Lecture 19. The Kadison-Singer problem

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1 Stanford University Spring 208 Math 233A: Non-constructive methods in combinatorics Instructor: Jan Vondrák Lecture date: March 3, 208 Scribe: Ben Plaut Lecture 9. The Kadison-Singer problem The Kadison-Singer problem originated in quantum mechanics in the 950s. It has been formulate in several equivalent forms. Here we state it in a form known as Weaver s conecture. Conecture There exist ɛ, δ > 0 such that given any vectors w, w 2...w m C n where w i 2 δ for all i m w iw T i = I there exists a partition of [m] into S and S 2 such that for all {, 2}, i S w i w i ɛ This holds independently of m, n N. Here i S w i wi is the operator norm, for a Hermitian matrix simply the maximum eigenvalue, so our goal will be to prove an upper bound on the maximum eigenvalue. Note that since w i wi are rank matrices that add up to the identity, we must have m n. The setup here is similar to that of Ramanuan graphs. For Ramanuan graphs, we wanted all of the eigenvalues of the adacency matrix to be in the set {d} { d} [ 2 d, 2 d ]. Here we are again looking to bound the size of the eigenvalues, and we will use a similar approach. 9. Setup for the proof of Conecture The specific statement we will prove is the following, which is actually stronger than Conecture. Theorem 9. Marcus, Spielman, Srivastava 3 For all α > 0 and vectors w, w 2...w m C n where w i 2 2 α for all i m w iw T i = I there exists a partition of [m] into S and S 2 such that for all {, 2}, i S w i w i 2 + 2α 2 This holds independently of m, n N.

2 Our strategy is as follows. We define random variables r...r m, each in C 2n, where either r = w 0, 0... meaning that the first n entries are the n entries of w, and the second n entries are all zeros, or r = 0, 0... w meaning that the first n entries are all zeros and the second n entries are the n entries of w. We will use r to refer to w 0, 0... and r 2 to refer to 0, 0... w. Each r is r with probability /2 and is r 2 with probability /2. Thus we have = r r = S w w S w w where the matrices on the right hand side are 2n 2n matrix where one n n quadrant is occupied by w w, and the rest of the entries are zeros. Thus r r = max w i wi, i S w i wi i S2 = and this is the quantity that we should upper-bound. The main technical theorem is the following: Theorem 9.2 Marcus, Spielman and Srivastava For independently random vectors r,..., r m in C n, if m E[r ir i ] = I and E [ r i 2 2] α, then with probability greater than 0, r r + α 2 = From here, Theorem 9. follows by setting r = 2 w 0 or 2 0 w with probability 2 2. Let us denote the first option by w and the second option by w 2. In the following, we prove Theorem 9.2. To bound the eigenvalues of m = r r, we consider the expected characteristic polynomial χ: χx = [ E detxi r...r m = ] r r As before, we fix the values of the first k vectors r,..., r k to be w σ,..., w σ k k, for some values of σ,..., σ k {, 2}. The remaining m k + vectors remain random. This yields the following polynomial: [ k ] χ σ,...,σ k x = E detxi w σ r k+,...,r w σ r r m = =k+ 2

3 We saw when studying Ramanuan graphs that this is an interlacing family of stable polynomials. We know from before that if the maximum root of χ is at most λ max, then there exists a choice of σ,..., σ m such that the maximum root of χ σ,...,σ m is also at most λ max. In the setting of Ramanuan graphs, the characteristic polynomial was the matching polynomial, so we already knew its maximum root. Here, we do not know the maximum root of χ if we did, we would be done, so we will have to do some more work to bound the maximum root. 9.2 Bounding the maximum root of χ Recall how we prove that χ is in fact stable. We will do this by showing that χ can be obtained applying a set of differential operations to a polynomial we already known to be stable: χx =... det xi + t E[r i ri ] t t 2 t m The matrix E[r i ri ] is positive semi-definite, so the polynomial det xi + m = t E[r i ri ] We show next that these differential operators preserve stability. = is stable. Lemma 9.3 Let z refer to the vector of variables z...z m. If a function fz, z 2...z m = fz is stable, then so is fz = fz f z. Proof: Recall that H = {z C : Imz > 0}. Suppose for sake of contradiction that fz f z were not stable: then there exists a root of fz f z in H m : that is, there exists α...α m H m such that fα...α m f α...α m = 0. Then fα...α m = f α...α m. Let gz be the univariate polynomial fz, α 2...α m. Then gα = fα...α m and g α = f α...α m, so we have gα = g α. If gz had a root α H, then α...α m H m would be a root of fz, z 2...z m, which contradicts the stability of fz, z 2...z m. Thus gz is stable. Thus we can write n gz = C z λ i where λ,..., λ n are the roots of gz. Therefore g z n gz = z λ i Examining the ratio of a function s derivative divided by the function will be a useful tool at multiple points in this lecture. Since gz is stable, we know that λ i H for all i. Thus for all z H, and all i, z λ i H, so z λ i H, where H is the lower half-plane. Therefore g z gz H, so g z gz But we know that g α = gα, so g α for all z. f gα =, which is a contradiction. Therefore fz z must be stable. Applying Lemma 9.3 repeatedly implies that χ is in fact stable. When t i = 0 for all i, we have det xi + m = t E[r i ri ] = detxi = x n, which has all of its roots at x = 0. If we could show that applying the t i operators doesn t increase the roots by much, that would give us an upper bound on the roots of χ, and we would be done. 3

4 9.3 A toy example Next, we examine a toy example: that of the univariate polynomial px = x n. This polynomial initially has all of its roots at x = 0. Applying d once gives us d px = x n nx n = x n x n which has one root at x = n. This is concerning, since applying the operator d ust once caused the maximum root to ump from 0 to n. The rest of the roots remain at x = 0. Applying the d operator again gives us d 2px = x n nx n nx n + nn x n 2 = x n 2 x 2 2nx + nn To compute the roots of this polynomial, we need to know the roots of x 2 2nx + nn. This can be done using the quadratic formula, and the resulting roots are n + n, and one at n n in addition to the n 2 roots at x = 0 from x n 2. This time, the maximum root increased only from n to n + n, which is a smaller ump than from 0 to n. Taking a wild guess, maybe the multiplicity of the roots influences the effect of the d on the maximum root? In the following, we design a way of relaxing the notion of maximum root in a way that takes multiplicity and proximity of other roots into account. 9.4 The barrier function Given a univariate polynomial px, we define the barrier function φ p x for x R by Thus φ p x = d log px φ p x = p x px = n z λ i where λ...λ n are the roots of p. Note the re-appearance of the ratio of a function s derivative divided by the function, or equivalently the derivative of the logarithm. For each α 0,, we define a set α max p by α max p = { x > maximum root of p : p x px < α } The intuition for Lemma 9.4 is that α max p doesn t change too much when we apply d. Lemma 9.4 If x α max p, then for δ = α, we have x + δ α maxp p. 4

5 Proof: We have φ p p x = d logp p x = d log px φ p x = d log px + log φ p x = φ p x φ p x φ p x Using our above computation of φ p p x, we have φ p p x + δ = φ p x + δ φ p x + δ φ p x + δ When x is larger than the maximum root, φ p x is decreasing and convex. Also, since x α max p, we have φ p x = p x px < α by assumption. Therefore φ p x + δ φ p x + δ φ p x + δ φ px + δ φ p x + δ φ p x 9. φ p x + δ φ p x + δ α 9.2 = φ p x + δ δφ p x + δ 9.3 φ p x 9.4 < α uses the fact that φ p x is decreasing when x is larger than the maximum root. 9. to 9.2 is because φ p x < α. 9.2 to 9.3 is by definition of δ. 9.3 to 9.4 is by convexity, and 9.4 to 9.5 is again because φ p x < α. Putting everything together gives us φ p p x + δ < α, so x + δ α max p p, as required. Recall our toy example of px = x n. Here φ p x = n/x, so n/α α max p. So by applying Lemma 9.4 m times, we know that the polynomial d m p has maximum root less than n α + m α. This holds for any α. If we choose α to be + m n, this implies that the maximum root of d is at most n + m 2. When m is smaller than n, this is at most 4n. 9.5 Back to the multivariate setting Recall that χx =... det xi + t t 2 t m Let A i = E[r i r i ]. The matrix A i is positive semi-definite. = t E[r i ri ] Theorem 9.5 If m A i = I, A i is positive semi-definite for all i, and T ra i α for all i, then the maximum root of χ when evaluated at t = 0 is at most + α 2. 5

6 First, we will do a variable substitution by letting z i = x + t i. Then det xi + m m t i A i = det x + t i A i = det z i A i Since z i is linear in t i, the operator z i is equivalent to t i. Therefore χx =... m det z i A i z m We will evaluate this at z i = x, which corresponds to t = 0. Let Qz...z m be defined by Also, for each k [m], let Qz...z m =... m det z i A zi i z m =x Q k z...z m =... m det z i A zi i z m k+ z m =x Thus Q k z...z m contains k differentiation operations. We will also use a multivariate version of the barrier function. We define the barrier function in direction φ pz by φ pz = log pz = z pz p z We say that w R m is above the roots of p real stable if t R m where t 0, we have pw + t > 0. Fixing z i R for each i, pz...z...z m has roots λ...λ n R. We can write To be continued... φ pz = n l= z λ l 6