Non-Commutative Subharmonic and Harmonic Polynomials

Transcription

1 Non-Commutative Subharmonic and Harmonic Polynomials by Joshua A. Hernandez Dr. J. William Helton, Honors Advisor UCSD Department of Mathematics June 9,

2 Contents 1 Introduction Non-Commutative Algebra in Engineering NC Polynomials NC-Symmetric Polynomials Matrix Positivity Non-Commutative Differentiation Non-Commutative Convexity Non-Commutative Laplacian and Subharmonicity Main Theorem (Two Variables) 8 4 Existence Proofs Product Rules for Derivatives Product Rule for DirD The Laplacian of a Product Formulas Involving Derivatives of γ d Harmonics degree > 2: Proof of Theorem 1 part (1) Subharmonics degree > 4: Proof of Theorem 1 (2)(3) Subharmonics of Odd Degree Classification when Degree is Four or Less The Matrix Representation Commutative Coefficients Non-commutative coefficients Convexity of p vs. Positivity of its Middle Matrix The Zeroes Lemma The Spanning Polynomials for Subharmonics The Laplacian Matrix Decomposition Conditions for Positivity Future Work 22 7 Acknowledgments 24 2

3 1 Introduction 1.1 Non-Commutative Algebra in Engineering Inequalities, involving polynomials in matrices and their inverses, and associated optimization problems have become very important in engineering. When the said polynomials are matrix convex, interior point methods apply directly. In the last few years, the approaches that have been proposed in the field of optimization and control theory based on linear matrix inequalities and semidefinite programming have become very important and promising, since the same framework can be used for a large set of problems. Matrix inequalities provide a nice setup for many engineering and related problems, and if they are convex the optimization problem is well behaved and interior point methods provide efficient algorithms which are effective on moderate sized problems. Unfortunately, the class of convex polynomials in matrix variables is very small - it has been proven, in fact, that they are all of degree two or less [HS04]. It is in our interest, then, to discover conditions similar to convexity, but not as restrictive, and to characterize these classes thoroughly. 2 NC Polynomials Definition 2.1 Non-Commutative Monomials A non-commutative monomial m of degree d on the indexed free variables x 1,..., x g is a product x a1 x a2 x ad of these variables, corresponding to a unique sequence of a i. A convenient convention is to write m = x w, where w is the d-tuple a 1,...,a d. We treat w as a word on the alphabet [g] ={1,..., g} and m, consequently, as a word on x = {x 1,..., x g }. Some word-conventions: w = d the length of w (w) i = a i the i th letter of w w T = a d,...,a 1 the transpose of w φ = the empty word (word of length zero) Finally, for any two words w 1 = a 1,..., a d1 and w 2 = b 1,..., b d2, we have w 1 w 2 = a 1,..., a d1,b 1,..., b d2, called w 1 concatenate w 2. 3

4 Definition 2.2 Non-Commutative Polynomials A non-commutative polynomial p(x) = p(x 1,...,x g ) is the sum of noncommutative monomials (on g variables) m, to which we associate scalar coefficients A m R ; m is a word ranging over a given set M p. The degree of p is equal to the length of the longest word in M p. p(x) = Example 2.1 A non-commutative polynomial m M p A m m. (1) p(x) =p(x 1,x 2 )=x 2 1x 2 x 1 +x 1 x 2 x 2 1+x 1 x 2 x 2 x 1 +7 (on commutative variables, this would be equivalent to 2x 2 1x 2 + 7). In this example, M p = { x 1,x 1,x 2,x 1, x 1,x 2,x 1,x 1, x 1,x 2, x 2,x 1,φ}. 2.1 NC-Symmetric Polynomials The transpose of a polynomial p, denoted p T, has the following properties: (1) (p T ) T = p (2) (p 1 + p 2 ) T = p T 1 + p T 2 (3) (αp) T = αp T (α R) (4) (p 1 p 2 ) T = p T 2 p T 1. In this paper, we shall consider only polynomials in symmetric variables. That is, we consider variables x i where x T i = x i. Transposition of a polynomial in such variables amounts to reversing the spelling of each of its terms. We know that Assuming that x T a 1 = x a1 (x a1...x ad 1 ) T =x ad 1...x a1,wehave (x a1...x ad ) T =x T a d (x a1...x ad 1 ) T =x ad (x ad 1...x a1 )=x ad...x a1. 4

5 Returning to our indexed-word notation, we find the identity (x w ) T = x wt. (3) Symmetric (or self-adjoint) polynomials are those that are equal to their transposes. We will consider this class of functions in our investigation. 2.2 Matrix Positivity Definition 2.3 Matrix Positive Symmetric Polynomials A symmetric polynomial p(x) =p(x 1,...,x g ) is considered matrix-positive over a domain D of square symmetric matrices X =(X 1,...,X g ) iff, for any X D, we have that p(x) is a positive-semidefinite matrix. Note that if p(x) = 1, then p(x) =I D. For instance: Example 2.2 p(x) =x x 2 g is matrix-positive. Substituting in symmetric matrices X i for x i, we have for any vector v v T (X X 2 g)v =v T X 2 1v+...+v T X 2 gv =(X 1 v) T (X 1 v)+...+(x g v) T (X g v) = X 1 v X g v 2 0, that is, p(x) is positive-semidefinite, for all X. matrix-positive. By definition p(x) is 2.3 Non-Commutative Differentiation For our non-commutative purposes, we take directional derivatives in x i with regard to an indeterminate direction parameter h [CHSY03]. DirD[p(x 1,...,x g ),x i,h]:= d dt [p(x 1,...,(x i +th),...,x g )] t=0. (4) We say that this is the directional deriative of p(x) =p(x 1,...,x g )onx i in the direction h. Example 2.3 The directional derivative 5

6 DirD[x 2 1x 2,x 1,h]= d[(x dt 1 + th) 2 x 2 ] t=0 = d dt [x2 1x 2 + thx 1 x 2 + tx 1 hx 2 + t 2 h 2 x 2 ] t=0 =[hx 1 x 2 + x 1 hx 2 +2th 2 x 2 ] t=0 = hx 1 x 2 + x 1 hx 2. As this example shows, the directional derivative of p on x i in the direction h is the sum of the terms produced by replacing one instance of x i with h. Lemma 2.1 Linearity of the directional derivative of NC polynomials DirD[ap(x)+bq(x),x i,h]=adird[p(x),x i,h]+bdird[q(x),x i,h]. Proof: DirD[ap(x)+bq(x),x i,h] = d dt [(ap(x)+bq(x))(x 1,...,(x i +th),...,x g )] t=0 = d dt [ap(x 1,...,(x i +th),...,x g )+bq(x 1,...,(x i +th),...,x g )] t=0. By linearity of the d dt operator, =(a d dt [p(x 1,...,(x i +th),...,x g )] + b d dt [q(x 1,...,(x i +th),...,x g )]) t=0 = a d dt [ap(x 1,...,(x i +th),...,x g )] t=0 + b d dt [bq(x 1,...,(x i +th),...,x g )] t=0 = a DirD[p(x),x i,h]+bdird[q(x),x i,h]. 2.4 Non-Commutative Convexity The non-commutative Hessian is defined as: NCHessian[p(x 1,...,x g ),{x 1,η 1 },...,{x g,η g }]:= d2 dt [p(x 1+tη 2 1,...,x g +t g η g )] t=0. Note that this is composed of several independent direction parameters, η i and that if p is a polynomial, then its Hessian is a polynomial in x and η which is homogeneous of degree 2 in η. Such notation is unique to this subsection. A non-commutative polynomial is considered convex wherever its Hessian is matrix-positive. Equivalently, 6

7 Definition 2.4 Geometrical NC Symmetric Convexity A polynomial p(x) = p(x 1,...,x d ) is convex over a convex domain D of ordered d-tuples X = (X 1,...,X d ) of square matrices if and only if, for every X and Y in D, ( ) 1( ) X+Y p(x)+p(y) p 2 2 is positive-semidefinite. It is proved in [HM98] that convexity is equivalent to geometric convexity. A crucial fact regarding these polynomials (proven by Helton, et. al [HS04]) is that they are all of degree two or less. The commutative analog of this directional Hessian is the quadratic function H ( p(x) ) η 1 η 1.. (7) η g where H ( p(x) ) is the Hessian matrix: x1 x 1 p(x) x1 x g p(x) (8) xgx1 p(x) xgxg p(x) If this Hessian is positive- (resp. negative-) semidefinite at some critical point (x 1,...,x g ), then f has a local minimum (resp. maximum) at that point, and f is said to be concave (resp. convex ) there. This corresponds, somewhat, to our intuitive physical notions of convexity. 2.5 Non-Commutative Laplacian and Subharmonicity The condition of subharmonicity is much less restrictive than that of convexity. It is determined by a polynomial s non-commutative Laplacian, which we define as: g Lap[p(x 1,...,x g ),h]:= DirD[DirD[p(x), {x i,h}],{x i,h}] (9) = i=1 η g g d 2 dt [p(x 1,...,(x 2 i +th),...,x g )] t=0. (10) i=1 7

8 Note that (9) defines Lap as a finite sum of compositions of linear functions (DirD); thus, it too must be linear. Analogous to the directional Laplacian, in commutative variables, is: where ( p(x) ) is the standard Laplacian, namely: ( p(x) ) h 2 (11) ( p(x) ) := g xi x i p(x). (12) i=1 A polynomial is considered harmonic if its Laplacian is zero, and subharmonic if its Laplacian is matrix-positive ( properly subharmonic is used to describe a polynomial which is subharmonic but not harmonic - that is, having a nonzero, matrix-positive Laplacian). 3 Main Theorem (Two Variables) For our special homogeneous polynomials on two variables, define where i is the imaginary number. γ := x 1 + ix 2 (13) Theorem 1 For d>2, the homogeneous NC symmetric polynomials in two symmetric variables which are (1) harmonic of degree d include the linear combinations of Re(γ d ) and Im(γ d ), (2) subharmonic of degree 2d, include the linear combinations: c 0 [Re(γ d )] 2 + c 1 Re(γ 2d )+c 2 Im(γ 2d ) = c 0 [Im(γ d )] 2 +(c 0 +c 1 ) Re(γ 2d )+c 2 Im(γ 2d ) (15) where c 0 0, 8

9 (3) subharmonic of odd degree do not exist. All NC symmetric harmonic polynomials of degree 2, and the subharmonic polynomials of degree 4 or less are listed: Degree 2: Spanning Polynomial A 1 x A 2 x 2 2 +A 3 (x 1 x 2 + x 2 x 1 ) Characteristics Region of Subharmonicity A 1 + A 2 > 0 Region of Harmonicity A 1 + A 2 =0 Degree 3: Spanning Polynomial A 1 (x 3 1 x 1 x 2 2 x 2 x 2 1) +A 2 x 2 x 1 x 2 + A 3 x 1 x 2 x 1 +A 4 (x 3 2 x 2 1x 2 x 2 x 2 1) Characteristics Region of Subharmonicity (A 1 + A 2 )x 1 +(A 3 +A 4 )x 2 >0 Region of Harmonicity A 1 + A 2 = A 3 + A 4 =0 1 Degree 4: Spanning Polynomial A 1 (x 4 1 x 2 1x 2 2 x 2 2x x 4 2) +A 2 x 1 x 2 2x 1 + A 3 x 2 x 2 1x 2 +A 4 (x 1 x 2 x 1 x 2 + x 2 x 1 x 2 x 1 ) +A 5 (x 2 2x 1 x 2 x 3 1x 2 x 2 x x 2 x 1 x 2 2) +A 6 (x 2 1x 2 x 1 + x 1 x 2 x 2 1 x 1 x 3 2 x 3 2x 1 ) Characteristics Region of Subharmonicity (A 1 + A 2 )(A 1 + A 3 ) > (A 1 + A 4 ) 2 +(A 5 +A 6 ) 2 A 1 +A 2 >0 or, equivalently, A 1 + A 3 > 0 Region of Harmonicity A 1 + A 2 = A 1 + A 3 = A 1 + A 4 =0 A 5 =A 6 =0 Conjecture 3.1 The subharmonic and harmonic polnomials are none other than those described above. 1 In the degree 3 case, note that the subharmonic region depends on x 1 and x 2. Therefore, there is no polynomial of degree three which is subharmonic over all values of x 1 and x 2 9

10 This is a result we believe we have proven but there was not time to write the proof out carefully in this thesis. 10

11 4 Existence Proofs 4.1 Product Rules for Derivatives Product Rule for DirD Lemma 4.1 The product rule for the directional derivative of NC polynomials is DirD[p 1 p 2,x i,h] = DirD[p 1,x i,h]p 2 + p 1 DirD[p 2,x i,h]. Proof: The directional derivative DirD[m, x i,h] of a product m = m 1 m 2 of non-commutative monomials m 1 and m 2 is the sum of terms produced by replacing one instance of x i in m by h. This sum can be divided into two parts: µ 1, the sum of terms whose h lie in the first m 1 letters, i.e. DirD[m 1,x i,h]m 2 µ 2, the sum of terms whose h lie in the last m 2 letters, i.e. m 1 DirD[m 2,x i,h]. Therefore DirD[m 1 m 2,x i,h] = DirD[m 1,x i,h]m 2 +m 1 DirD[m 2,x i,h]. Noting the linearity of the directional derivative, we can extend this product rule to the product of any two non-commutative polynomials p 1 and p 2. [( )( DirD[p 1 p 2,x i,h] = DirD A m1 m 1 m 1 M p1 = A m1 A m2 DirD[m 1 m 2,x i,h] m 1 M p1 m 2 M p2 = m 2 M p2 A m2 m 2 ) ],x i,h A m1 A m2 DirD[m 1,x i,h]m 2 +A m1 A m2 m 1 DirD[m 2 x i,h] m 1 M p1 m 2 M p2 [ ] [ ] = DirD A m1 m 1,x i,h A m2 m 2 + A m1 m 1 DirD A m2 m 2,x i,h m 1 M p1 m 2 M p2 m 1 M p1 m 2 M p2 = DirD[p 1,x i,h]p 2 +p 1 DirD[p 2,x i,h]. (19) 11

12 4.1.2 The Laplacian of a Product Lemma 4.2 The product rule for the Laplacian of NC polynomials is g ( Lap[p 1 p 2,h] = Lap[p 1,h]p 2 +p 1 Lap[p 2,h]+2 DirD[p1,x i,h] DirD[p 2,x i,h] ). Proof: Lap[p 1 p 2,h]= = = i=1 g DirD[DirD[p 1 p 2,x i,h],x i,h] i=1 g DirD [p 1 DirD[p 2,x i,h] + DirD[p 1,x i,h]p 2,x i,h] ) i=1 g ( p1 DirD[DirD[p 2,x i,h],x i,h] + DirD[DirD[p 1,x i,h],x i,h]p 2 i=1 + 2 DirD[p 1,x i,h],dird[p 2,x i,h] ) = Lap[p 1,h]p 2 +p 1 Lap[p 2,h] Formulas Involving Derivatives of γ d Recall γ := x 1 + ix 2. g ( DirD[p1,x i,h] DirD[p 2,x i,h] ). Lemma 4.3 The derivatives of of γ d exhibit the following symmetries. and Proof: It is easily seen that i=1 DirD[Re(γ d ),x 1,h] = DirD[Im(γ d ),x 2,h] DirD[Re(γ d ),x 2,h]= DirD[Im(γ d ),x 1,h]. DirD[Re(γ),x 1,h] = DirD[Im(γ),x 2,h]=h DirD[Im(γ),x 1,h]= DirD[Re(γ),x 2,h]=0. 12

13 Assume that DirD[Re(γ d 1 ),x 1,h] = DirD[Im(γ d 1 ),x 2,h] DirD[Im(γ d 1 ),x 1,1] = DirD[Re(γ d 1 ),x 2,h]. Then DirD[ Re(γ d ),x 1,h] = DirD[x 1 Re(γ d 1 ) x 2 Im(γ d 1 ),x 1,h] =x 1 DirD[Re(γ d 1 ),x 1,h]+hRe(γ d 1 ) x 2 DirD[Im(γ d 1 ),x 1,h] DirD[Im(γ d ),x 2,h] = DirD[x 1 Im(γ d 1 )+x 2 Re(γ d 1 ),x 2,h] =x 1 DirD[Im(γ d 1 ),x 2,h]+x 2 DirD[Re(γ d 1 ),x 2,h]+hRe(γ d 1 ), so DirD[Re(γ d ),x 1,h] = DirD[Im(γ d ),x 2,h] (23) which satisfies the first half of our inductive hypothesis. For the next half compute DirD[ Re(γ d ),x 2,h] = DirD[x 1 Re(γ d 1 ) x 2 Im(γ d 1 ),x 2,h] =x 1 DirD[Re(γ d 1 ),x 2,h] x 2 DirD[Im(γ d 1 ),x 2,h] him(γ d 1 ) DirD[Im(γ d ),x 1,h] = DirD[x 1 Im(γ d 1 )+x 2 Re(γ d 1 ),x 1,h] =x 1 DirD[Im(γ d 1 ),x 1,h]+hIm(γ d 1 )+x 2 DirD[Re(γ d 1 ),x 2,h], so DirD[Re(γ d ),x 2,h]= DirD[Im(γ d ),x 1,h]. (24) Also note that γ = γ T and therefore that γ d =(γ d ) T.So ( Re(γ d ) ) T ( = Re((γ d ) T ) = Re(γ d ) and Im(γ d ) ) T = Im((γ d ) T ) = Im(γ d ). 4.3 Harmonics degree > 2: Proof of Theorem 1 part (1) Since no double-substitutions can be made on words of length 1, Lap[Re(γ),h] = Lap[x 1,h] = 0 and Lap[Im(γ),h] = Lap[x 2,h]=0. (26) 13

14 Now, assume that Lap[Re(γ d 1 ),h] = Lap[Im(γ d 1 ),h]=0. (27) Pushing ahead, Re(γ d ) = Re((x 1 + ix 2 ) γ d 1 )=x 1 Re(γ d 1 ) x 2 Im(γ d 1 ) (28) Im(γ d ) = Im((x 1 + ix 2 ) γ d 1 )=x 1 Im(γ d 1 )+x 2 Re(γ d 1 ). (29) Applying our product rule to (28): Lap[Re(γ d ),h] = Lap[x 1 Re(γ d 1 ),h] Lap[x 2 Im(γ d 1 ),h] =x 1 Lap[Re(γ d 1 ),h] + Lap[x 1,h]Re(γ d 1 ) + 2 DirD[x 1,x 1,h] DirD[Re(γ d 1 ),x 1,h] + 2 DirD[x 1,x 2,h] DirD[Re(γ d 1 ),x 2,h] x 2 Lap[Im(γ d 1 ),h] Lap[x 2,h]Im(γ d 1 ) 2 DirD[x 2,x 1,h] DirD[Im(γ d 1 ),x 1,h] 2 DirD[x 2,x 2,h] DirD[Im(γ d 1 ),x 2,h]. Use (26) and (27) to obtain that the Lap[] terms are 0, and that cross partials are 0 to get Lap[Re(γ d ),h]=hdird[re(γ d 1 ),x 1,h] hdird[im(γ d 1 ),x 2,h]. By symmetry Lemma 4.3, this means Re(Lap[γ d,h]) = Lap[Re(γ d ),h]=0. (30) By a similar argument, applying the product rule to (29), Im(Lap[γ d,h]) = Lap[Im(γ d ),h]=0. (31) Harmonic Polynomials are further discussed in [McA04]. 4.4 Subharmonics degree > 4: Proof of Theorem 1 (2)(3) matrix-positive for all d. 14

15 By the product rule for the Laplacian Lap[ ( Re(γ d ) ) 2,h] = Lap[Re(γ d ),h]re(γ d ) + Re(γ d ) Lap[Re(γ d ),h] +2 ( (DirD[Re(γ d ),x 1,h]) 2 + (DirD[Re(γ d ),x 2,h]) 2) =2 ( (DirD[Re(γ d ),x 1,h]) 2 + (DirD[Re(γ d ),x 2,h]) 2) which is a sum of squares (as Re(γ d ) is symmetric). Thus we have shown ( Re(γ d ) ) 2 ( is subharmonic. The same argument applies to Im(γ d ) ) 2. By Helton s [Hel02], we know that a matrix positive polynomial must be expressible as a sum of squares, and therefore must have even degree. Since the directional Laplacian preserves degree, subharmonic polynomials, having matrix-positive Laplacians, must have even degree. Now we prove the formula (15) relating subharmonics. We use from the elementary result γ 2d = (Re(γ d )+iim(γ d )) 2 = (Re(γ d )) 2 (Im(γ d )) 2 + i(re(γ d )Im(γ d ) + Im(γ d ) Re(γ d )). Therefore Re(γ 2d )= ( Re(γ d ) ) 2 ( Im(γ d ) ) 2. So [ c 0 Re(γ d ) ] 2 + c1 Re(γ 2d )+c 2 Im(γ 2d ) = c 0 [ Im(γ d ) ] 2 +(c0 +c 1 ) (Re(γ 2d )) + c 2 Im(γ 2d ). 4.5 Subharmonics of Odd Degree To see that there are no subharmonics of odd degree note that the Laplacian of an odd degree polynomial is itself an odd degree polynomial. However, by [Hel02] this polynomial being positive makes it a sum of squares. Any sum of squares has even degree, thereby producing a contradiction. 5 Classification when Degree is Four or Less 5.1 The Matrix Representation Each polynomial quadratic in the variables x 1,...,x g can be expressed as a product q T Pq, where P is independent of all x i. We shall call P the middle 15

16 matrix and q the border vector for this representation Commutative Coefficients Consider the generalized quadratic polynomial Q(x) = g g A ij x i x j i=1 j=1 with scalar coefficients A ij. Define vector q = q(x) and matrix P such that then q T Pq = g x i i=1 j=1 p ij = A ij and q =(x 1,...,x g ), (34) g P ij x j = g i=1 j=1 g x i A ij x j = g i=1 j=1 g A ij x i x j = Q(x). It is easy to see that when Q(x) is symmetric (i.e. when the coefficient of x i x j equals that of x j x i ) P must also be symmetric: p ij = A ij = A ji = p ji. Example 5.1 The Matrix Representation of the general quadratic, g =2 ( ) ( A B x1 x 2 C D )( x1 x 2 ) =Ax 2 1 +Bx 1 x 2 +Cx 2 x 1 +Dx Non-commutative coefficients The generalized quadratic polynomial with non-commutative coefficients is a trifle more complicated. However, in this paper, we will be concerned mostly with the case g = 1. That is, we will need to extract only one letter, h, from the middle matrix P. The general polynomial quadratic in h, with non-commutative coefficients is written n n Q(h) = A ij (hm i ) T c ij (hm j ) i=1 j=1 here m i are the terms applied outside the target letters h, and c ij, are the inside terms and A ij are scalars. These terms do not necessarily commute, 16

17 so they can not be grouped together and simplified as above. Define A as the N-by-N matrix whose i, j th element is A ij c ij, and define q as Then we see that n n q T Pq = (q) T i P ij (q) j = i=1 j=1 q = h(m 1,m 2,..., m N ). n i=1 j=1 n A ij, (hm i ) T c ij (hm j ). Example 5.2 A matrix representation with non-commutative coefficients 3 x 1 hx 2 2hx 1 + hx 1 x 2 x 1 h h 2 x hx 3 2x 1 hx x 1 x 2 hx 2 hx 2 x 1 T h x 1 x 2 x h = hx 1 0 3x hx 1 hx 2 x x 2 0 hx 2 x 1 = qt Pq. hx 2 2 x 3 2x hx 2 2 It can be easily verified that Q is symmetric if and only if the matrix P is symmetric Convexity of p vs. Positivity of its Middle Matrix Given the polynomial Q(x, h) which is quadratic in h there may be several q T Pq representations of Q. However, once q is chosen, with entries which are monomials that do not repeat, the middle matrix P is uniquely determined (up to adding on rows and columns of zeroes). Furthermore, the polynomial Q in x and h is matrix positive if and only if P (x) is a positive semidefinite matrix whenever symmetric matrices are substituted for x. Thus checking matrix positivity of Q is equivalent to checking matrix positivity of P. For discussion and proofs of all this, see [CHSY03]. 5.2 The Zeroes Lemma The following is useful in our analysis of subharmonics. Lemma 5.1 Let A R N N be any symmetric matrix with entries in some domain R of polynomials on free variables with real coefficents. If there exists some diagonal entry a ii =0and corresponding off-diagonal entries a ij = a T ji 0, then A is not matrix-positive semidefinite 17

18 Proof: Let e i and e j be standard basis vectors for R N (i.e. e T i Ae j =a ij ) and define v := β 1 e i + β 2 e j where β 1,β 2 R. Then, v T Av =((a ij + a ji )β 1 + a jj β 2 )β 2 =(2a ij β 1 + a jj β 2 )β 2 Given β 2 > 0, we can choose β 1 such that is either matrix-positive or -negative. 2 β 1 a ij β 2 + β 2 a jj β 2 This lemma is very useful when applied to our matrix representation of the Laplacian of a symmetric polynomial on free variables. Note that we can express the Laplacian, Lap(p, h), of any polynomial p(x) as: Lap[p, h] = n n A ij (hm i ) T c ij (hm j ) i=1 j=1 and that Lap[p, h] =q T Pq, where p ij = A ij c ij and For any i,if then q =(hm 1,..., hm n ) T. A ii (hm i ) T c ii (hm i )=0, p ii = A ii c ii =0. By the Zeroes Lemma, either P is not positive-semidefinite (Lap[p, h] is not matrix-positive) or in the latter case, p ij = p ji = 0 for 1 j n A ij (hm i ) T c ij (hm j )=A ji (hm j ) T c ji (hm i ) = 0 for 1 j n. 18

19 5.3 The Spanning Polynomials for Subharmonics We begin with a basis for the set of degree 4 homogeneous polynomials in symmetric free variables p = A 1 x 4 1x A 2 (x 3 1x 2 + x 2 x 3 1) + A 3 (x 2 1x 2 x 1 + x 1 x 2 x 2 1)+A 4 (x 2 1x 2 2+x 2 2x 2 1) +A 5 (x 1 x 2 x 1 x 2 +x 2 x 1 x 2 x 1 )+A 6 x 1 x 2 2x 1 We calculate the Laplacian of p: +A 7 (x 1 x 3 2+x 3 2x 1 )+A 8 x 2 x 2 1x 2 +A 9 (x 2 x 1 x 2 2+x 2 2x 1 x 2 )+A 10 x A 1 (h 2 x hx 1 hx 1 + hx 2 1h + x 1 h 2 x 1 + x 1 hx 1 h + x 2 1h 2 ) +2A 2 (h 2 x 1 x 2 +hx 1 hx 2 + x 1 h 2 x 2 + x 2 h 2 x 1 + x 2 hx 1 h + x 2 x 1 h 2 ) +2A 3 (h 2 x 2 x 1 +hx 1 x 2 h + hx 2 hx 1 + hx 2 x 1 h + x 1 hx 2 h + x 1 x 2 h 2 ) +2A 4 (h 2 x 2 1 +h 2 x 2 2 +x 2 1h 2 +x 2 2h 2 ) +2A 5 (hx 1 hx 1 + hx 2 hx 2 + x 1 hx 1 h + x 2 hx 2 h)+ 2A 6 (hx 2 2h + x 1 h 2 x 1 ) +2A 7 (h 2 x 2 x 1 +hx 2 hx 1 + x 1 h 2 x 2 + x 1 hx 2 h + x 1 x 2 h 2 + x 2 h 2 x 1 )+ 2A 8 (hx 2 1h + x 2 h 2 x 2 ) +2A 9 (h 2 x 1 x 2 +hx 1 hx 2 + hx 1 x 2 h + hx 2 x 1 h + x 2 hx 1 h + x 2 x 1 h 2 ) +2A 10 (h 2 x hx 2 hx 2 + hx 2 2h + x 2 h 2 x 2 + x 2 hx 2 h + x 2 2h 2 ) The Laplacian Matrix Decomposition The directional Laplacian is quadratic in h, and so can be represented by the matrix product q T Pq, where q and P/2 are ( h x1 h x 2 h x 2 1h x 1 x 2 h x 2 x 1 h x 2 2h ) T and 19

20 (A 1 + A 8 ) x 2 1 +(A 6+A 10 ) x 2 2 +(A 3 +A 9 )(x 1 x 2 +x 2 x 1 ) (A 1 +A 5 )x 1 (A 2 +A 9 )x 1 A 1 +A 4 A 3 +A 7 A 2 +A 9 A 4 +A 10 +(A 3 +A 7 )x 2 +(A 5 +A 10 ) x 2 (A 1 + A 5 ) x 1 +(A 3 +A 7 )x 2 A 1 +A 6 A 2 +A (A 2 +A 9 )x 1 +(A 5 +A 10 ) x 2 A 2 + A 7 A 8 + A A 1 +A A 3 +A A 2 +A A 4 +A respectively. We wish to find the conditions on {A 1,...,A 10 } that ensure that this matrix is positive-semidefinite. By the Zeroes Lemma, the zeroes on the last four diagonals force all entries in the last four rows or columns to be zero. This corresponds to the following assignments: A 4 = A 1, A 10 = A 1, A 9 = A 2, A 7 = A 3. (44) Applying these conditions to the matrix above, and ignoring the zeroed-out rows and columns, we have: (A 1 + A 8 ) x 2 1 +(A 1+A 6 )x 2 2 +(A 2 A 3 )(x 2 x 1 x 1 x 2 ) (A 1 +A 5 )x 1 (A 1 +A 5 )x 2 (A 1 +A 5 )x 1 A 1 +A 6 A 2 A 3 (A 1 +A 5 )x 2 A 2 A 3 A 1 +A 8. This matrix can be simplified by substitution of recurring pairs by single letters: G = A 1 + A 6, H = A 1 +A 8 J = A 2 A 3, K = A 1 +A 5. Now we have a matrix small enough that we can find its LDL T (Cholesky) decomposition Hx 2 1 J(x 1 x 2 + x 2 x 1 )+Gx 2 2 Kx 1 Kx 2 Kx 1 G J. Kx 2 J H 20

21 5.3.2 Conditions for Positivity It is now possible to discover the exact conditions for matrix positivity. Each diagonal entry of D must be equal to or greater than zero for the Laplacian to be matrix-positive). This will reveal the values of G, H, J and K such that the above is positive-semidefinite. G H J2 G Hx J(x 1 x 2 + x 2 x 1 )+Gx 2 2 K2 x 2 1 G ( JKx1 G +Kx 2)( JKx 1 H J2 G G +Kx 2) And so we are left with three inequalities, which must be satisfied for matrix positivity: G>0, Hx J(x 1 x 2 + x 2 x 1 )+Gx 2 2 K2 x 2 1 G H J2 G > 0, ( JKx1 G + Kx 2)( JKx 1 G + Kx 2) H J2 G > 0. The last condition involves x 1 and x 2, and therefore must itself be converted into a matrix to reveal the conditions on G, H, J and K under which it is positive. [CHSY03] G H2K2 K2 (G H2 )J 2 J HK2 (G H2 H + J H + HK2 (G H2 J )J G H2 J J )J J K2.. Again we perform the LDL T decomposition: G K2 0 H J2 JG 0 H J2 K 2 Although the inequality (H J2 G K2 )G2 G ) 2 (J JK2 (H J2 G )G G K2 H J2 G. ( J 2 K 2 J JK 2 K2 H (H J 2 )G2 G G G 21 (H J2 G )G ) 2 K2 H J2 G > 0 (45)

22 is quite complicated, we can simplify it some by multiplying it by expressions which are known to be positive, such as: G, H J 2 G, and G K2 H J 2 G which we encountered earlier. This gives a polynomial inequality equivalent to (45), which, after some simplification, gives us: (GH J 2 K 2 ) 2 > 0. Which, considering only the case of all real coefficients, is rather vacuous, informing us only that GH J 2 K 2 0. Bringing all our inequalities together (simplifying each as we did above), we notice that two are redundant : (I) G>0, (II) GH > J 2, (III) GH > J 2 +K 2, (IV ) GH H 2 +K 2 as (III) implies (II) and (IV). Therefore, we conclude that the set of polynomials making the Laplacian matrix positive is exactly those of the form: f =A 1 (x 4 1 x 2 1x 2 2 x 2 2x x 4 2) (46) + A 2 (x 3 1x 2 + x 2 x 3 1 x 2 x 1 x 2 2 x 2 2x 1 x 2 ) + A 3 (x 2 1x 2 x 1 + x 1 x 2 x 2 1 x 1 x 3 2 x 3 2x 1 ) + A 5 (x 1 x 2 x 1 x 2 + x 2 x 1 x 2 x 1 ) + A 6 x 1 x 2 2x 1 + A 8 x 2 x 2 1x 2 with coefficients satisfying the inequalities: (A 1 + A 8 )(A 1 + A 6 ) > (A 3 A 2 ) 2 +(A 1 +A 5 ) 2 (47) and A 1 + A 8 > 0 (or, equivalently A 1 + A 6 > 0). (48) 6 Future Work The subharmonic polynomials may serve as substitutes for convex polynomials in special cases. Currently, our group is looking at other weakenings 22

23 of convexity, including quasiconvexity, in which the number of negative eigenvalues are tolerably small. The proofs in progress that Re(γ d ) and Im(γ d ) (resp. [Re(γ d )] 2 and [Im(γ d )] 2 ) are the only harmonic (resp. subharmonic) polynomials of degree d (resp. 2d when d>2) relies on an observation on the words generated by substituting pairs of identical letters (as with the Laplacian). These proofs should follow shortly. 23

24 7 Acknowledgments Thanks to Jan Bitmead for her support and for organizing the Honors Program, Thanks to Danny MacAllaster for the recursion for the harmonics, and Thanks to everybody in the lab: Nick Slinglend, John Shopple and Mark Stankus, and Thanks especially to John Farina for looking over the manuscript a couple of times Thanks to the Math Department for reviewing this paper, and Very much thanks to Professor Helton, who was there to help, from the very beginning, right down to the end. Thanks to the McNair Foundation and to Professor Helton s NSF grant for finanacial support. Thanks go from Professor Helton to Professor Roger Howe for very helpful conversations about the classical commutative analog of the noncommutative results here. 24

25 References [CHSY03] Juan F. Camino, J. W. Helton, R.E. Skelton, and Jieping Ye. Matrix inequalities: a symbolic procedure to determine convexity automatically. Integral Equations Operator Theory, 46(4): , [Hel02] [HM98] [HS04] [McA04] J. W. Helton. Positive noncommutative polynomials are sums of squares. Ann. of Math. (2), 156(2): , J. W. Helton and Orlando Merino. Classical Control Using H Methods. SIAM, J. W. Helton and McCullough Scott Skelton, R. E. Convex noncommutative polynomials have degree two or less. Siam J. Matrix Anal. Appl, 25(4): , Daniel P. McAllaster. Homogeneous harmonic noncommutative polynomials from degree 3 to 7 (includes a recurrence relation for higher degrees). dmcallas/papers.html, July