we obtain

$\displaystyle (\frac{d}{dt} A(t)) A(t)^{-1} + A(t) \frac{d}{dt} A(t)^{-1} = 0$

and hence

$\displaystyle \frac{d}{dt} A(t)^{-1} = - A(t)^{-1} (\frac{d}{dt} A(t)) A(t)^{-1}$

(this identity may also be easily derived from (6)). One can then use the fundamental theorem of calculus to obtain variants of (6), for instance by using the curve ${\gamma(t) = (1-t) A + tB}$ we arrive at

$\displaystyle B^{-1} - A^{-1} = \int_0^1 ((1-t)A + tB)^{-1} (A-B) ((1-t)A + tB)^{-1}\ dt$

assuming that the curve stays entirely within the set of invertible matrices. While this identity may seem more complicated than (6), it is more symmetric, which conveys some advantages. For instance, using this identity it is easy to see that if ${A, B}$ are positive definite with ${A>B}$ in the sense of positive definite matrices (that is, ${A-B}$ is positive definite), then ${B^{-1} > A^{-1}}$. (Try to prove this using (6) instead!)

Exercise 6 If ${A}$ is an invertible ${n \times n}$ matrix and ${u, v}$ are ${n \times 1}$ vectors, establish the Sherman-Morrison formula

$\displaystyle (A + t uv^T)^{-1} = A^{-1} - \frac{t}{1 + t v^T A^{-1} u} A^{-1} uv^T A^{-1}$

whenever ${t}$ is a scalar such that ${1 + t v^T A^{-1} u}$ is non-zero. (See also this previous blog post for more discussion of these sorts of identities.)

One can use the Cauchy integral formula to extend these identities to other functions of matrices. For instance, if ${F: {\bf C} \rightarrow {\bf C}}$ is an entire function, and ${\gamma}$ is a counterclockwise contour that goes around the spectrum of both ${H_0}$ and ${H_0+V}$, then we have

$\displaystyle F(H_0+V) = \frac{-1}{2\pi i} \int_\gamma F(z) R_V(z)\ dz$

and similarly

$\displaystyle F(H_0) = \frac{-1}{2\pi i} \int_\gamma F(z) R_0(z)\ dz$

and hence by (7) one has

$\displaystyle F(H_0+V) - F(H_0) = \frac{1}{2\pi i} \int_\gamma F(z) R_V(z) V F_0(z)\ dz;$

similarly, if ${H(t)}$ depends on ${t}$ in a continuously differentiable fashion, then

$\displaystyle \frac{d}{dt} F(H(t)) = \frac{1}{2\pi i} \int_\gamma F(z) (H(t) - zI)^{-1} H'(t) (z) (H(t)-zI)^{-1}\ dz$

as long as ${\gamma}$ goes around the spectrum of ${H(t)}$.

Exercise 7 If ${H(t)}$ is an ${n \times n}$ matrix depending continuously differentiably on ${t}$, and ${F: {\bf C} \rightarrow {\bf C}}$ is an entire function, establish the tracial chain rule

$\displaystyle \frac{d}{dt} \hbox{tr} F(H(t)) = \hbox{tr}(F'(H(t)) H'(t)).$

In a similar vein, given that the logarithm function is the antiderivative of the reciprocal, one can express the matrix logarithm ${\log A}$ of a positive definite matrix by the fundamental theorem of calculus identity

$\displaystyle \log A = \int_0^\infty (I + sI)^{-1} - (A + sI)^{-1}\ ds$

(with the constant term ${(I+tI)^{-1}}$ needed to prevent a logarithmic divergence in the integral). Differentiating, we see that if ${A(t)}$ is a family of positive definite matrices depending continuously on ${t}$, that

$\displaystyle \frac{d}{dt} \log A(t) = \int_0^\infty (A(t) + sI)^{-1} A'(t) (A(t)+sI)^{-1}\ dt.$

This can be used for instance to show that ${\log}$ is a monotone increasing function, in the sense that ${\log A> \log B}$ whenever ${A > B > 0}$ in the sense of positive definite matrices. One can of course integrate this formula to obtain some formulae for the difference ${\log A - \log B}$ of the logarithm of two positive definite matrices ${A,B}$.

To compare the square root ${A^{1/2} - B^{1/2}}$ of two positive definite matrices ${A,B}$ is trickier; there are multiple ways to proceed. One approach is to use contour integration as before (but one has to take some care to avoid branch cuts of the square root). Another to express the square root in terms of exponentials via the formula

$\displaystyle A^{1/2} = \frac{1}{\Gamma(-1/2)} \int_0^\infty (e^{-tA} - I) t^{-1/2} \frac{dt}{t}$

where ${\Gamma}$ is the gamma function; this formula can be verified by first diagonalising ${A}$ to reduce to the scalar case and using the definition of the Gamma function. Then one has

$\displaystyle A^{1/2} - B^{1/2} = \frac{1}{\Gamma(-1/2)} \int_0^\infty (e^{-tA} - e^{-tB}) t^{-1/2} \frac{dt}{t}$

and one can use some of the previous identities to control ${e^{-tA} - e^{-tB}}$. This is pretty messy though. A third way to proceed is via implicit differentiation. If for instance ${A(t)}$ is a family of positive definite matrices depending continuously differentiably on ${t}$, we can differentiate the identity

$\displaystyle A(t)^{1/2} A(t)^{1/2} = A(t)$

to obtain

$\displaystyle A(t)^{1/2} \frac{d}{dt} A(t)^{1/2} + (\frac{d}{dt} A(t)^{1/2}) A(t)^{1/2} = \frac{d}{dt} A(t).$

This can for instance be solved using Exercise 5 to obtain

$\displaystyle \frac{d}{dt} A(t)^{1/2} = \int_0^\infty e^{-sA(t)^{1/2}} A'(t) e^{-sA(t)^{1/2}}\ ds$

and this can in turn be integrated to obtain a formula for ${A^{1/2} - B^{1/2}}$. This is again a rather messy formula, but it does at least demonstrate that the square root is a monotone increasing function on positive definite matrices: ${A > B > 0}$ implies ${A^{1/2} > B^{1/2} > 0}$.

Several of the above identities for matrices can be (carefully) extended to operators on Hilbert spaces provided that they are sufficiently well behaved (in particular, if they have a good functional calculus, and if various spectral hypotheses are obeyed). We will not attempt to do so here, however.