One of the basic tools in modern combinatorics is the probabilistic method, introduced by Erdos, in which a deterministic solution to a given problem is shown to exist by constructing a *random* candidate for a solution, and showing that this candidate solves all the requirements of the problem with positive probability. When the problem requires a real-valued statistic to be suitably large or suitably small, the following trivial observation is often employed:

Proposition 1 (Comparison with mean)Let be a random real-valued variable, whose mean (orfirst moment) is finite. Thenwith positive probability, and

with positive probability.

This proposition is usually applied in conjunction with a computation of the first moment , in which case this version of the probabilistic method becomes an instance of the *first moment method*. (For comparison with other moment methods, such as the second moment method, exponential moment method, and zeroth moment method, see Chapter 1 of my book with Van Vu. For a general discussion of the probabilistic method, see the book by Alon and Spencer of the same name.)

As a typical example in random matrix theory, if one wanted to understand how small or how large the operator norm of a random matrix could be, one might first try to compute the expected operator norm and then apply Proposition 1; see this previous blog post for examples of this strategy (and related strategies, based on comparing with more tractable expressions such as the moments ). (In this blog post, all matrices are complex-valued.)

Recently, in their proof of the Kadison-Singer conjecture (and also in their earlier paper on Ramanujan graphs), Marcus, Spielman, and Srivastava introduced an striking new variant of the first moment method, suited in particular for controlling the operator norm of a Hermitian positive semi-definite matrix . Such matrices have non-negative real eigenvalues, and so in this case is just the largest eigenvalue of . Traditionally, one tries to control the eigenvalues through averaged statistics such as moments or Stieltjes transforms ; again, see this previous blog post. Here we use as short-hand for , where is the identity matrix. Marcus, Spielman, and Srivastava instead rely on the interpretation of the eigenvalues of as the roots of the characteristic polynomial of , thus

where is the largest real root of a non-zero polynomial . (In our applications, we will only ever apply to polynomials that have at least one real root, but for sake of completeness let us set if has no real roots.)

Prior to the work of Marcus, Spielman, and Srivastava, I think it is safe to say that the conventional wisdom in random matrix theory was that the representation (1) of the operator norm was not particularly useful, due to the highly non-linear nature of both the characteristic polynomial map and the maximum root map . (Although, as pointed out to me by Adam Marcus, some related ideas have occurred in graph theory rather than random matrix theory, for instance in the theory of the matching polynomial of a graph.) For instance, a fact as basic as the triangle inequality is extremely difficult to establish through (1). Nevertheless, it turns out that for certain special types of random matrices (particularly those in which a typical instance of this ensemble has a simple relationship to “adjacent” matrices in this ensemble), the polynomials enjoy an extremely rich structure (in particular, they lie in families of real stable polynomials, and hence enjoy good combinatorial interlacing properties) that can be surprisingly useful. In particular, Marcus, Spielman, and Srivastava established the following nonlinear variant of Proposition 1:

Proposition 2 (Comparison with mean)Let . Let be a random matrix, which is the sum of independent Hermitian rank one matrices , each taking a finite number of values. Thenwith positive probability, and

with positive probability.

We prove this proposition below the fold. The hypothesis that each only takes finitely many values is technical and can likely be relaxed substantially, but we will not need to do so here. Despite the superficial similarity with Proposition 1, the proof of Proposition 2 is quite nonlinear; in particular, one needs the interlacing properties of real stable polynomials to proceed. Another key ingredient in the proof is the observation that while the determinant of a matrix generally behaves in a nonlinar fashion on the underlying matrix , it becomes (affine-)linear when one considers rank one perturbations, and so depends in an affine-multilinear fashion on the . More precisely, we have the following deterministic formula, also proven below the fold:

Proposition 3 (Deterministic multilinearisation formula)Let be the sum of deterministic rank one matrices . Then we havefor all , where the

mixed characteristic polynomialof any matrices (not necessarily rank one) is given by the formula

Among other things, this formula gives a useful representation of the mean characteristic polynomial :

Corollary 4 (Random multilinearisation formula)Let be the sum of jointly independent rank one matrices . Then we have

*Proof:* For fixed , the expression is a polynomial combination of the , while the differential operator is a linear combination of differential operators for . As a consequence, we may expand (3) as a linear combination of terms, each of which is a multilinear combination of for some . Taking expectations of both sides of (2) and using the joint independence of the , we obtain the claim.

In view of Proposition 2, we can now hope to control the operator norm of certain special types of random matrices (and specifically, the sum of independent Hermitian positive semi-definite rank one matrices) by first controlling the mean of the random characteristic polynomial . Pursuing this philosophy, Marcus, Spielman, and Srivastava establish the following result, which they then use to prove the Kadison-Singer conjecture:

Theorem 5 (Marcus-Spielman-Srivastava theorem)Let . Let be jointly independent random vectors in , with each taking a finite number of values. Suppose that we have the normalisationwhere we are using the convention that is the identity matrix whenever necessary. Suppose also that we have the smallness condition

for some and all . Then one has

Note that the upper bound in (5) must be at least (by taking to be deterministic) and also must be at least (by taking the to always have magnitude at least ). Thus the bound in (5) is asymptotically tight both in the regime and in the regime ; the latter regime will be particularly useful for applications to Kadison-Singer. It should also be noted that if one uses more traditional random matrix theory methods (based on tools such as Proposition 1, as well as more sophisticated variants of these tools, such as the concentration of measure results of Rudelson and Ahlswede-Winter), one obtains a bound of with high probability, which is insufficient for the application to the Kadison-Singer problem; see this article of Tropp. Thus, Theorem 5 obtains a sharper bound, at the cost of trading in “high probability” for “positive probability”.

In the paper of Marcus, Spielman and Srivastava, Theorem 5 is used to deduce a conjecture of Weaver, which was already known to imply the Kadison-Singer conjecture; actually, a slight modification of their argument gives the paving conjecture of Kadison and Singer, from which the original Kadison-Singer conjecture may be readily deduced. We give these implications below the fold. (See also this survey article for some background on the Kadison-Singer problem.)

Let us now summarise how Theorem 5 is proven. In the spirit of semi-definite programming, we rephrase the above theorem in terms of the rank one Hermitian positive semi-definite matrices :

Theorem 6 (Marcus-Spielman-Srivastava theorem again)Let be jointly independent random rank one Hermitian positive semi-definite matrices such that the sum has meanand such that

for some and all . Then one has

with positive probability.

In view of (1) and Proposition 2, this theorem follows from the following control on the mean characteristic polynomial:

Theorem 7 (Control of mean characteristic polynomial)Let be jointly independent random rank one Hermitian positive semi-definite matrices such that the sum has meanand such that

for some and all . Then one has

This result is proven using the multilinearisation formula (Corollary 4) and some convexity properties of real stable polynomials; we give the proof below the fold.

Thanks to Adam Marcus, Assaf Naor and Sorin Popa for many useful explanations on various aspects of the Kadison-Singer problem.

** — 1. Multilinearisation formula — **

In this section we prove Proposition 3. The key observation is a standard one:

Lemma 8 (Determinant is linear with respect to rank one perturbations)Let be matrices, with rank one. Then the polynomial is affine-linear (i.e. it is of the form for some coefficients ).

*Proof:* When is the identity matrix, this follows from the Sylvester determinant identity , which shows in particular that for any rank one matrix . As the determinant is multiplicative, the claim then follows when is any invertible matrix; as the invertible matrices are dense in the space of all matrices, we conclude the general case of the lemma. (One can also prove this lemma via cofactor expansion, after changing variables to make or a basis vector.)

Iterating this, we obtain

Corollary 9 (Determinant is multilinear with respect to rank one perturbations)Let be matrices, with rank one. Then the polynomial is affine-multilinear, that is to say it is of the formfor some coefficients .

*Proof:* By Lemma 8, the polynomial is affine-linear in each of the variables separately (if one freezes the other variables ). The claim follows.

Now observe that if a one-variable polynomial is affine-linear, then it can be recovered from its order Taylor expansion at the origin:

More generally, if a polynomial is affine-multilinear in the sense of Corollary 9, then it can be recovered from an “order Taylor expansion” at the origin by the formula

Applying this to the polynomial in Corollary 9, we conclude that

Setting and , we obtain Proposition 3 as required.

** — 2. Real stable polynomials and interlacing — **

Now we prove Proposition 2. The key observation is that almost all of the polynomials one needs to work with here are of a special class, namely the *real stable polynomials*.

Definition 10A polynomial is stable if it has no zeroes in the region (in particular, is non-zero). A polynomial isreal stableif it is stable and all of its coefficients are real.

Observe from conjugation symmetry that a real stable polynomial has no zeroes in either of the two regions and . In particular, a one-variable real polynomial is stable if and only if it only has real zeroes. The notion of stability is however more complicated in higher dimensions. See this survey of Wagner for a more thorough discussion of the property of real stability.

To me, the property of being real stable is somewhat similar to properties such as log-convexity or total positivity; these are properties that are only obeyed by a minority of objects being studied, but in many key cases, these properties are available and can be very profitable to exploit. In practice, real stable polynomials get to enjoy the combined benefit of real analysis methods (e.g. the intermediate value theorem) with complex analysis methods (e.g. Rouche’s theorem), because of the guarantee that zeroes will be real.

A basic example of a real stable polynomial is the characteristic polynomial of a Hermitian matrix; this is a rephrasing of the fundamental fact that the eigenvalues of a Hermitian matrix are all real. There is a multidimensional version of this example, which will be the primary source of real stability for us:

Lemma 11If are positive semi-definite Hermitian matrices, then the polynomial is real stable.

*Proof:* It is clear that this polynomial is real (note that it takes real values for real inputs). To show that it is stable, observe that if have positive imaginary part, then the skew-adjoint part of is strictly positive definite, and so is non-singular (since the quadratic form is non-degenerate).

Stable polynomials are also closed under restriction and under differential operators such as .

Lemma 12Let be a real stable polynomial of variables.

- (Restriction) If and is real, then the polynomial is a real stable polynomial of variables, unless it is identically zero.
- (Differentiation) For any real , the polynomial is a real stable polynomial of variables.
Similarly for permutations of the variables.

*Proof:* The final claim is clear, as the property of real stability is unchanged under permutation of the variables.

We first prove the restriction claim. The polynomial is clearly real. If we have a zero with having positive imaginary part, but does not vanish identically, then by the stable nature of zeroes (see, e.g., Rouche’s theorem) we will also have a nearby zero for any sufficiently close to , with arbitrarily close to . In particular, by setting have positive imaginary part, we contradict the stability of .

Now we turn to the differentiation claim. Again the real nature of is clear. To prove stability, it suffices after freezing the first variables to show that is stable whenever is stable. We may of course take non-zero. But a stable polynomial of one variable may be factored as for some roots with non-positive imaginary part. Since

we see that if has positive imaginary part, then the imaginary part of has the opposite sign to , and is in particular non-zero; since has no zeroes in the upper half-plane, neither does .

Combining these facts with (3), we immediately obtain:

Corollary 13 (Mixed characteristic polynomials are real stable)If are positive semi-definite matrices, then is real stable. In other words, all the zeroes of are real.

Now, we relate stability with “convexity” properties of the largest root. Note in general that if is a convex combination of two polynomials , that there is no requirement that lie between and . For instance, if and , then and , but . Note also that are stable in this case. However the situation improves when one also assumes that all convex combinations of are stable:

Lemma 14 (Comparison with mean)Let be real stable polynomials of one variable that are monic of the same degree. Suppose that is also real stable for every . Then for every , lies between and .

*Proof:* We may of course assume . Without loss of generality let us assume , so our task is to show that

The second inequality is easy: as are monic, is positive for and is positive for , and so is positive for , giving the claim.

The first inequality is trickier. Suppose it fails; then has no zero in . We have already seen that this polynomial is positive at , thus it must also be positive at . In other words, is positive at . Since already had a zero at and was positive to the right of this zero, we conclude that has at least two zeroes (counting multiplicity) to right of . On the other hand, for sufficiently close to , has no zeroes to the right of , and all intermediate convex combinations cannot vanish at . Thus, at some intermediate value of , the number of real zeroes of to the right of must acquire a jump discontinuity (e.g. two real zeroes colliding and then disappearing, or vice versa), but this contradicts Rouche’s theorem and real stability.

One can control other roots than the maximal roots of by these arguments, and show that these polynomials have a common interlacing polynomial of one lower degree, but we will not need these stronger results here.

Specialising this to mixed characteristic polynomials (which are always monic of degree ), we conclude

Corollary 15 (Comparison with mean)Let be jointly independent random rank one positive semi-definite Hermitian matrices, with each taking finitely many values. For any and any fixed choice of , we havewith positive probability, and

with positive probability.

*Proof:* For fixed , is a convex combination of all the values of that occur with positive probability, and hence (by the affine-multilinear nature of ) is a convex combination of the over the same range of . On the other hand, by Corollary 13, all such convex combinations are real stable. Iterating Lemma 14, we see that must lie in the convex hull of the , and the claim follows.

Iterating the above corollary for and using Proposition 3, we obtain Proposition 2 as desired.

** — 3. Convexity — **

In order to establish Theorem 7, we will need some convexity properties of real stable polynomials , or more precisely of the log-derivatives of such polynomials, when one is “above” all the zeroes. We begin with the one-dimensional case:

Lemma 16 (One-dimensional convexity)Let be a real stable polynomial of one variable, and let be the logarithmic derivative (which is defined away from the zeroes of ). Then whenever and . In particular, is positive, decreasing, and strictly convex to the right of .

*Proof:* If we factor , where are the real roots, then . From direct computation we have

Since is positive when , the claim follows.

Now we obtain a higher dimensional version of this lemma. If is a real stable polynomial, we say that a point lies *above* the roots of if has no zeroes on the orthant .

Lemma 17 (Multi-dimensional convexity)Let be a real stable polynomial of variables, let , and let be the logarithmic derivative in the direction. Then whenever and lies above the roots of . In particular, the function is non-negative, non-increasing, and convex for non-negative .

*Proof:* For or , the claim follows from Lemma 16 after freezing all the other variables, so we may assume that and . By freezing all the other variables and relabeling, we may assume that , , and , thus is a real stable polynomial, lies above the roots of , and the task is to show that

for all . Using the variables to denote differentiation instead of , the left-hand side is equal to

so it suffices to show that is non-decreasing in the direction. By continuity, it suffices to do this for generic (thus, we may exclude a finite number of exceptional if we wish).

For fixed , the polynomial is real stable, and thus has real roots, which we denote as (counting multiplicity). For generic , the number of roots does not depend on , and the can be chosen to vary smoothly in (though we do not require the to be distinct). We then have

so it suffices to show that each of the is a non-increasing function of . But if lies above the roots of , then the all lie below ; so it suffices to show that the are all generically non-increasing. But if this were not the case, then (as is generically smooth), would have a positive derivative at some ; setting , then would have a Taylor expansion

for near . But from the inverse function theorem, this would mean that there are zeroes of near with both strictly positive, contradicting stability, and the claim follows.

Now we begin the proof of Theorem 7. We abbreviate as . We start with Proposition 3 to write

From the hypothesis , we may rewrite

and so after translating the by , we have

Replacing each by their mean, we reduce to showing the following deterministic estimate:

Theorem 18Let be Hermitian positive semi-definite matrices such thatfor some and all . Let be the polynomial

Then lies above the roots of .

We will prove this theorem as follows. First observe that lies above the roots of for any , since is positive semi-definite whenever , and so is non-singular in this region. We now wish to pass from to by successive application of the derivative operators . An initial step in this direction is the following elementary consequence of the monotonicity properties from Lemma 17.

Lemma 19Let be a real stable polynomial, and suppose that lies above the roots of . Suppose that , with the stability boundThen also lies above the roots of .

*Proof:* By the monotonicity properties in Lemma 17, for all above (by which we mean that for all ). On the other hand, as is non-zero in this region, can only vanish when is equal to , and so lies above the roots of as claimed.

Unfortunately, this lemma does not directly lend itself to iteration, because the lemma provides no control on . To fix this problem, we need the following more complicated variant of Lemma 19:

Lemma 20Let be a real stable polynomial, and suppose that lies above the roots of . Suppose that , with the stability boundfor some . Then lies above the roots of , and furthermore one has the monotonicity

*Proof:* From Lemma 19 we already know that lies above the roots of , and so does also. Now we prove (9). Above the roots of , we factorise

and thus

Applying , we conclude that

The claim (9) is then equivalent to

On the other hand, by Lemma 17, is convex for , and thus

But

and this quantity is non-positive at thanks to Lemma 17. Thus it suffices to show that

But by Lemma 17 and (8) one has

and the claim follows.

We can iterate this to obtain

Corollary 21Let be a real stable polynomial, and suppose that lies above the roots of . Suppose that

*Proof:* For each , set and . By Lemma 20 induction we see that for each , lies above the roots of with

for all . Setting , we obtain the claim.

Finally, we prove Theorem 18. By definition, we have

and hence on taking and using the Jacobi formula we conclude that

In particular, for any , we may use (6), (7) to conclude that

As remarked before, lies above the zeroes of . Thus, if we choose such that

then by Corollary 21, we conclude that lies above the zeroes of . To optimise subject to the constraint (11), we set and , and Theorem 18 follows.

** — 4. Weaver, Akemann-Anderson, and paving — **

Theorem 5 is probabilistic in nature, but by specialising to a specific type of random vector (and by taking advantage of the freedom to enlarge the vector space , since the bounds are independent of the dimension ), we obtain the following deterministic consequence:

Theorem 22 (Weaver-type bound)Let be integers and be real numbers. Suppose that are such that for all andfor every unit vector (i.e., is a –tight frame), then there exists a partition into components such that

*Proof:* Let notation and the hypotheses be as in the theorem. Define the independent random vectors in the -dimensional tensor product of and by setting

for each , where are sampled uniformly and independently from . From (12) one has

so a brief calculation shows that

(Here is the identity transformation on .) Also, we have

for all . Applying Theorem 5, we conclude that

with positive probability. In particular, this occurs for at least one realisation of the . Setting to be those for which for each , we conclude that

for ; testing this against a unit vector we conclude that

and the claim follows after a little arithmetic.

For a given , the Weaver conjecture asserts for the existence of a such that the above theorem holds with the upper bound in (13) replaced by for some absolute constant . This is easily seen to follow from (13) by taking large enough. For instance (as noted by Marcus, Spielman, and Srivastava), if one takes then the upper bound in (13) is , giving the case of Weaver’s conjecture, which was already known to imply the Kadison-Singer conjecture (see e.g. this paper of Bice for an ultrafilter-based proof of this fact). Note also that in the large limit (keeping fixed), the bound in (13) approaches , which is the optimal value (as can be seen by considering the case where each has the maximal magnitude of ); this will be important later.

We can dualise Theorem 22 as follows, giving a bound of Akemann-Anderson type:

Corollary 23 (Paving conjecture for projections)Let , and let be an orthogonal projection matrix, that is to say an idempotent Hermitian matrix. Suppose that we have the bound for all and some . Then, for any , there exists a partition of the identity into diagonal projections (i.e. the are diagonal matrices with s and s on the diagonal, that sum to the identity matrix) such thatfor all .

*Proof:* Let be the range of , then are vectors in with

Also, for any unit vector , we have

Applying Theorem 22 with , , , we can find a partition such that

for all unit vectors and . If we let be the diagonal projection , then

and so

for all in . Thus

and thus on squaring

as desired.

This is a special case of the Kadison-Singer paving conjecture, but we may use a number of standard maneuvers to amplify this to the full paving conjecture, loosely following the papers of Popa and of Akemann-Weaver. First, by using the “dilation trick“, we may relax the hypothesis of being a projection:

Corollary 24 (Paving conjecture for positive semi-definite matrices)Let , and let be a Hermitian positive semi-definite matrix with the bound for some . Suppose that we have the bound for all and some . Then, for any , there exists a partition of the identity into diagonal projections such thatfor all .

*Proof:* By scaling we may set . The matrix is no longer a projection matrix, but by the spectral theorem and the hypotheses on , we still have a decomposition of the form

where are an orthonormal basis for , and for all .

Now we embed in for some large . If is large enough (depending on ), we may find orthonormal vectors with disjoint supports, such that each of the have coefficients of magnitude at most . We may then form the dilated vectors

which are an orthonormal system in , and the dilated operator

which is an orthogonal projection on . This can be viewed as a matrix, whose top left minor is . By construction, all the diagonal entries of this matrix are at most . Applying Corollary 23, we may find a partition of the identity matrix on into diagonal projections such that

for all . If we let be the component of that maps to (i.e. the top left minor), then the are diagonal matrices summing to the identity, and we have

for all , as required.

Now we shift by a multiple of the identity:

Corollary 25 (One-sided paving for Hermitian matrices)Let , and let be a Hermitian matrix with vanishing diagonal, whose eigenvalues lie between and for some . Then, for any , there exists a partition of the identity into diagonal projections such that for each , the eigenvalues of lie between and .

*Proof:* Apply Corollary 24 with replaced by , replaced by , and set equal to .

We combine this result with its reflection:

Corollary 26 (Paving conjecture for Hermitian matrices)Let , and let be a Hermitian matrix with vanishing diagonal, with for some . Then, for any , there exists a partition of the identity into diagonal projections such thatfor all .

*Proof:* The eigenvalues of lie between and . By Corollary 25, we may find a partition into diagonal projections such that each have eigenvalues between and ; of course these matrices also have eigenvalues between and and have zero diagonal. Thus, for each , we may apply Corollary 25 to to find a further splitting into diagonal projections such that lie between and . The eigenvalues of are then bounded in magnitude by , and the claim follows by using the partition .

Finally, we remove the Hermitian condition:

Corollary 27 (Paving conjecture for finite matrices)Let , and let be a matrix with vanishing diagonal, with for some . Then, for any , there exists a partition of the identity into diagonal projections such thatfor all .

*Proof:* By applying the preceding corollary to the Hermitian matrices and , which both have operator norm at most , we may find partitions and into diagonal projections such that

and

for all . In particular,

(note that commute). The claim then follows from the triangle inequality.

Because the bound here is independent of the dimension , we may use a standard compactness argument to handle infinite matrices:

Corollary 28 (Paving conjecture for infinite matrices)Let be an infinite matrix with vanishing diagonal, with for some . Then, for any , there exists a partition of the identity into diagonal projections such thatfor all .

*Proof:* For each , let be the top minor of , extended by zero to an infinite matrix. This matrix has zeroes on the diagonal and has operator norm at most . Applying Corollary 27 to the minor of and then extending, we may find a partition into diagonal projections such that

By the Arzela-Ascoli theorem (or the sequential Banach-Alaoglu theorem), we may pass to a subsequence such that each of the converges pointwise to a limit (i.e. each coefficient of the matrix converges as to the corresponding coefficient of . As the are diagonal projections, the are as well; since , we see on taking pointwise limits that .

Now let be unit vectors in with finite support. Then for sufficiently large , and similarly . As and have finite support, we also have on the support of for sufficiently large , and hence

for sufficiently large . By (15) and the Cauchy-Schwarz inequality, we conclude that

taking suprema over all unit vectors we obtain

for all as required.

** — 5. The Kadison-Singer problem — **

To turn to the original formulation of the Kadison-Singer problem, we need some terminology from -algebras. We will work with two specific algebras: the space of bounded linear operators on (or equivalently, infinite matrices with bounded operator norm); and the subspace of diagonal matrices in . These are both examples of -algebras, but we will not need the specific definition of a -algebra in what follows.

A *state* on is a bounded linear functional on with , such that for all positive semi-definite . A state is *pure* if it cannot be represented as a convex combination of two other states. Define a state (and a pure state ) similarly.

The Kadison-Singer problem, now solved in the affirmative by Marcus, Spielman, and Srivastava, states the following:

Theorem 29 (Kadison-Singer problem)Let be a pure state. Then there is a unique state that extends . (In particular, this state is automatically pure).

Existence of is easy: set , where is the matrix with all off-diagonal entries set to zero. The hard part is uniqueness. To solve this problem, let us first understand the structure of pure states on .

Let , and let be a diagonal projection matrix. Since and are both positive semi-definite, we have , and hence . We in fact claim that either or . Indeed, suppose that for some . Then we can write , where and . It is easy to see that are both states on , contradicting the hypothesis that is pure.

Remark 1With only a little more effort, this argument shows that the pure states of are in one-to-one correspondence with the ultrafilters of , but we will not need to use the finer properties of ultrafilters here.

Now we show uniqueness. Let be a state extending . Let ; our task is to show that . By subtracting off , we may assume that has zero diagonal. Our task is now to show that .

Let . By Corollary 28 (with sufficiently large), we may find a partition into diagonal projections such that

Recall that is either or ; since , we have for exactly one , say , and for all other .

Suppose that is such that . Since is a Hermitian semi-definite inner product on , we have the Cauchy-Schwarz inequality

since , we conclude that for all and ; similarly . We conclude that , and hence by (16) we have

where is the operator norm of . Since may be arbitrarily small, we conclude that as required.

## 27 comments

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4 November, 2013 at 6:43 pm

AnonymousYour first few references to the authors misspell Nikhil Srivastava’s last name as Srinivasta.

[Corrected, thanks. -T.]4 November, 2013 at 8:05 pm

DPIn Theorems 6 and 7 the expected value is missing for the trace, and in Theorem 7, the last statement does not agree with the one in Theorem 6. Also, above Theorem 7, the reference is to Proposition 2.

[Corrected, thanks – T.]5 November, 2013 at 12:11 pm

COA few small corrections: Parentheses before 1-\lambda are missing on lines 4, 7 and 8 in the proof of Lemma 14. In Corollary 15, matrices are independent and the inequalities are not strict. In the proof of Corollary 15 – iterating Lemma 14. In the proof of Lemma 17 (first line), the claim follows from Lemma 16.

[Corrected, thanks – T.]6 November, 2013 at 10:28 pm

Orr ShalitThanks for the beautiful post.

A pure state is one that cannot be written as a *convex* combination of other states; one may write some pure states as linear combinations of other states (second paragraph of last section).

[Corrected, thanks – T.]7 November, 2013 at 12:25 pm

Felix V.Dear Prof. Tao,

thanks for the beautiful post.

I have a question regarding Lemma 12: It seems to me that real stable polynomials are NOT invariant under restriction. Consider the polynomial p(z_1, z_2) = z_2^2. According to the definition, p is real stable (for every zero, we have z_2 = 0, i.e. (z_1, z_2) is no member of the “upper half plane”).

But if we substitute z_2 = t := 0, then we get p(z_1, t) = 0, which is NOT real stable.

(The application of Rouches Theorem in the proof fails if the polynomial p(z_1, \dots, z_{m-2}, z, t) vanishes identical as a function of z.)

I am sorry if I am making some silly mistake.

Best regards,

Felix V.

[Clarified that the restriction property only holds when the restricted polynomial is not identically zero. (In the applications, all the polynomials are monic in at least one of the variables and so are not identically zero.) -T.]8 November, 2013 at 6:02 am

DPz_2^2 is not real stable, for example, it has a root (i, 0).

8 November, 2013 at 10:42 am

DPDear Terry,

I think your clarification is a bit confusing and unnecessary. The polynomial can not become identically zero, because the same argument by Rouche’s theorem would imply that the original polynomial was not stable, wouldn’t it?

8 November, 2013 at 11:10 am

DPSorry, you are right, I was confused.

14 November, 2013 at 6:44 am

Felix V.Thank you very much for the clarification.

So an alternative argument (avoiding Rouches’ theorem) would be that p(\cdot, …, \cdot, t + i/n) converges locally uniformly to p(\cdot, …, \cdot, t) and has no zeros on the upper half plane, so that (a multidimensional form of) Hurwitz’ theorem (http://en.wikipedia.org/wiki/Hurwitz%27s_theorem_%28complex_analysis%29#Applications ) implies that p(\cdot, …, \cdot, t) vanishes identically or has no zeros on the upper half plane.

I have another question regarding the proof of lemma 17. You state that by excluding finitely many points, the zeros can be smoothly parameterized. Can you give me a hint on where to look that up or how to show it?

It is clear that the claim follows by the implicit function theorem as long as \frac{\partial}{\partial z_1} p(z_1, z_2) \neq 0 for all zeros z_2 of p(z_1, \cdot), but I don’t see why this should be true for all but finitely many points.

Am I missing something?

Best regards,

Felix V.

14 November, 2013 at 1:37 pm

Felix V.In my last comment I meant \frac{\partial}{\partial x_2} instead of x_1. This is then of course the same as to say that all zeros have multiplicity one.

As you explicitly state that you do NOT exclude this case (after all, this could happen for ALL x1), I am even more at a loss.

Any help would be appreciated.

Best regards,

Felix V.

14 November, 2013 at 2:22 pm

Terence TaoHere I am using the fact that an irreducible algebraic curve is smooth at all but finitely many points. (Indeed, if the curve is given by for some irreducible P, then the partial derivatives of P are coprime to P, so by Bezout’s theorem there are only finitely many points where P and both vanish.) For similar reasons, there are only finitely many points where the gradient is horizontal, unless the curve is a vertical line (which can be excluded here).

In this setting, P need not be irreducible, but it can be broken up into irreducible components, which again by Bezout only meet each other in finitely many places. (It is possible for components to appear with multiplicity, but then one can just take a single instance of each component, which will still be smooth away from finitely many points.)

p.s. one can certainly use Hurwitz’s theorem here in place of Rouche’s theorem, but the former theorem is often presented as a corollary of the latter, so the arguments are not terribly different from each other.

17 November, 2013 at 2:17 am

Felix V.Thank you very much. That helped a lot.

2 December, 2013 at 10:55 am

Felix V.Dear Prof. Tao,

here are a few minor (supposed) mistakes that I found:

In Lemma 12 “the same sign as t” should be “the opposite sign of t”.

In Corollary 15 in the expression “we see that X must lie in the convex hull of the Y” (for appropiate values of X,Y) X and Y should have been interchanged.

In Lemma 16 you mean k! instead of (k-1)! and (x-y_i)^(k+1) instead of (x – y_i)^k.

In Lemma 20 the “convexity estimate” is incorrect. On the right hand side you have to write \partial_j \Phi_q^i (x + \delta e_j) instead of \partial_j Phi_q^i (x). (This replacement has also to be made in the next formula).

In the proof of Theorem 22, the very beginning of the last formula (\Vert) can be removed. In the line before that: “j=1, \dots, r” instead of “j=1,2”.

In the statement of corollary 23: “0s and 1s on the identity” should have been “0s and 1s on the diagonal”.

Minor improvement: In the statement of Corollary 27 it should be possible to replace both 4s by 2s, as you are using the triangle inequality on P = (P + P*)/2 + (P – P*)/2. This can consequently also be done in Corollary 28.

Thanks again for the nice post.

Best regards,

Felix V.

[Corrected, thanks – T.]8 November, 2013 at 12:29 pm

The Kadison-Singer Conjecture has beed Proved by Adam Marcus, Dan Spielman, and Nikhil Srivastava | Combinatorics and more[…] Updates: A very nice post on the relation of the Kadison-Singer Conjecture to foundation of quantum mechanics is in this post in Bryan Roberts‘ blog Soul Physics. Here is a very nice post on the mathematics of the conjecture with ten interesting comments on the proof by Orr Shalit, and another nice post on Yemon Choi’s blog. Nov 4, 2013: A new post appeared on Terry tao’s blog, Real stable polynomials and the Kadison Singer Problem. […]

8 November, 2013 at 5:48 pm

MustafaThere is a typo in Corollary 26: you repeat the “then for r \geq 1.”

[Corrected, thanks – T.]13 November, 2013 at 2:39 am

TRINH Tuan PhongDear Prof. Tao,

May I have a question? I wonder whether “x=(x_1,…,x_m) lies above a root z=(z_1,…,z_m) of the polynomial p” in your note means x_j\geq z_j for some j?

Best regards,

TRINH Tuan Phong

[For all j, not some j; see text before Lemma 17, and also the proof of Lemma 19. -T.]14 November, 2013 at 4:15 am

TRINH Tuan PhongThank you so much for your answer.

Best regards,

T.T.Phong

13 November, 2013 at 5:29 pm

Peter ToblerHi – hope you don’t mind me raising this minor point – the guy’s name is Srivastava

[Corrected, thanks – T.]17 November, 2013 at 9:42 am

Pete CasazzaThere is a trivial way to see that paving is equivalent to paving projections with small diagonal. As Teri noted, we may assume T is self-adjoint. Also, we clearly may assume \|T\| \le 1. Now form the square matrix

Since R^2=I, P= 1/2(I+R) is a projection with diagonal 1/2 and we can pave

T if and only if we can pave P. Iterating this, we can reduce to diagonal to

1/2^n with the same conclusion.

18 November, 2013 at 3:38 pm

omar abouraIn theorem 7: $ maxroot E p $ should be $ maxroot (E p) $.

In “…are all geerically non-increasing”, geerically should be generically.

At the end of the proof of Corollary 27, “…commute” should be “…commute)”.

In Corollary 27 and Corollary 28, “for any $r\geq 1$, for any $r\geq 1$” should be “for any $r\geq 1$”.

[Corrected, thanks – T.]17 April, 2014 at 10:14 pm

Anonymoushttp://www.siam.org/prizes/sponsored/polya.php

17 May, 2014 at 10:05 am

Daniel Spielman talks at HUJI – thoughts | Noncommutative Analysis[…] KS2 conjecture, which implies a positive solution to Kadison-Singer (the full story been worked out to expository perfection on Tao’s […]

19 September, 2014 at 5:39 am

Fall 2014: Roots of polynomials and probabilistic applications | Stochastic Analysis Seminar[…] Notes by T. Tao […]

13 October, 2014 at 11:10 am

André SchemaitatIn Corollary 28 it is stated that P Q_j u = P^{(m_l)} Q_j u for large l since Q_j u has finite support. Why should this be true ? I.e. if P has many non-zero elements in its rows than P Q_j u won’t even be a vector of finite support.

[Oops, this equality should be restricted to the support of . I’ve updated the text accordingly. -T.]3 November, 2014 at 8:37 pm

Lecture 5. Proof of Kadison-Singer (3) | Stochastic Analysis Seminar[…] of the proof of the Kadison-Singer theorem has largely followed the approach from the blog post by T. Tao, which simplifies some of the arguments in the original paper by A. W. Marcus, D. A. Spielman, and […]

6 January, 2015 at 5:45 am

Another one bites the dust (actually many of them) | Noncommutative Analysis[…] there is Terry Tao’s post on this subject that I read and […]

8 April, 2015 at 10:25 am

jacobcarruthReblogged this on jacobcarruth.