Operator norms of words formed from positivedefinite

Transcription

1 Electronic Journal of Linear Algebra Volume 18 Volume 18 (009) Article 009 Operator norms of words formed from positivedefinite matrices Stephen W Drury drury@mathmcgillca Follow this and additional wors at: Recommended Citation Drury, Stephen W (009), "Operator norms of words formed from positive-definite matrices", Electronic Journal of Linear Algebra, Volume 18 DOI: This Article is brought to you for free and open access by Wyoming Scholars Repository It has been accepted for inclusion in Electronic Journal of Linear Algebra by an authorized editor of Wyoming Scholars Repository For more information, please contact scholcom@uwyoedu

2 Volume 18, pp 13-0, January 009 OPERATOR NORMS OF WORDS FORMED FROM POSITIVE-DEFINITE MATRICES SW DRURY Abstract Let α 1,α,,α n,β 1,β,,β n be strictly positive reals with α 1 + α + + α n = β 1 + β + + β n = s In this paper, the inequality A α 1 B β 1 A α A αn B βn AB s when A and B are positive-definite matrices is studied Related questions are also studied Key words Positive-definite matrix, Matrix power, Operator norm, Matrix words AMS subect classifications 15A45 1 Introduction Let A and B be positive-definite matrices of the same shape and let W be a word formed from A and B containing pa s and qb s Then we may as whether necessarily W A p B q Here we have denoted the operator norm For example we have (AB) A B for a nonnegative integer To see this we assume without loss of generality that we have the normalization A B = 1, which amounts to A B A L I or B L A From this it follows that B L A since the map u u 1 is matrix monotone The reader should consult [1] for facts about matrix monotone functions We can then deduce that AB 1 and thence (AB) 1 This method, which we will call the monotonicity tric, wors in many situations, for example A B A A 3 B We normalize as before and obtain B 4 L A 6 and A 6 L B 4 whence AB A 4 B A L AB B 8 3 B A = AB 4 3 A L AA A = I as required The monotonicity tric also shows that A B AB A B A A 7 B 6 However, it does not wor for all words An example where it fails is A BAB Moregenerally, andafter replacingthe matricesa andb by suitable positive powers we may pose the following question Let α 1,α,,α n,β 1,β,,β n be strictly Received by the editors July 31, 008 Accepted for publication December 9, 008 Handling Editor: Harm Bart Department of Mathematics and Statistics, McGill University, 805 Sherbrooe Street West, Montreal, Canada H3A K6 13

3 Volume 18, pp 13-0, January SW Drury positive reals with α 1 + α + + α n = β 1 + β + + β n = s When is it necessarily true that (11) A α1 B β1 A α A αn B βn AB s for all pairs of positive-definite matrices A and B of the same shape? Colloquially, we may state the main result of this article by saying that if (11)can be proved by the monotonicity tric, then it is true and otherwise it is false (for some choice of A and B) 0 β n 1 β n 0 0 α n β n 1 α n + β β n 1 s 0 β n 1 + β n α n 1 α α n 1 + β n 1 + β n s 0 β + + β n α 3 α n 1 α 1 + α + β + + β n s 0 α + + α n β β n 1 β 1 + α + + α n s 0 β β n α α n 1 α 1 0 Table 1 The Main Theorem Our main result is the following Theorem 1 The inequality (11) is true for all pairs of positive-definite matrices A and B of the same shape if and only if the inequalities in the left-hand column of Table 1 are satisfied We remar that the hypothesis that α 1,α,,α n,β 1,β,,β n are strictly positive is unfortunately necessary For example, it is easy to see that AB A B = (AB)(BA)(AB) AB 3 Now consider AB A 0 B 0 A B Then the left inequality in 0 β + β 3 α 3 1 fails since β + β 3 α 3 = 1 Proof Let W = A α1 B β1 A α A αn B βn, then it is routine to chec that the inequalities necessary for the application of the monotonicity tric applied to WW are precisely those in the left-hand column of Table 1 If one attempts to apply the monotonicity tric applied to W W, then one gets the same inequalities in an equivalent form and in the reverse order For the reverse implication, we need to show that if (11)is true for all pairs of positive-definite matrices A and B then the inequalities in the left-hand column of Table 1 are satisfied To tacle the right-hand inequalities, we use infinitessimal methods Let B = diag(b 1,,b d )where b = e p for =1,,dLetP =diag(p 1,,p d )and A(t) = exp( P tq)where Q is a d d matrix to be specified later Then A(0)= B 1 and

4 Volume 18, pp 13-0, January 009 Operator Norms of Words Formed From Positive-Definite Matrices 15 we have [ e A(t) α = B α αp e αp ] + t q, + O(t ) p p, where the quotient e αp e αp is interpreted as a divided difference (evaluating p p to αe αp )in case p = p NowletHbean arbitrary positive-definite d d matrix and define Q by q, = p p e p e p h, then it follows that A(t) = B th +O(t ) L B for t 0 sufficiently small Let W (t) =A(t) α1 B β1 A(t) α A(t) αn B βn,thenw (0)= I and so that by hypothesis W (t) 1for0 t small and therefore A calculation shows that [ ( d n W (t)w (t) = h, dt t=0 d W (t)w (t) L 0 dt t=0 l=1 b α l b b α l b ( ) )] b γ l bα l γ l + b γ l bα l γ l where the γ l are linear combinations of α 1,α,,α n,β 1,β,,β n to be specified later It follows that the matrix M given by m, = n l=1 b α l b b α l b ( ) b γ l bα l γ l + b γ l bα l γ l is a Schur multiplier of positive-definite matrices to positive-semidefinite matrices and therefore M is itself positive-semidefinite However, the expression b 1 b 1 m, which is also positive-semidefinite is a function of b /b for all choices of positive (b ) and hence arises as a continuous positive-definite function on the group (0, )with multiplication as the group operation Written as a positive-definite function on the line this is d sinh( α lu ϕ(u) = ) ( ) αl γ l cosh u sinh(u) l=1 It is well-nown [3] that such a function has to be bounded, in fact we must have ϕ(u) ϕ(0) However, each of the terms in the sum is nonnegative and it follows that α l + α l γ l 1,

5 Volume 18, pp 13-0, January SW Drury or equivalently α l γ l 1 and γ l 1 for all l =1,,,nWehaveγ 1 =0sothat0 1andα 1 1, γ = β 1 α 1 so that β 1 α 1 1andα 1 + α β 1 1 etc obtaining the n 1 right-hand inequalities from the left-hand column of Table 1 reading upwards Next we tacle the left-hand inequalities in the left-hand column of Table 1 and these have been rewritten in an equivalent form in the right-hand column of the table The first and last inequalities are therefore evident A continuity argument allows us to assume that (11)is true for all pairs of positive-semidefinite matrices A and B Let us define p p 0 A = 0 p p 0 p p and B = p p with p small and positive to be determined Then we have x 1 p x x 1 p x 0 A x = 0 x 1 p x x 1 p x and B x = x 1 p x x 1 p x 0, 0 x 1 p x x 1 p x provided that x>0, but not for x =0 Then p p 3 p 3 (AB)(AB) = p 3 p +p 4 p +p 4 p 3 p +p 4 p +p 4, It follows that AB = p (1+p + ) p 4 +p = O(p) Clearly W has ran two and it follows that 1 tr(w W ) W tr(w W ) leading to tr(w W )=O(p s ) Also, with p chosen small and positive, all the terms comprising tr(w W )are nonnegative and hence each individual term in the expansion of tr(w W )is O(p s ) There are two types of inequalities remaining to be discussed Case 1 The inequality β β + α α n s for =1,,n 1 We consider the individual term zz where z = A α1 1,1 Bβ1 1,1 Aα 1,1 Bβ 1, Aα +1,3 B β +1 3,3 B βn 3,3 = Cpβ1+ +β +α α n

6 Volume 18, pp 13-0, January 009 Operator Norms of Words Formed From Positive-Definite Matrices 17 Since z = O(p s )as p 0, we get the desired inequality Case The inequality α α + β + + β n s for =,,n 1 We consider the individual term zz where z = A α1 3,3 Bβ1 3,3 Bβ 1 3,3 A α 3, Bβ,1 Aα +1 1,1 B β +1 1,1 B βn 1,1 = Cpα1+ +α +β + +β n Since z = O(p s )as p 0, we get the desired inequality 3 Remars and Comments In light of the strict positivity assumption, it is instructive to observe that there is a corresponding result for words of odd length 0 α n 1 α n 0 0 β n 1 α n 1 α α n 1 + β n 1 s 0 α n 1 + α n β n 1 1 α n 1 + α n + β β n s 0 β + + β n 1 α 3 α n 1 α 1 + α + β + + β n 1 s 0 α + + α n β β n 1 1 β 1 + α + + α n s 0 β β n 1 α α n 1 α 1 0 Table Theorem 31 Let α 1,α,,α n,β 1,β,,β n 1 be strictly positive reals with α 1 + α + + α n = β 1 + β + + β n 1 = s Then A α1 B β1 A α B βn 1 A αn AB s is true for all pairs of positive-definite matrices A and B of the same shape if and only if the inequalities in the left-hand column of Table are satisfied We omit the proof since it is very similar to the proof of Theorem 1 A case of some interest is Theorem 1 for n =,α 1 = β = x, α = β 1 =y, ie when does A x B y A y B x AB x+y hold for all A and B positive-definite? An equivalent formulation is to as when we have A y B x A y L B y A x B y given that A and B are positive-definite and that B L A Theorem 1 gives that this holds if and only if either of the following conditions holds y =0and0 x 1, 0 x 1and 1 x y 1 (1 + x) Note that the region 0 <x 1, 0 <y< 1 x is excluded This brings into sharp focus the necessity of insisting on strict positivity in Theorems 1 and 31

7 Volume 18, pp 13-0, January SW Drury It is amusing that one can give a precise answer to the infinitessimal question in this case Precisely, this ass whether (31) d ( B y A(t) x B y A(t) y B x A(t) x) L 0 dt t=0 given that A(0)= B and d dt t=0a(t) L 0 Following the method of Theorem 1, we conclude that infinitessimal question is satisfied if and only if the function ϕ given by ϕ(u) = sinh(xu) sinh((x y)u) sinh(u) is positive-definite on R According to Bochner s Theorem [3], the continuous function ϕ is positive-definite if and only if it is the Fourier transform of a positive measure According to [, page 1148] the inverse Fourier transform of u sinh(xu) is ξ sinh(u) π sin(πx) in case x < 1andisapointmassat0incasex = ±1 cosh(πξ)+cos(πx) y x -0 Fig 31 Region for the infinitessimal question Theorem 3 Let x and y be real Then the infinitessimal question (31) is satisfied if and only if all of the following hold x y 1,

8 Volume 18, pp 13-0, January 009 Operator Norms of Words Formed From Positive-Definite Matrices 19 x 1, sin(πx) sin(π(x y)), y x 1 (tan π arctan ( πx )) The region of validity is depicted in Figure 31 and includes an area where x<0 and an area where y<0 Proof Case x < 1 and x y < 1 It follows from Bochner s Theorem that we must have that for all ξ π sin(πx) cosh(πξ)+cos(πx) π sin(π(x y)) cosh(πξ)+cos(π(x y)) This holds for all ξ if and only if it holds for ξ = 0 and also in the limit as ξ The two conditions boil down to ( πx ) ( ) π(x y) tan tan and sin(πx) sin(π(x y)) Case x =1and x y =1 We see that if x = 1, then necessarily x =1orelse ϕ cannot be nonnegative The two points (x, y) =(1, 1)and (1, 0)corresponding to x y = 1 are both admissible Case x =1and x y < 1 Then we must have x =1and π sin(π(x y)) cosh(πξ)+cos(π(x y)) 0 for all ξ which occurs for x =1and 1 y<1 Case x < 1 and x y =1 Then we must have x y = 1 for else the point mass will have a negative sign We also require for all ξ that π sin(πx) cosh(πξ)+cos(πx) 0 which occurs for 0 x<1 Some additional wor is required to show that these constraints reduce to the ones given in the theorem REFERENCES [1] William F Donoghue, Jr Monotone Matrix Functions and Analytic Continuation Die Grundlehren der mathematischen Wissenschaften, Band 07 Springer-Verlag, New Yor- Heidelberg, 1974

9 Volume 18, pp 13-0, January SW Drury [] I S Gradshteyn and I M Ryzhi Table of Integrals, Series, and Products Corrected and enlarged edition edited by Alan Jeffrey Academic Press, New Yor-London-Toronto, Ont, 1980 [3] Walter Rudin Fourier Analysis on Groups Interscience, New Yor, 196