Real-rooted polynomials and interlacing families

Transcription

1 page.1 Real-rooted polynomials and interlacing families Adam W. Marcus Yale University Crisply, LLC October 10, 2013

2 page.2 2/50 Joint work with: Dan Spielman Yale University Nikhil Srivastava Microsoft Research, India My involvement supported by: National Science Foundation Mathematical Sciences Postdoctoral Research Fellowship

3 page.3 Because not everyone can be a combinatorist 3/50 Some notation 1 [k] will denote the set of integers {1,..., k}. 2 for a set S, 2 S will denote the set of all subsets of S 3 For a set S, ( S k) will denote the subsets of S that are size k 4 For a set S, a S = i S a i

4 page.4 Outline Introduction 4/50 1 Introduction 2 Multivariate Extensions Real Stable Polynomials Hyperbolic Polynomials 3 Roots 4 Interlacing families

5 page.5 Understanding distributions Introduction 5/50 Question: How do we normally understand distributions?

6 page.6 Understanding distributions Introduction 5/50 Question: How do we normally understand distributions? Answer: Use things like norms (moments) and transforms.

7 page.7 Understanding distributions Introduction 5/50 Question: How do we normally understand distributions? Answer: Use things like norms (moments) and transforms. Question: How do we compare distributions?

8 page.8 Understanding distributions Question: How do we normally understand distributions? Answer: Use things like norms (moments) and transforms. Question: How do we compare distributions? Answers: Use inequalities.

9 page.9 Finite distributions Introduction 6/50 Finite distributions are the same.

10 page.10 Finite distributions Introduction 6/50 Finite distributions are the same. That said, there can be advantages to different encodings of the distribution.

11 page.11 Finite distributions Introduction 6/50 Finite distributions are the same. That said, there can be advantages to different encodings of the distribution. Example: Generating Functions Given a sequence a 0, a 1,... the ordinary generating function is the formal power series a 0 + a 1 x + a 2 x 2 + = a i x i i So the sequence 1, 1, 1,... can be encoded as 1 1 x.

12 page.12 Finite distributions Introduction 6/50 Finite distributions are the same. That said, there can be advantages to different encodings of the distribution. Example: Generating Functions Given a sequence a 0, a 1,... the ordinary generating function is the formal power series a 0 + a 1 x + a 2 x 2 + = a i x i i So the sequence 1, 1, 1,... can be encoded as 1 1 x. Advantage: we can use power series arithmetic to combine sequences and get new ones.

13 page.13 Polynomials Polynomials can be used in a similar way: Given values a 1,..., a d, we can encode them as p(x) = (x a 1 )(x a 2 )... (x a d ) = i (x a i )

14 page.14 Polynomials Polynomials can be used in a similar way: Given values a 1,..., a d, we can encode them as p(x) = (x a 1 )(x a 2 )... (x a d ) = i (x a i ) Advantage: we can use our knowledge of polynomials to help understand the distributions. To see that this is non-trivial, try to come up with a non-polynomial way of getting distribution that is produced by p (x).

15 page.15 Example: Finite L p -norms Given real numbers a 1,..., a d, define the k th power sum P k (a 1,..., a d ) = a k a k d Used all the time in harmonic analysis.

16 page.16 Example: Finite L p -norms Given real numbers a 1,..., a d, define the k th power sum P k (a 1,..., a d ) = a k a k d Used all the time in harmonic analysis. Also define the k th elementary symmetric polynomial E k (a 1,..., a d ) = a S = S ( [d] k ) S [d] S =k i S a i

17 page.17 Example: Finite L p -norms Given real numbers a 1,..., a d, define the k th power sum P k (a 1,..., a d ) = a k a k d Used all the time in harmonic analysis. Also define the k th elementary symmetric polynomial E k (a 1,..., a d ) = a S = Examples: 1 E 0 (a 1,..., a n ) = 1 S ( [d] k ) 2 E 1 (a 1,..., a d ) = a 1 + a a n 3 E d (a 1,..., a d ) = a 1 a 2... a d 4 E k (a 1,..., a d ) = 0 for k > d S [d] S =k i S a i

18 page.18 Newton Identities Introduction 9/50 These are connected by Newton s identities ke k (x 1,..., x d ) = k ( 1) i 1 E k i (x 1,..., x d )P i (x 1,..., x d ) i=1

19 page.19 Newton Identities Introduction 9/50 These are connected by Newton s identities ke k (x 1,..., x d ) = k ( 1) i 1 E k i (x 1,..., x d )P i (x 1,..., x d ) i=1 Elementary symmetric functions can serve similar role as p-norms.

20 page.20 Newton Identities These are connected by Newton s identities ke k (x 1,..., x d ) = k ( 1) i 1 E k i (x 1,..., x d )P i (x 1,..., x d ) i=1 Elementary symmetric functions can serve similar role as p-norms. Can we get inequalities like we do with p-norms?

21 page.21 Why real-rooted polynomials? Regular generating functions can have arbitrary coefficients. Obviously arbitrary does not have any added structure.

22 page.22 Why real-rooted polynomials? Regular generating functions can have arbitrary coefficients. Obviously arbitrary does not have any added structure. By maintaining real-rootedness, we are also maintaining structure.

23 page.23 Why real-rooted polynomials? Introduction 10/50 Regular generating functions can have arbitrary coefficients. Obviously arbitrary does not have any added structure. By maintaining real-rootedness, we are also maintaining structure. This allows us to get inequalities that would not be true in the general case.

24 page.24 Example: Bernoulli Random Variables Introduction 11/50 Exercise: Let X 1,..., X d be a collection of independent Bernoulli random variables with P [X i = 0] = p i and P [X i = 1] = 1 p i and let X = i X i. What is P [X = k]?

25 page.25 Example: Bernoulli Random Variables Introduction 11/50 Exercise: Let X 1,..., X d be a collection of independent Bernoulli random variables with P [X i = 0] = p i and P [X i = 1] = 1 p i and let X = i X i. What is P [X = k]? For each X i create the (generating) polynomial y i (x) = p i x + (1 p i ). Then m d Y (x) = y i (x) = x d k P [X = k] i=1 k=0

26 page.26 Example: Bernoulli Random Variables Exercise: Let X 1,..., X d be a collection of independent Bernoulli random variables with P [X i = 0] = p i and P [X i = 1] = 1 p i and let X = i X i. What is P [X = k]? For each X i create the (generating) polynomial y i (x) = p i x + (1 p i ). Then m d Y (x) = y i (x) = x d k P [X = k] i=1 Coefficients act like Gaussians? k=0

27 page.27 Making this more precise Recall the quadratic formula: if p(x) = Ax 2 + 2Bx + C then p(x) = (x R 1 )(x R 2 ) where R 1 = B B 2 AC A and R 2 = B + B 2 AC A

28 page.28 Making this more precise Recall the quadratic formula: if p(x) = Ax 2 + 2Bx + C then p(x) = (x R 1 )(x R 2 ) where R 1 = B B 2 AC A and R 2 = B + B 2 AC A In particular, if R 1 and R 2 are real numbers, B 2 AC.

29 page.29 Making this more precise Recall the quadratic formula: if p(x) = Ax 2 + 2Bx + C then p(x) = (x R 1 )(x R 2 ) where R 1 = B B 2 AC A and R 2 = B + B 2 AC A In particular, if R 1 and R 2 are real numbers, B 2 AC. Extends to (using derivatives!) Theorem (Newton Inequalities) Let p(x) = ) i ai x i be a degree-d polynomial with all real roots. Then ( d i a 2 i a i 1 a i+1

30 page.30 Outline Multivariate Extensions 13/50 1 Introduction 2 Multivariate Extensions Real Stable Polynomials Hyperbolic Polynomials 3 Roots 4 Interlacing families

31 page.31 Two extensions Real-rooted polynomials are somewhat restricted by the fact that they are polynomials (you only get one x). There are two well-studied multivariate extensions of real-rooted polynomials.

32 page.32 Two extensions Real-rooted polynomials are somewhat restricted by the fact that they are polynomials (you only get one x). There are two well-studied multivariate extensions of real-rooted polynomials. Actually somewhat equivalent, but theory is more developed in different contexts. 1 Real stable polynomials: better construction properties 2 Hyperbolic polynomials: more refined convexity properties

33 Multivariate Extensions 14/50 page.33 Two extensions Real-rooted polynomials are somewhat restricted by the fact that they are polynomials (you only get one x). There are two well-studied multivariate extensions of real-rooted polynomials. Actually somewhat equivalent, but theory is more developed in different contexts. 1 Real stable polynomials: better construction properties 2 Hyperbolic polynomials: more refined convexity properties Important trait: both are closed under diagonalization (z 1,..., z d ) (x, x,..., x) Allows us to start with one of these and prove things about univariate polynomials

34 page.34 What do I mean by construction properties? Multivariate Extensions 15/50 The issue with real-rooted polynomials is that it is hard to see how to get from one to another.

35 page.35 Parking garage phenomenon Multivariate Extensions 16/50 The issue with real-rooted polynomials is that it is hard to see how to get from one to another. Unless you consider them to be a projection of higher dimensional objects.

36 page.36 Real Stable Polynomials Multivariate Extensions 17/50 Real stable polynomials are useful in this scenario. A polynomial p is called real stable if all coefficients are real and p(x 1,..., x d ) 0 whenever I(x i ) > 0 for all i (there are no zeroes where all coordinates are in the upper half plane).

37 page.37 Real Stable Polynomials Multivariate Extensions 17/50 Real stable polynomials are useful in this scenario. A polynomial p is called real stable if all coefficients are real and p(x 1,..., x d ) 0 whenever I(x i ) > 0 for all i (there are no zeroes where all coordinates are in the upper half plane). Univariate polynomials are real-rooted if and only if they are real stable.

38 page.38 Closure Multivariate Extensions 18/50 Stable polynomials have nice closure properties: For f (z 1, z 2,... z n ) real stable, the following operations preserve stability: 1 Permutation: for σ Σ n, f f (z σ(1),..., z σ(n) ). 2 Scaling: for a > 0, f f (az 1, z 2,..., z n ). 3 Diagonalization: f f (z 1, z 1, z 3..., z n ) 4 Specialization: for I(a) 0, f f (a, z 2,..., z n ). 5 Inversion: if deg z1 (f ) = d, f (z 1 ) d f ( 1/z 1, z 2,..., z d ) 6 Translation: f g(t, z 1,..., z d ) = f (z 1 + t, z 2,..., z n ) 7 Differentiation: f f / z 1

39 page.39 Example: Permanents The permanent of a d d square matrix A is defined as Multivariate Extensions 19/50 perm(a) = d A(i, σ(i)) σ Σ d i=1

40 page.40 Example: Permanents The permanent of a d d square matrix A is defined as Multivariate Extensions 19/50 perm(a) = d A(i, σ(i)) σ Σ d i=1 Similar to the determinant but without the annoying minuses.

41 page.41 Example: Permanents The permanent of a d d square matrix A is defined as Multivariate Extensions 19/50 perm(a) = d A(i, σ(i)) σ Σ d i=1 Similar to the determinant but without the annoying minuses. We can use real stable polynomials to help understand permanents: ( d d ) Q(t 1,..., t d ) = a i t i j=1 is a real stable polynomial, and therefore so is perm(a) = i=1 d t 1,... t d Q(t 1,..., t d ).

42 page.42 Linear Transformations For a linear differential operator T = c α,β z α β α,β N n define its Weyl polynomial F T (z, w) = c α,β z α w β α,β N n

43 page.43 Linear Transformations For a linear differential operator T = c α,β z α β α,β N n define its Weyl polynomial F T (z, w) = c α,β z α w β α,β N n Call T stability preserving if T [p] is real stable for all real stable p.

44 page.44 Linear Transformations Multivariate Extensions 20/50 For a linear differential operator T = c α,β z α β α,β N n define its Weyl polynomial F T (z, w) = c α,β z α w β α,β N n Call T stability preserving if T [p] is real stable for all real stable p. Theorem (Borcea and Brändén) T is stability preserving if and only if F T (z, w) is real stable

45 page.45 Negative association Let µ be a probability distribution on 2 [n].

46 page.46 Negative association Let µ be a probability distribution on 2 [n]. A function f is called non-increasing (non-decreasing) if S T f (S) ( )f (T )

47 page.47 Negative association Let µ be a probability distribution on 2 [n]. A function f is called non-increasing (non-decreasing) if S T f (S) ( )f (T ) Set functions f and g are called disjoint if f (S) = f (S A) and g(s) = g(s A) for some A (the variables that f uses and g uses are disjoint subsets of [n]). µ is said to be negatively associated if for all disjoint non-increasing functions f, g E [f (X )g(x )] E [f (X )] E [g(x )]

48 page.48 Negative association Let µ be a probability distribution on 2 [n]. A function f is called non-increasing (non-decreasing) if S T f (S) ( )f (T ) Set functions f and g are called disjoint if f (S) = f (S A) and g(s) = g(s A) for some A (the variables that f uses and g uses are disjoint subsets of [n]). µ is said to be negatively associated if for all disjoint non-increasing functions f, g E [f (X )g(x )] E [f (X )] E [g(x )] Example: Uniform distribution on spanning trees of a graph

49 page.49 Characterization of NA Theorem (Brändén) A distribution µ is negatively associated if and only if the polynomial G µ = E [x S] = P [X = S] x i S 2 [n] i S is real stable.

50 page.50 Characterization of NA Theorem (Brändén) A distribution µ is negatively associated if and only if the polynomial G µ = E [x S] = P [X = S] x i S 2 [n] i S is real stable. Implies tight concentration and convex-geometry-type inequalities: Lemma Let µ be negatively associated and let f i be non-decreasing functions. Then E f i (X i ) E [f i (X i )] i [n] i [n]

51 page.51 Hyperbolic polynomials Let x = (x 1,..., x d ). A multivariate homogeneous polynomial p( x) is said to be hyperbolic in direction e if 1 p( e) 0 2 The univariate polynomial q y (t) = p( y + t e) is real-rooted for all y R d If this makes no intuitive sense, wait for the picture.

52 page.52 Hyperbolic polynomials Multivariate Extensions 23/50 Let x = (x 1,..., x d ). A multivariate homogeneous polynomial p( x) is said to be hyperbolic in direction e if 1 p( e) 0 2 The univariate polynomial q y (t) = p( y + t e) is real-rooted for all y R d If this makes no intuitive sense, wait for the picture. Examples: 1 p(x, y) = xy is hyperbolic in the direction (1, 1) 2 E k ( x) is hyperbolic in the direction (1,..., 1) 3 p(x, y, z) = x 2 y 2 z 2 is hyperbolic in the direction (1, 0, 0) 4 For Hermitian A, p(a) = det [A] is hyperbolic in the direction I

53 Multivariate Extensions 24/50 page.53 Zero surfaces Theorem (Helton Vinnikov) The zero surfaces of a degree d hyperbolic polynomial form d 2 nested ovaloids and a pseudo-hyperplane (if d is odd)

54 page.54 Characteristic Polynomials Hyperbolic polynomials can be viewed as a generalization of characteristic polynomials (for matrices).

55 page.55 Characteristic Polynomials Hyperbolic polynomials can be viewed as a generalization of characteristic polynomials (for matrices). Multivariate Extensions 25/50 Any hyperbolic polynomial can be factored: p( x + t e) = p( e) j (t + λ j ( e, x)) with λ = (λ 1,..., λ d ) ordered as λ 1... λ d.

56 page.56 Characteristic Polynomials Hyperbolic polynomials can be viewed as a generalization of characteristic polynomials (for matrices). Multivariate Extensions 25/50 Any hyperbolic polynomial can be factored: p( x + t e) = p( e) j (t + λ j ( e, x)) with λ = (λ 1,..., λ d ) ordered as λ 1... λ d. The λ j are called eigenvalues. When p( x) = det [ x] and e = I, then p( x + t e) = det [ti + x] = i (t + λ i ) which is (the negative of) the matrix version of eigenvalues.

57 page.57 Hyperbolicity Cones The hyperbolicity cone of p (with respect to e) is the set and is denoted Λ ++ (p, e). { x : λ 1 ( x) > 0}

58 page.58 Hyperbolicity Cones Multivariate Extensions 26/50 The hyperbolicity cone of p (with respect to e) is the set { x : λ 1 ( x) > 0} and is denoted Λ ++ (p, e). Theorem (Gårding) Let p be hyperbolic in the direction e. Then 1 p is hyperbolic in the direction e for all e Λ ++ (p, e) 2 e Λ ++ (p, e) e Λ ++ (p, e ) 3 D e p is hyperbolic in direction e and Λ ++ (p, e) Λ ++ (D e p, e) 4 Λ ++ (p, e) is convex Inequalities exploit this convexity.

59 page.59 Some examples Theorem (Bauschke, Güler, Lewis, Sendov) Let p be degree-d and hyperbolic in the direction e and let f : R d [, + ] be convex and symmetric. Let µ( x) = λ( e, x) Then f µ is convex.

60 page.60 Some examples Theorem (Bauschke, Güler, Lewis, Sendov) Let p be degree-d and hyperbolic in the direction e and let f : R d [, + ] be convex and symmetric. Let µ( x) = λ( e, x) Then f µ is convex. Theorem (Kummer, Plaumann, Vinzant) Let p be degree-d, square-free, and hyperbolic in the direction e and let h be any degree-(d 1) polynomial such that for all x p( x) = 0 D e p( x) h( x) 0 Then D e p h p D e h (everywhere).

61 page.61 Lax conjecture Multivariate Extensions 28/50 Theorem (Lewis, Parrilo, Ramana) Let h(x, y, z) be degree-d and hyperbolic in the direction (e 1, e 2, e 3 ) such that h(e 1, e 2, e 3 ) = 1. Then there exist symmetric d d matrices A, B, C such that e 1 A + e 2 B + e 3 C = I and h(x, y, z) = det [xa + yb + zc].

62 page.62 Lax conjecture Multivariate Extensions 28/50 Theorem (Lewis, Parrilo, Ramana) Let h(x, y, z) be degree-d and hyperbolic in the direction (e 1, e 2, e 3 ) such that h(e 1, e 2, e 3 ) = 1. Then there exist symmetric d d matrices A, B, C such that e 1 A + e 2 B + e 3 C = I and h(x, y, z) = det [xa + yb + zc]. Uses theory developed by Helton and Vinnikov.

63 page.63 Lax conjecture Multivariate Extensions 28/50 Theorem (Lewis, Parrilo, Ramana) Let h(x, y, z) be degree-d and hyperbolic in the direction (e 1, e 2, e 3 ) such that h(e 1, e 2, e 3 ) = 1. Then there exist symmetric d d matrices A, B, C such that e 1 A + e 2 B + e 3 C = I and h(x, y, z) = det [xa + yb + zc]. Uses theory developed by Helton and Vinnikov. Not true for more than 3 variables.

64 page.64 More inequalities Multivariate Extensions 29/50 Let p( x) be degree-d and hyperbolic in direction e. Let a 1,..., a d Λ ++ (p, e). Denoting the d th directional derivative of p in the directions a 1,..., a d as D d p[ a 1,..., a d ], we have: Theorem (Gårding) D d p( x)[ a 1,..., a d ] d! d p( a i ) 1/d i=1 Define Cap(p) = inf{p(t 1 a 1,..., t d a d ) i t i = 1, t i > 0} Theorem (Gurvits) D d p( x)[ a 1,..., a d ] d! Cap(p) d d

65 page.65 Equivalence From stable to hyperbolic: Lemma Let p(x 1,..., x n ) be a polynomial. Then p is stable if and only if y d p(x 1 /y,..., x n /y) is hyperbolic with respect to (1,..., 1, 0).

66 page.66 Equivalence From stable to hyperbolic: Lemma Let p(x 1,..., x n ) be a polynomial. Then p is stable if and only if y d p(x 1 /y,..., x n /y) is hyperbolic with respect to (1,..., 1, 0). From hyperbolic to stable: Lemma (Borcea and Brändén) Let h( x) be a degree-d homogeneous polynomial, and let a, b be such that h( a)h( b) 0. The following are equivalent 1 h is hyperbolic with respect to a, and b Λ ++ (h, a). 2 The bivariate polynomial f (s, t) = h( x + s a + t b) is real stable for all x.

67 page.67 Outline Roots 31/50 1 Introduction 2 Multivariate Extensions Real Stable Polynomials Hyperbolic Polynomials 3 Roots 4 Interlacing families

68 page.68 Roots Roots 32/50 So far we have seen that having the property of real roots (or one of the multivariate extensions) gives interesting inequalities on the coefficients and the values.

69 page.69 Roots Roots 32/50 So far we have seen that having the property of real roots (or one of the multivariate extensions) gives interesting inequalities on the coefficients and the values. We have still not talked about the roots themselves.

70 page.70 Roots Roots 32/50 So far we have seen that having the property of real roots (or one of the multivariate extensions) gives interesting inequalities on the coefficients and the values. We have still not talked about the roots themselves. For good reason... The technology we have seen so far can be used to prove the realness of roots, but not the location of them!

71 page.71 Some motivation Let s begin with a simple question. You have a symmetric d d matrix A and you add a rank-1 matrix uu T. What happens to the eigenvalues?

72 page.72 Some motivation Let s begin with a simple question. You have a symmetric d d matrix A and you add a rank-1 matrix uu T. What happens to the eigenvalues? Some trivial observations: The eigenvalues have to go up The amount λ i goes up should depend on u, v i Traces add Averaging over all possible u should move all eigenvalues the same amount Let s try looking at this through the frames of polynomials.

73 page.73 Linear algebra review Lemma (Matrix Determinant Lemma) Let A be an invertible d d matrix and let u, v R d. Then [ ] det A + uv T = det [A] (1 + u T A 1 v) Lemma (Spectral Decomposition) Let A be a d d symmetric matrix. Then there exists real numbers λ 1,..., λ d and an orthonormal basis v 1,..., v d such that A = i λ i v i v T i

74 page.74 The characteristic polynomial The characteristic polynomial of a d d matrix A is the polynomial Roots 35/50 χ A (x) = det [xi A]

75 page.75 The characteristic polynomial The characteristic polynomial of a d d matrix A is the polynomial Roots 35/50 If the spectral decomposition of A is χ A (x) = det [xi A] A = i λ i v i v T i then d χ A (x) = (x λ i ) i=1 and so the spectral decomposition of xi A is xi A = i (x λ i )v i v T i

76 page.76 Adding a rank-1 operator Roots 36/50 Using the matrix determinant lemma, we can see what happens to the eigenvalues of a matrix when a rank-1 operator is added: [ ] ( ) det xi (A + uu T ) = det [xi A] 1 u T (xi A) 1 u Decompose xi A = i (x λ i )v i v T i where the v i are orthonormal. Then (xi A) 1 = i ( 1 x λ i ) v i v T i

77 page.77 New roots Roots 37/50 So we have ( [ ] det xi (A + uu T ) = det [xi A] 1 i ) v i, u 2 x λ i

78 page.78 New roots Roots 37/50 So we have ( [ ] det xi (A + uu T ) = det [xi A] 1 i ) v i, u 2 x λ i The roots of det [ xi (A + uu T ) ] are either roots of det [xi A] or solutions to v i, u 2 = 1 x λ i i

79 page.79 New roots Roots 37/50 So we have ( [ ] det xi (A + uu T ) = det [xi A] 1 i ) v i, u 2 x λ i The roots of det [ xi (A + uu T ) ] are either roots of det [xi A] or solutions to v i, u 2 = 1 x λ i i Let s assume that v i, u 2 > 0 for all i and that all λ i are distinct.

80 page.80 Digging deeper Consider the equation f (x) := i v i, u 2 x λ i = 1

81 page.81 Digging deeper Consider the equation f (x) := i v i, u 2 x λ i = 1 Notice that for all i, lim x λ i f (x) = and lim x λ + i f (x) = + so by continuity, f (x) = 1 must have a solution between λ i and λ i+1.

82 page.82 Digging deeper Consider the equation f (x) := i v i, u 2 x λ i = 1 Notice that for all i, lim x λ i f (x) = and lim x λ + i f (x) = + so by continuity, f (x) = 1 must have a solution between λ i and λ i+1. In other words, if λ i are the new eigenvalues, we have λ 1 λ 1 λ 2 λ d 1 λ d 1 λ d λ d This phenomenon is known as interlacing (and is definitely not a trivial observation).

83 page.83 Billiard Balls Roots 39/50 This has interesting consequences. Consider adding the operator αuu T for different α R.

84 page.84 Billiard Balls Roots 39/50 This has interesting consequences. Consider adding the operator αuu T for different α R. As α increases, the only eigenvalue that has room to move is the top one.

85 page.85 Billiard Balls Roots 39/50 This has interesting consequences. Consider adding the operator αuu T for different α R. As α increases, the only eigenvalue that has room to move is the top one. As a result, we cannot hope to understand the maximal eigenvalue by only keeping track of the maximum eigenvalue. And (as we will see tomorrow) understanding maximum eigenvalues seems to be a useful thing to be able do.

86 page.86 Outline Interlacing families 40/50 1 Introduction 2 Multivariate Extensions Real Stable Polynomials Hyperbolic Polynomials 3 Roots 4 Interlacing families

87 page.87 Adding randomness We have seen that polynomials can help us understand the process of adding rank-1 operators. But what if we add a random rank-1 operator?

88 page.88 Adding randomness We have seen that polynomials can help us understand the process of adding rank-1 operators. But what if we add a random rank-1 operator? One thing we can do is look at what happens in expectation.

89 page.89 Adding randomness We have seen that polynomials can help us understand the process of adding rank-1 operators. But what if we add a random rank-1 operator? One thing we can do is look at what happens in expectation. As before, let s look at this through the frames of polynomials.

90 page.90 In expectation Interlacing families 42/50 Assume we have an operator (matrix) A and we add a random operator to it that takes the values {uu T, vv T } uniformly.

91 page.91 In expectation Interlacing families 42/50 Assume we have an operator (matrix) A and we add a random operator to it that takes the values {uu T, vv T } uniformly. We will (perhaps naively) consider the polynomial Why is this naive? p(x) = 1 2 χ A+vv T (x) χ A+uu T (x)

92 page.92 In expectation Interlacing families 42/50 Assume we have an operator (matrix) A and we add a random operator to it that takes the values {uu T, vv T } uniformly. We will (perhaps naively) consider the polynomial Why is this naive? p(x) = 1 2 χ A+vv T (x) χ A+uu T (x) Adding polynomials is a function of the coefficients and we are interested in the roots. In general, it is easy to get the coefficients from the roots but hard to get the roots from the coefficients.

93 page.93 But wait... Interlacing families 43/50 But we have already seen that we can say something about the case of adding rank-1 operators (interlacing). Can we do something similar here?

94 page.94 But wait... Interlacing families 43/50 But we have already seen that we can say something about the case of adding rank-1 operators (interlacing). Can we do something similar here? Let s formalize the interlacing property we saw before.

95 page.95 Interlacing polynomials Interlacing families 44/50 Let p be a degree-n real-rooted polynomial and q a degree-(n 1) real-rooted polynomial p(x) = n n 1 (x α i ) and q(x) = (x β i ) i=1 with α 1 α n and β 1 β n 1 i=1

96 page.96 Interlacing polynomials Interlacing families 44/50 Let p be a degree-n real-rooted polynomial and q a degree-(n 1) real-rooted polynomial p(x) = n n 1 (x α i ) and q(x) = (x β i ) i=1 with α 1 α n and β 1 β n 1 We say q interlaces p if α 1 β 1 α 2 α d 1 β n 1 α n. Think: The roots of q separate the roots of p i=1

97 page.97 Interlacing polynomials Interlacing families 44/50 Let p be a degree-n real-rooted polynomial and q a degree-(n 1) real-rooted polynomial p(x) = n n 1 (x α i ) and q(x) = (x β i ) i=1 with α 1 α n and β 1 β n 1 We say q interlaces p if α 1 β 1 α 2 α d 1 β n 1 α n. Think: The roots of q separate the roots of p Example 1: p (x) interlaces p(x) Example 2: If p has no multiple roots (and largest root R), then let q = p/(x R). Then q(x + ɛ) interlaces p(x) i=1

98 page.98 Common Interlacers We say that two degree n polynomials p and r have a common interlacer if there exists a q such that q interlaces both p and r simultaneously. Think: the roots of q split up R into n intervals, each of which contains exactly one root of p and one root of r

99 page.99 Common Interlacers We say that two degree n polynomials p and r have a common interlacer if there exists a q such that q interlaces both p and r simultaneously. Think: the roots of q split up R into n intervals, each of which contains exactly one root of p and one root of r Example 1: If p has no multiple roots, then p(x) and p(x) + ɛ have a common interlacer (p (x)) Example 2: If p has no multiple roots, then p(x) and p(x + ɛ) have a common interlacer (p (x))

100 page.100 Common Interlacers We say that two degree n polynomials p and r have a common interlacer if there exists a q such that q interlaces both p and r simultaneously. Think: the roots of q split up R into n intervals, each of which contains exactly one root of p and one root of r Example 1: If p has no multiple roots, then p(x) and p(x) + ɛ have a common interlacer (p (x)) Example 2: If p has no multiple roots, then p(x) and p(x + ɛ) have a common interlacer (p (x)) How does this help?

101 Interlacing families 46/50 page.101 A lemma Lemma Let f and g be monic polynomials. Assume there exists a point c R such that f and g each has exactly one real root larger than c (call these the extreme roots ). Then the largest real root of f + g lies between these extreme roots.

102 page.102 A lemma Lemma Let f and g be monic polynomials. Assume there exists a point c R such that f and g each has exactly one real root larger than c (call these the extreme roots ). Then the largest real root of f + g lies between these extreme roots. Proof. By picture Note: if f and g have a common interlacer (say q), then setting c = q d 1 satisfies the lemma! Interlacing families 46/50

103 page.103 Without c to anchor

104 page.104 Without c to anchor

105 page.105 So what can we say? Recall our goal was to understand the roots of p(x) = 1 2 χ A+vv T (x) χ A+uu T (x) = 1 2 q(x) r(x)

106 page.106 So what can we say? Recall our goal was to understand the roots of Interlacing families 48/50 p(x) = 1 2 χ A+vv T (x) χ A+uu T (x) = 1 2 q(x) r(x) We will say that {p, q, r} form an interlacing family if 1 p, q and r are all real rooted 2 q and r have a common interlacer

107 page.107 So what can we say? Recall our goal was to understand the roots of Interlacing families 48/50 p(x) = 1 2 χ A+vv T (x) χ A+uu T (x) = 1 2 q(x) r(x) We will say that {p, q, r} form an interlacing family if 1 p, q and r are all real rooted 2 q and r have a common interlacer Corollary If {p, q, r} forms an interlacing family, then there exists an assignment of our random variable (either uu T or vv T ) such that the largest root of the resulting polynomial is at most the largest root of p(x) (the expected polynomial).

108 page.108 Interlacing for free Fortunately, the interlacing follows directly from a well-known lemma: Lemma (Fisk, among others) Let f, g be polynomials of the same degree such that every λf + (1 λ)g is real-rooted for all λ [0, 1]. Then f and g have a common interlacer.

109 page.109 Interlacing for free Fortunately, the interlacing follows directly from a well-known lemma: Lemma (Fisk, among others) Let f, g be polynomials of the same degree such that every λf + (1 λ)g is real-rooted for all λ [0, 1]. Then f and g have a common interlacer. Recall (again) our equation p(x) = 1 2 χ A+vv T (x) χ A+uu T (x)

110 page.110 Interlacing for free Interlacing families 49/50 Fortunately, the interlacing follows directly from a well-known lemma: Lemma (Fisk, among others) Let f, g be polynomials of the same degree such that every λf + (1 λ)g is real-rooted for all λ [0, 1]. Then f and g have a common interlacer. Recall (again) our equation p(x) = 1 2 χ A+vv T (x) χ A+uu T (x) If we could show that p(x) = λχ A+vv T (x) + (1 λ)χ A+uu T (x) for all λ [0, 1], then we would get the interlacing for free.

111 page.111 Full Circle Interlacing families 50/50 So what have we accomplished?

112 page.112 Full Circle Interlacing families 50/50 So what have we accomplished? We saw (from previous results) that real-rootedness gave inequalities on coefficients of polynomials (making them useful generating functions).

113 page.113 Full Circle Interlacing families 50/50 So what have we accomplished? We saw (from previous results) that real-rootedness gave inequalities on coefficients of polynomials (making them useful generating functions). We also saw (from previous results) that real-rootedness gave inequalities on the values of polynomials.

114 page.114 Full Circle Interlacing families 50/50 So what have we accomplished? We saw (from previous results) that real-rootedness gave inequalities on coefficients of polynomials (making them useful generating functions). We also saw (from previous results) that real-rootedness gave inequalities on the values of polynomials. But now (via interlacing families) we can use real-rootedness to relate the roots of expected polynomials with the possible realizations of the polynomial!

115 page.115 Full Circle So what have we accomplished? We saw (from previous results) that real-rootedness gave inequalities on coefficients of polynomials (making them useful generating functions). We also saw (from previous results) that real-rootedness gave inequalities on the values of polynomials. But now (via interlacing families) we can use real-rootedness to relate the roots of expected polynomials with the possible realizations of the polynomial! How that could be useful remains to be seen...