Interlacing Families and their Applications

Interlacing Families and their Applications Guide: Prof. Ajit Iqbal Singh INSA Honorary Scientist Indian Statistical Institute, Delhi

Student: Sagar Shrivastava National Institute of Science Education and Research, Bhubaneshwar

I started my summer project under the Guidance of Prof. Ajit Iqbal Singh ( INSA Honorary Scientist ) on 12th May, 2014 for a duration of 6 weeks (i.e. till 21st June, 2014 ). I specially focused on the research papers authored by Adam Marcus, Daniel Spielman and Nikhil Srivastava [MSS13b] and some others authored by Kathy Driver [DJM08]. Here with Prof. Singh’s help, I was able to brush up a lot of things in various topics I had studied before (like complex analysis, metric spaces, theory of differentiation, measure theory , real analysis etc.) and also got to know a lot about the research going out nowadays in Mathematics. From the research papers [MSS13a] I was able to learn a new method developed to give a solution of the Kadison-Singer problem which is called as interlacing families. In Kathy Driver’s paper [Dri09] my main focus was to get some examples of interlacing families which as mentioned in her paper are easy enough to find in orthogonal polynomials. Also to get a brief introduction to Orthogonal Polynomials, I read the first 2 chapters of the book An Introduction to Orthogonal Polynomials by T.S. Chihara [Chi11] and some others too [SSS59]. Along with this, my project at ISI was very fruitful in the sense that I could attend many seminars here, especially I would like to mention the two seminars taken by Prof. R. Bhatia on Positive Semi-definite Matrices and because of those seminars, I was able to relate some Matrix Analysis (of Hankel Matrices) to Orthogonal Polynomial Systems. Along with me, two other students, Priya Tayal and Rachna Maurya (Science Academies Summer Fellow ) were also doing their project under Prof Singh’s guidance. Priya’s topic was “ Fact that homeomorphisms of the unit circle with dense orbits turn out to be essentially rotations by an irrational (multiple of 2π ) and operators associated with such rotations ”, and Rachna’s topic was “ Rootless Matrices and Spectral Theorem” In all, I am very grateful to KVPY for allowing me opportunity like this every year to understand new things, meet new teachers, other students who like me are interested in research. 1

Orthogonal Polynomials Basics To define orthogonality, we need a non trivial vector space, with a well defined inner product. For this, consider the vector space of all continuous functions (from R to C). This vector spce is in-fact a ring with the usual operations. Now there are many ways to define an inner product in this vector space. We will use the linear functional L (called as the moment functional ) to do this. Now it is possible to consider a weight function w(x) which is non-negative and integrable on an interval (a, b), also w(x) > 0 on a sufficiently large subset(set of non zero measure) of (a, b). With this it is possible to define L as: Z

b

L[f ] =

f (x)w(x)dx a

then µn = L[xn ], n ∈ N and hf, gi = L[f.g] By this definition L is a linear functional and we can define it for all polynomials (using the linearity of L), if we know the value of the sequence {µn }∞ n=0 and with polynomials defined, it can be defined for all continuous functions as polynomials are dense in the set of continuous functions. There is a more general case where we define L without using w(x) but only n a sequence of complex numbers {µn }∞ n=0 and defining L[x ] = µn .

Orthogonal Polynomial System Definition. A sequence of polynomials {Pn (x)}∞ n=0 is called an Orthogonal Polynomial System (OPS) w.r.t. the moment functional L if ∀m, n ∈ N0 1. Pn (x) is a polynomial of degree n. 2. L[Pn (x)Pm (x)] = 0 for m 6= n 3. L[Pn2 (x)] 6= 0, ∀n ∈ N0 Now it should be noted that not all sequences {µn }∞ n=0 give rise to a sequence of Orthogonal polynomials. Theorem. Let L be the moment functional and {Pn (x)}∞ n=0 is a sequence of polynomials where deg(Pn (x) = n, then the following are equivalent

{Pn (x)}∞ n=0 is an OPS w.r.t. L L[π(x)Pn (x)] = 0 for every polynomial π(x) of degree m¡n and L[π(x)Pn (x) 6= 0 if m = n L[xm Pn (x)] = kn δmn where kn 6= 0 2

Theorem. Let {Pn (x)}∞ n=0 be an OPS w.r.t. L, then for every polynomial π(x) of degree n, π(x) =

n X

ck Pk (x),

ck =

k=0

L[π(x)Pn (x)] L[Pn2 (x)

. Theorem. If {Pn (x)}∞ n=0 is an OPS w.r.t. L, then each Pn (x) is uniquely determined upto an arbitrary non-zero factor. That is if {Qn (x)}∞ n=0 is another OPS w.r.t. L, then ∃cn 6= 0∀n ∈ N0 such that Qn (x) = cn Pn (x) .

∆n = det(µi+j )ni,j=0

µ0 µ1 · · · µ1 µ2 · · · = det . .. .. . . . . µn µn+1 · · ·

µn µn+1 .. . µ2n

Theorem. Let L be the moment functional with the moment sequence {µn }∞ n=0 . An OPS for L exists ⇔ ∆n 6= 0 , n ∈ N0 . Theorem. Let {Pn (x)}∞ n=0 be an OPS w.r.t. L. Then for any polynomial πn (x) of degree n L[πn (x)Pn (x)] = an L[xn Pn (x)] =

an kn ∆n , ∆n−1

∆−1 := 1

where an denotes the leading coefficient of πn (x) and kn denotes the leading coefficient of Pn (x). Definition. A moment functional L is called Positive-definite if L[π(x)] > 0 for every polynomial π(x) that is not identically zero and is non-negative for all real x. Theorem. Let L be positive-definite. Then L has real moments and a corresponding OPS consisting of real polynomials exists. Theorem. Let π(x) be a polynomial that is not identically zero and is nonnegative for all real x.Then there are real polynomials p(x) and q(x) such that π(x) = p2 (x) + q 2 (x) Theorem. L is positive-definite iff all its moments are real and ∆n > 0 , n ∈ N0 . 3

The Fundamental Recurrence Formula Definition. A moment functional L is called Quasi-definite if ∆n 6= 0 , n ∈ N0 Theorem. Let L be a quasi-definite moment functional and let {Pn (x)}∞ n=0 be the corresponding monic OPS. Then there exists constants cn and λn 6= 0 such that Pn (x) = (x − cn )Pn−1 (x) − λn Pn−2 (x), n ∈ N Also we have that: 1. λn+1 =

L[Pn2 (x)] ∆n−2 ∆n = 2 L[Pn−1 (x)] ∆2n−1

2. L[Pn2 (x)] = λ1 λ2 . . . λn+1 provided we define λ1 = µ0 = ∆0 3. cn =

2 (x)] L[xPn−1 2 (x)] L[Pn−1

4. The coefficient of xn−1 is −(c1 + c2 + . . . + cn )

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The Kadison - Singer Conjecture Introduction The Kadison-Singer problem was one of the central open questions in operator theory, until its recent solution by Marcus, Spielman and Srivastava [ref]. The importance of the problem is underscored by its connections to numerous diverse areas of mathematics. The Kadison-Singer problem has many equivalent forms, some of which are much easier to understand. In particular, there several problems involving finite-dimensional matrices that are equivalent to the Kadison-Singer problem, and can be easily understood by anyone with basic understanding of linear algebra. A concise statement of the Kadison-Singer problem is: Problem. Does every pure state on the algebra of bounded diagonal operators on (the complex Banach space l2 ) have a unique extension to a state on the algebra of all bounded operators on l2 ?

Two dimensional Example In the 2 diminsional case, the algebra of “bounded diagonal operators on l2 ” simply becomes the algebra of diagonal 2 × 2 matrices over the complex numbers a 0 , a, d ∈ C D2 := M = 0 d A linear functional on D2 is simply a map f : D2 → C with f (M ) = f (a, d) = αa + δd and α, δ ∈ C. A state is a linear function f satisfying f (I) = 1, So we must have α = 1 − δ f (M ) must be real and non-negative whenever M is positive semidefinite (i.e., whenever a,d are both real and non-negative). So we must have α is real and α ∈ [0, 1].

A pure state is a state f satisfying f cannot be written as a non-trivial convex combination of two different states. So we must have either α = 0 or α = 1.

So the only pure states on D2 are f (M ) = a and f (M ) = d In this two-dimensional example, the algebra of “bounded operators on l2 ” simply becomes the algebra of two-dimensional matrices over the complex numbers

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2×2

C

:=

a b M= , a, b, c, d ∈ C c d

A linear functional on C2×2 is a function g(M ) = g(a, b, c, d) = αa + βb + γc + δd, α, β, γ, δ ∈ C Letting G = αγ βδ , we can write g(M ) = tr(GM ). A state on C2×2 is a linear functional g satisfying: g(I) = 1 g(M ) must be real and non-negative whenever M is Hermitian, positive semi-definite.

The first condition is equivalent to tr(G) = 1. The second condition implies that v ∗ Gv = tr(Gvv ∗ ) = g(vv ∗ ) ≥ 0 for all v ∈ C2 , which implies that G is Hermitian and positive semi-definite. Conversely, any function g(M ) = tr(GM ) with G complex, positive semi-definite and tr(G) = 1 is a state on C2×2 . A state g on C2×2 is an extension of a state f on D2 if g(M ) = f (M ) ∀M ∈ D2 . (i.e.,g(a, 0, 0, d) = f (a, d)). Every state f on D2 has a canonical extension to a state g on C2×2 obtained simply by defining g(M ) = g(a, b, c, d) = f (a, d) (since positive semi-definite matrices have non-negative diagonals). Let us consider the pure state f (M ) = a. Consider any linear functional g(M ) = tr(GM ) that is a state on C2×2 and is an extension of f. Then we 1 β must have G = β 0 for some β ∈ C, and G positive semi-definite. As the diagonal entries of G are non-negative, G is positive semi-definite if and only if det(G) ≥ 0. As det(G) = −ββ, this is non-negative only when β = 0. Thus g(M ) = a is the unique state on C2×2 that is an extension of f (M ) = a. Similarly, f (M ) = d has a unique extension to a state on C2×2 . So we may conclude that the two-dimensional analog of the Kadison-Singer problem is true.

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Interlacing Families and other Methods The following are my own ideas except for the one to which the citation has been made. Lemma. If α1 6= α2 ( and assume w.l.o.g. α1 < α2 , then there exists at most one natural numbers m for some fixed j ∈ N, such that α1 +

1 1 = α2 + m m+j

Proof. Assume to the contrary, that there are more than one natural numbers m,n such that α1 +

1 1 1 1 = α2 + , α1 + = α2 + m m+j n n+j

⇒ α1 − α2 = ⇒ ⇒

1 1 1 1 − = − m+j m n+j n

−j −j = m(m + j) n(n + j)

1 1 = m(m + j) n(n + j)

(∵ j 6= 0)

⇒ m2 + mj = n2 + nj ⇒ m2 − n2 = (n − m)j ⇒ −(m + n)(n − m) = (n − m)j ⇒ −(m + n) = j ⇒ j < 0 ⇒⇐

(∵ n 6= m) (∵ j ∈ N)

Theorem. Given a real polynomial f (x) having real roots, it is possible to construct a sequence of polynomials Pk (x) such that limk→∞ Pk (x) = f (x) uniformly on a compact subset of R and Pk (x) have all distinct real roots. Q Proof. Let f (x) = a ni=1 (x − αi ), define Pk (x) as: n Y Pk (x) = a x − αi − i=1

1 N0 + k + i

(N0 ∈ N)

where the choice of N0 depends on f (x). From the lemma, we know that there is at most one integer mij satisfying αi +

1 1 = αj + mij + i mij + j 7

The number of such pairs of αi , αj is finite ( as the polynomial has a finite degree), so we get that ∃N0 ∈ N s.t. N0 > mij ∀i, j, i 6= j . Now αi +

1 1 6= αj + , 1 ≤ i < j ≤ n, l ∈ N N0 + l + i N0 + l + j

Hence Pk (x) has simple roots. Also 1 lim αi + = αi k→∞ N0 + k + i Since the roots of Pk (x) converge to the roots of f (x) and we know that the coefficients of a polynomial are a symmetric function of their roots, we have that the coefficients of Pk (x)P converges to the respective coefficients f (x). For any polynomial q(x) = ni=0 ai xi , if there is a bound on x (i.e. |x| ≤ η, η > 1) we have that | q(x) |= |

n X i=0

i

ai x | ≤

n X

i

|ai ||x| ≤ (

i=0

n X i=0

n X |ai |)|x| ≤ ( |ai |)η n n

i=0

If we take η n to be a weight, we get a very nice norm on polynomials Pof degree n over a compact subset of R, so we have our norm as kq(x)k = ni=0 |ai |. Coming back to our proof, take γ ∈ R+ s.t. γ > |αi |, 1 ≤ i ≤ n and η = γ +1 Using the norm as defined above, we see that that Pk (x) converges to f (x) uniformly on any compact super set of [−η, η], which complete the proof . Definition (Interlacing). We say that a real rooted polynomial Q Qn g(x) = a0 n−1 (x − α ) interlaces a real rooted polynomial f (x) = b i 0 i=1 i=1 (x − βi ) if β1 ≤ α1 ≤ β2 ≤ α2 ≤ . . . ≤ αn−1 ≤ βn We say that polynomials f1 , f2 , . . . , fn have a common interlacing if there is a polynomial g so that g interlaces fi for each i. Theorem. Take polynomials f (x) and g(x) of degree n with positive leading coefficients such that, for every λ ∈ [0, 1], the polynomial λf (x)+(1−λ)g(x) have n real roots. Then f(x) and g(x) have a common interlacing.[Mix14] Proof. We only treat the special case where the 2n roots of f(x) and g(x) are all distinct (for the other cases we can use the theorem above to get a sequence of polynomials all of degree n and with distinct roots converging to f(x) uniformly in some compact set in R). By assumption, f(x) and g(x) each have n real roots, which form 2n points on the real line, and their complement in the real line forms 2n+1 open intervals. Viewing f and g as functions over the real line, note that their signs are constant over each of these open intervals (this follows from the intermediate value theorem). In 8

fact, ordering these intervals from right to left as {Ik }2n k=0 , then since the polynomials have positive leading coefficients, both f and g are positive on I0 . Next, the largest of the 2n roots belongs to either f(x) or g(x) (but not both), and so f and g have opposite sign over I1 . Continuing in this way, f and g have common sign over even-indexed intervals and opposite sign over odd-indexed intervals. As such, addition gives that for any λ > 0, every real root of λf (x)+(1−λ)g(x) must lie in an odd-indexed interval. Moreover, the (real) roots are continuous in λ, and so the boundary of each odd-indexed interval had better match a root of f (x) with a root of g(x). Picking a point from each of the even intervals then produces a common interlacing of f(x) and g(x). Corollary. Take a finite collection of polynomials {fi (x)}i of degree n with positive leading coefficients such that every convex combination of these polynomials has n real roots. Then the polynomials {fi (x)}i have a common interlacing. Proof. By the above theorem, any pair has a common interlacing. With this in mind, suppose {fi (x)}i does not have a common interlacing. Then there is some i,j,k such that the kth largest root of fi (x) is smaller than the (k+1)th largest root of fj (x), contradicting the fact that fi (x) and fj (x) have a common interlacing. Theorem. Take a finite collection of polynomials {fi (x)}i , of same degree n, all real coefficients, a positive leading coefficient, and {fi (x)}i has a common interlacing.Then every convex combination of these polynomials has n real roots. Proof. Now given that there exists a common interlacing implies that all the roots are real for all fi (x). Let the roots of fi (x) be {βi,j }nj=1 such that βi,1 ≤ βi,2 ≤ . . . ≤ βi,n . P It is enough to give the proof for the polynomial F (x) = ki=1 fi (x) as the proof is only dependent on the non-negativity of the leading coefficient of each polynomial fi (x). Since we have a common interlacing, we get that max βi,j ≤ min βi,j+1 i

i

Let us say γj = maxi βi,j and ηj = mini βi,j . Then we get the relation : η1 ≤ γ1 ≤ η2 ≤ . . . ≤ ηn ≤ γn ≤ ηn+1 where ηn+1 is some real number greater than γn Claim : fi (ηk ) ≤ 0

if n-k+1 odd

≥0

if n-k+1 even 9

Proof. We prove this by induction on k. Now fi (ηn+1 ) > 0 (∵ fi have positive leading coefficient.) Also if y ∈ (βi,n−1 , βi,n ) ⇒ fi (y) < 0 (as fi (y) > 0 ⇒ ∃ another root of fi in (βi,n−1 , βi,n ) (By IVT) which goes against the hypothesis) Now ηn ∈ (βi,n−1 , βi,n ) ⇒ fi (ηn ) ≤ 0 Let the hypothesis of the induction be true for some l ≤ n + 1, i.e. fi (ηk ) ≤ 0

if n-k+1 odd

≥0

if n-k+1 even

for all l ≤ k ≤ n + 1 Now ηl ∈ (βi,l−1 , βi,l ) and by IVT we have that the sign of the polynomial fi is the same for all y ∈ (βi,l−1 , βi,l ). Again by IVT we have that the sign changes in the interval (βi,l−2 , βi,l−1 ), also ηl−1 ∈ (βi,l−2 , βi,l−1 ), hence if fi (ηl ) ≤ 0 ⇒ fi (ηl−1 ) ≥ 0 and fi (ηl ) ≥ 0 ⇒ fi (ηl−1 ) ≤ 0 which completes our proof for the claim. P We define fλ = ni=1 λi fP i as the convex combination of the polynomials n {fi (x)}i (where λ ∈ R s.t. ni=1 λi = 1, λi ≥ 0). So Using the Claim we get that fλ (ηk ) ≤ 0

if n-k+1 odd

≥0

if n-k+1 even

as multiplication by a positive scalar does not change the sign of a polynomial in an interval. Also if two polynomials have the same sign at some value , so will be the sign of their sum. Hence we get it for a convex combination. which implies by IVT that ∀λ, fλ has a root in the intervals P (ηk , ηk+1 ) k= 1,2, . . . n. Hence fλ has n real roots for all λ ∈ Rn s.t. ni=1 λi = 1, λi ≥ 0

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Bibliography [Chi11]

Theodore S Chihara. An introduction to orthogonal polynomials. Courier Dover Publications, 2011.

[DJM08] Kathy Driver, Kerstin Jordaan, and Norbert Mbuyi. Interlacing of the zeros of jacobi polynomials with different parameters. Numerical Algorithms, 49(1-4):143–152, 2008. [Dri09]

Kathy Driver. A note on the interlacing of zeros and orthogonality. Journal of Approximation Theory, 161(2):508–510, 2009.

[Mix14]

Dustin G. Mixon. An introduction to interlacing families, August 2014.

[MSS13a] Adam Marcus, Daniel A Spielman, and Nikhil Srivastava. Interlacing families i: Bipartite ramanujan graphs of all degrees. In Foundations of Computer Science (FOCS), 2013 IEEE 54th Annual Symposium on, pages 529–537. IEEE, 2013. [MSS13b] Adam Marcus, Daniel A Spielman, and Nikhil Srivastava. Interlacing families ii: Mixed characteristic polynomials and the kadison-singer problem. arXiv preprint arXiv:1306.3969, 2013. [SSS59]

Gabor Szeg¨ o, G´ abor Szegö, and Gábor Szegö. Orthogonal polynomials, volume 23. American Mathematical Society New York, 1959.

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