Norms and CB Norms of Jordan Elementary Operators

Transcription

1 arxiv:math/ v1 [math.oa] 7 Aug 003 Norms and CB Norms of Jordan Elementary Operators Richard M. Timoney February 1, 008 Abstract We establish lower bounds for norms and CB-norms of elementary operators on B(H). Our main result concerns the operator T a,b x = axb + bxa and we show T a,b a b, proving a conjecture of M. Mathieu. We also establish some other results and formulae for and T a,b for special cases. Our results are related to a problem of M. Mathieu [13, 14] asking whether T a,b c a b holds in general with c = 1. We prove this in Theorem 6 below. In [14] the inequality is established for c = /3 and the best known result to date is c = ( 1) as shown in [17, 5, 11]. There are simple examples which show that c cannot be greater than 1 in general and there are results which prove the inequality with c = 1 in special cases. The case a = a and b = b is shown in [1] where it is deduced from = T a,b under these hypotheses. The equality of the the CB norm and the operator norm of T a,b also holds if a, b are commuting normal operators. See section 3 below for references. A result for c = 1 is shown in [] under the assumption that a+zb a for all z C. In more general contexts similar results (with varying values of c) are shown in [6, 5]. As this manuscript was being written we learned of another proof of the main result ([4]), using rather different methods. Thanks are due to M. Mathieu for drawing our attention to this reference. Acknowledgement. Part of this work was done during a visit by the author to the University of Edinburgh in the autumn of 00. A significant impetus to the work arose from discussions with Bojan Magajna and Aleksej Turnšek during a visit to Ljubljana in March 003 and the author is very grateful to them for that. Thanks also to P. Legiša for finding the reference to [10] below. 1

2 1 Preliminaries We call T: B(H) B(H) an elementary operator if T has a representation T(x) = l a i xb i with a i, b i B(H) for each i. We cite [1] for an exposition of many of the known results on (more general) elementary operators and for other concepts we cite a number of treatises on operator spaces including [8, 15, 7]. In particular we will use the completely bounded (or CB) norm T cb of an elementary operator, the operator norm T and the estimate in terms of the Haagerup tensor product norm T T cb l a h i b i. We recall that the Haagerup norm of an element w B(H) B(H) (of the algebraic tensor product) is defined by k w h = inf a i a k i b i b i where the infimum is over all representations w = k a i b i. Moreover this infimum is achieved with both k-tuples (a 1, a,..., a k ) and (b 1, b,..., b k ) linearly independent. Throughout H denotes a (complex) Hilbert space and B(H) the algebra of bounded linear operators on H. For x in the class of Hilbert-Schmidt operators on H we denote the Hilbert-Schmidt norm by x (so that x = trace x x). Lower bounds Lemma 1 Given linearly independent a, b B(H), we can find c 1, c B(H), δ 1, δ > 0 and z C \ {0} so that a b + b a = c 1 c 1 + c c, c 1 = (za + z 1 b)/, c = i(za z 1 b)/ and a b + b a h = δ 1 c 1 c 1 + δ c c = δ 1 1 c 1c 1 + δ 1 c c. Proof. We know from general facts cited above that the Haagerup norm infimum for w = a b+b a is realised via a representation w = a 1 b 1 +a b. Moreover, by scaling a i to λa i and b i to λ 1 b i for a suitable λ we can arrange that w h = a 1 a 1 + a a = b 1 b 1 + b b.

3 We adopt a convenient matrix notation w = [a, b] [b, a] t = [a 1, a ] [b 1, b ] t for the two tensor product expressions above (t for transpose) and note that all possible (linearly independent) representations of w take the form w = [a 1, a ] [b 1, b ]t = ([a 1, a ]α) (α 1 [b 1, b ] t ) for a invertible scalar matrix α. We use the transpose notation also for the linear operation on the tensor product that sends a 1 b 1 to b 1 a 1. Then we have w = w t = [b 1, b ] [a 1, a ] t = ([a 1, a ]α) ([b 1, b ](α 1 ) t ) t. From [b 1, b ] = [a 1, a ]α and [a 1, a ]α t = [b 1, b ] together with linear independence we get α = α t symmetric. We can now express α = u u t where u is a unitary matrix and is a diagonal matrix with positive diagonal entries δ1 1, δ 1 ([10, Takagi s factorisation, 4.4.4] see also the problems on pages 1, 17 in [10]). Take [a 1, a ] = [a 1, a ]u, [b 1, b ] = [b 1, b ](u 1 ) t so that w = [a 1, a ] [b 1, b ]t, and w h = (a 1 )(a 1 ) + (a )(a ) = (b 1 ) (b 1 ) + (b ) (b ) [a 1, a ] = [a 1, a ]u = [a 1, a ]α(u 1 ) t = [b 1, b ](u 1 ) t = [b 1, b ]. In other words, a i δ 1 i = b i (i = 1, ). We now take c i = δ i b i and we then have w = c 1 c 1 + c c together with w h = δ 1 c 1 c 1 + δ c c = δ 1 1 c 1 c 1 + δ 1 c c. It remains to relate c 1, c to a, b as claimed. If we put a = (c 1 ic )/ and b = (c 1 + ic )/ we have w = a b + b a = [a, b ] [b, a ] t = [a, b] [b, a] t. An easy argument shows that there is z C with either a = za and b = z 1 b or else a = z 1 b and b = za. The first case is exactly as required but for the second case we need to swap the roles of c 1 and c. 3

4 Theorem Assume that H is two-dimensional and a, b B(H). Let T a,b (x) = axb + bxa. Then a b. Proof. In the case where a, b are linearly dependent (a = λb, say, T a,b x = λaxa) we know T cb = T = a b a b. So we deal only with the case of independent a, b. We first apply Lemma 1, = a b+b a h and the fact that the norm of a positive matrix (the max of the eigenvalues) is at least half the trace to get 1 1 ( δ1 c 1 + δ c ) ( δ 1 1 c 1 + δ 1 c ) We deduce 1 ( (δ1 + δ1 1 4 ) c 1 + (δ + δ 1 ) c ) 1 ( c1 + c ) = 1 trace (c 1c 1 + c c ) = 1 trace ( (za) (za) + (z 1 b) (z 1 b) ) = 1 ( za + z 1 b ) za z 1 b = a b. Corollary 3 ([11], Theorem.1) For a, b B(H) (H arbitrary) a b. Proof. We can reduce the proof to the case where H is two-dimensional by the argument given in [11, Theorem.1] (take unit vectors ξ, η H where aξ a ε and bη b ε; consider T qap,qbp where p is a projection onto the span of ξ, η and q a projection onto the span of aξ, bη). In two dimensions the result follows from Theorem. Proposition 4 If a, b B(C ) are symmetric matrices, then = T a,b = inf x>0 xaa + (1/x)bb 4

5 Proof. Now c 1, c obtained from Lemma 1 are symmetric matrices. Using c i = c i = the complex conjugate matrix we have δ 1 1 c 1 c 1 + δ 1 c c = δ 1 1 c 1c 1 + δ 1 c c = δ 1 1 c 1 c 1 + δ 1 c 1 c Thus δ 1 + δ1 1 c 1 c 1 + c c = c 1c 1 + c c c 1 c 1 + δ + δ 1 c c so that the infimum in the Haagerup tensor norm is attained with δ 1 = δ = 1. We thus have = inf z aa + z bb z and the desired formula for (taking x = z ). From [18] we know that the convex hulls of the following two sets of matrices intersect {[ ] c1 c W l = 1ξ, ξ c c 1ξ, ξ c 1 c ξ, ξ c c : ξ H, ξ = 1, ξ, ξ ( ) } c i c i ξ, ξ = T a,b cb, (1) W r = {[ ] c 1 c 1 η, η c c 1 η, η c 1 c η, η c c : η H, η = 1, η, η ( ) } c ic i η, η = T a,b cb. () Moreover the equality = T a,b holds if and only if the sets themselves intersect. For either of the sets (say W l ) to consist of more than one element, the hermitian operator concerned must have a double eigenvalue of the maximum eigenvalue, which means that (taking the case W l ) c i c i is a multiple of the identity matrix. But then by complex conjugation and symmetry c i c i is the same multiple of the identity. In the case when W l (and W r by the symmetry) are singletons, we have = T a,b and using the following lemma, we can complete the proof for the other case. 5

6 Lemma 5 If c 1, c B(C ) are symmetric and satisfy c 1 c 1 + c c = a multiple of the identity matrix, there exists u unitary so that either uc 1 u t and uc u t are both diagonal (t for transpose) or uc 1 u t = with λ > 0, β > 0, ζ = 1. ( λ 0 0 λ ), uc u t = ( ζα ζβ ζβ ζᾱ Proof. We can find u so that uc 1 u t is diagonal (with positive entries, [10, 4.4.4]). We can replace c i by uc i u t (i = 1, ) and assume without loss of generality that c 1 is diagonal. Then c c is diagonal, which means that the rows of c are orthogonal. An easy analysis shows that either c is diagonal or is a multiple (of modulus one) of a matrix of the form ( ) α β β ᾱ The relation satisfied by c 1 and c dictates that c 1 is a multiple of the identity in the latter case. Proof. (of Proposition 4, completed). Invoking the lemma and the fact that S(x) = ut(u t xu)u t has the same norm as T, and the same CB norm, we can reduce to the case where c 1, c generate a commutative C algebra. In this case the fact that S cb = S is known (see references in section 3). Theorem 6 If a, b B(H) and T a,b (x) = axb + bxa. Then T a,b a b. More generally, the same inequality holds if A is a prime C -algebra, a, b are in the multiplier algebra of A and T a,b : A A is T a,b (x) = axb + bxa. Proof. As shown in [14] and [11, Theorem.1], the essential case is the case where A = B(H) and H = C is -dimensional. We show in this case that T a,b a b a b and so we can assume a = b = 1 (a, b B(C )). There exists u, v unitary so that uav is a diagonal matrix with diagonal entries 1, λ, 0 λ 1. Replacing T by S(x) = ut(vxu)v we can assume that a = ( λ ), b = ( b11 b 1 b 1 b ). ) 6

7 By multiplying b by a scalar of modulus 1 we can assume that b 1 = b 1. Multiplying both a and b by a diagonal unitary u with diagonal entries 1 and b1 / b 1 (that is, replacing T by S(x) = ut(xu)) we can assume also that b 1 = b 1. Now consider T t (x) = T(x t ) t = axb t + b t xa and with T s (x) = 1 (T(x) + T t(x)) = axb s + b s xa b s = 1 (b + bt ) = ( ) b11 s 1 s 1 b, s 1 = b 1 + b 1. We claim that T s 1 and this will prove the theorem because T t = T and so T s T. To show T s 1 we invoke Proposition 4 and show T s cb 1. Note 1 b s = b 1 (b 1 b 1 ) 1, b s b s = ( b11 + s 1 s 1 (b 11 + b ) s 1 ( b 11 + b ) b + s 1 ) and write µ i = b ii + s 1 (i = 1, ) for the diagonal entries. Now consider a unit vector ξ = (ξ 1, ξ ) C. Then xaa + (1/x)b s b s (xaa + (1/x)b s b s )ξ, ξ = x aa ξ, ξ + (1/x) b s b sξ, ξ aa ξ, ξ b s b sξ, ξ and we claim that there is a point in the joint numerical range W = {(x, y) = ( aa ξ, ξ, b s b sξ, ξ ) : ξ = 1} R which is also on (or above) the hyperbola xy = 1/4. Verifying the claim will complete the proof. We assume from now on that λ = 0, as this is the hardest case (smallest aa ξ, ξ ). Being the joint numerical range of two hermitian operators (or the numerical range of the single operator aa + ib s b s ), W is a convex set in the plane. In fact, because the space is -dimensional, W is either a straight line (in the case where the two operators commute, that is s 1 (b 11 + b ) = 0) or else an ellipse (together with its interior) [3, I.6.]. The ellipse touches 7

8 the vertical lines x = 0 and x = 1 at the points (0, µ ) and (1, µ 1). Hence the centre of the ellipse is at the midpoint (x 0, y 0 ) = (1/, (1/)(µ 1 + µ )) = (1/, (1/)( b 11 + b ) + s 1) = (1/, (1/) b s ). In the case where we have a line and not a genuine ellipse, either s 1 = 0 (then the midpoint is (1/, 1/) and so on the hyperbola) or b 11 = b and the line is horizontal (at y = (1/) b s 1/4 and so also meets the hyperbola). If b 11 b, then the point (x, y) = (1, µ 1 ) on the ellipse already satisfies 4xy 1 and so we assume that b > b 11. For the genuine ellipse case we write its equation in the form α 11 (x x 0 ) + α 1 (x x 0 )(y y 0 ) + (y y 0 ) + β = 0. (3) Using the information that the ellipse has a vertical tangent at (0, µ ) and its intersection with the line x = 1/ is the line segment {(1/, y) : y y 0 s 1 b 11 + b } (take ξ with ξ 1 = 1/ ), we can solve for the coefficients α 1 = µ µ 1 = b b 11 α 11 β = s 1 b 11 + b (4) = ( b 11 b ) + 4s 1 b 11 + b = α 1 4β We can rewrite the equation in the form (α 1 (x x 0 ) + (y y 0 )) 4β(x x 0 ) + β = 0 and so we can parametrise the ellipse via x = x 0 + (1/) sinω (5) y = y 0 (1/)α 1 sin ω + β cosω = (1/)( b 11 + b ) + s 1 (1/)( b b 11 ) sin ω +s 1 b 11 + b cosω (6) (0 ω π.) We look for ω [0, π/] where 4xy 1. We use b 11 + b b b 11 = ǫ 1 (say) and represent for convenience b 11 + b = cos θ (0 θ < π/). Note 4s 1 (b 1 b 1 ), s 1 (1/)(b 1 b 1 ) = 1 b s, 4s 1 1 cos θ and s 1 (1/) sinθ. Moreover b + b 11 cosθ. Thus y (1/) + (1/) cos θ + ǫ 1 (sin θ cosω cosθ sin ω) (7) Choose ω = tan 1 ((1/ ) tanθ), sin ω = sin θ/ sin θ + cos θ and ( ) sin θ 4xy 1 + (1/ + (1/) cos θ) cos θ 8

9 Remark 7 With some additional effort, we can adapt the proof above to establish the lower bound T a,b a b for the case a, b B(C ) (and thus get a stronger result than Theorem ). It seems that this does not follow from the methods used in [4]. Proof. A sketch of the additional details follows. We assume by symmetry that a / a b / b and normalise a = 1, b = 1 as before. This time we cannot assume λ = 0, but we note that det b λ /(1 + λ ) (for example, take b = ub 0 v where u, v are unitary and b 0 is diagonal with diagonal entries 1/ 1 + µ and µ/ 1 + µ, 1 µ λ ). In this case the ellipse will have vertical tangents at x = λ and x = 1 and will be centered at (x 0, y 0 ) = ((1+ λ )/, (1/) b s ). The equation (3) of the ellipse now has α 1 = b b 11 1 λ, β as in (4) and α 11 = α 1 4β/(1 λ ). We can rewrite the equation of the ellipse as (α 1 (x x 0 ) + (y y 0 )) and then we can parametrise via 4β (1 λ ) (x x 0) + β = 0 x = (1/)(1 + λ ) + (1/)(1 λ ) sin ω (8) (in place of (5)) and (6) as before. We now seek a point (x, y) on the ellipse where 4xy 1 + λ. To dispose of the case b 11 b we show 4y λ (and this also deals with the case where the ellipse degenerates into a line). Using b = 1, 4y 0 = b s = (b 1 b 1 ) = 1 + ( b 11 + b + b 1 b 1 ) λ 1 + b 11 b b 1 b λ 1 + λ. When ǫ 1 = b b 11 > 0 we choose the same ω as before. From the lower bound (7) and (8) we get the desired 4xy 1 + λ if we have cos θ λ /(1 + λ 4 ). For the remaining case note that y b 11 + b + s 1 = ( b 11 + b ) + b 1 b det b and the resulting y 1/ + λ /(1 + λ ) is a better lower bound that (7) when cos θ < λ /(1 + λ ). In this situation we do get 4xy 1 + λ. All eventualities are now covered because λ /(1 + λ 4 ) λ /(1 + λ ). 9

10 3 Commuting cases We consider now some cases where we can find relatively explicit formulae for T a,b. These may shed some light on the difficulty of finding any explicit formula for the norm of a general elementary operator. One may consider the Haagerup formula for the CB norm as an explicit formula, though we shall observe that this is not so simple to compute even in the simplest cases. The equality of the CB norm and the operator norm of T a,b holds if a, b are commuting normal operators. This appears already in the unpublished [9]. A significant part of the argument from [9] is published in [1, 5.4] and the remaining part uses the fact that all states on a commutative C*-algebra are vector states. (By the Putnam-Fuglede theorem the C*-algebra generated by commuting normal operators is commutative.) See also [16, Theorem.1] for a more general result on bimodule homomorphisms. Another proof (with slightly weaker hypotheses) is in [18]. We deal here only with H of dimension. Proposition 8 If H is two-dimensional and a, b B(H) commute, then = T a,b. Proof. We can find an orthonormal basis of H so that a and b both have upper triangular ( ) matrices. If a, b are diagonal, then they generate a commutative C*-subalgebra of B(H) and in this case that = a b + b a h = T a,b (see above). Now c 1, c obtained from Lemma 1 are also commuting upper triangular matrices. As used already in (1) (), from [18] we know that the convex hulls of the two sets of matrices intersect. In this case the sets are as not quite as before. Each c i should be replaced by δ i c i in the definition of W l and by 1/ δ i c i for W r. Moreover the equality = T a,b holds if and only if the sets themselves intersect. For either of the sets (say W l ) to consist of more than one element, the hermitian operator concerned must have a double eigenvalue of the maximum eigenvalue, which means that (taking the case W l ) δ i c i c i is a multiple of the identity matrix. But the following lemma asserts that this cannot happen unless δ 1 c 1 and δ c are simultaneously diagonalisable (the case where we know the result). So W l and W r have one element each, they intersect and the result follows. 10

11 Lemma 9 If a 1, a are commuting elements of B(H) with H of dimension and if a 1 a 1 + a a is a multiple of the identity, then a 1, a generate a commutative *-subalgebra of B(H). Proof. In a suitable orthonormal basis for H we can represent a 1, a as upper triangular matrices [ ] [ ] x1 y a 1 = 1 x y, a 0 z = 1 0 z and then the condition for them to commute is y 1 (x z ) = y (x 1 z 1 ). (For later reference we call this value ρ). So if y 1 = 0, then either y also zero (both matrices diagonal and we are done) or else x 1 = z 1 and a 1 = x 1 I is a multiple of the identity. But then a a is a multiple of the identity and this forces y = 0 (both diagonal again). In the case when y 1 and y are both nonzero, we compute [ ] a 1 a 1 + a a = x1 + y 1 + x + y y 1 z 1 + y z ȳ 1 z 1 + ȳ z z 1 + z Thus we have y 1 z 1 + y z = 0, which implies (z 1, z ) = ω(ȳ, ȳ 1 ) for some ω C. We also have equality of the two diagonal entries of the above matrix which gives us x 1 + x = ( ω 1)( y 1 + y ) Now x 1 = ρ/y + z 1 = ρ/y + ωȳ and x = ρ/y 1 ωȳ 1, yielding ρ + ωȳ y + ρ ωȳ 1 y 1 = ( ω 1)( y 1 + y ). and hence the impossible condition ρ ( y 1 + y ) = ( y 1 + y ) Example 10 Consider T a,b acting on B(C ) with a, b diagonal matrices. Then c 1, c in Lemma 1 are also diagonal and we can see then directly that c 1 c 1 + c c 1 ( δ 1c 1 c 1 + δ c c + δ 1 1 c 1 c 1 + δ 1 c c ) so that the Haagerup norm is minimised with δ 1 = δ = 1. Also c 1 c 1 + c c = z aa + z bb and so the Haagerup norm is the minimum of this. Say the diagonal entries are λ 1, λ for a and µ 1, µ for b. Normalising a and b to have norm one, we can assume max( λ 1, λ ) = 1 and max( µ 1, µ ) = 11

12 1. If they both attain the maximum at the same index then it is easy to see that T a,b = = a b. If not, assume by symmetry that λ 1 = 1 = µ and that µ 1 λ. The Haagerup norm is then the minimum value of the maximum of two functions, and can be computed by elementary means. It gives the norm (the same as the CB norm in this case) as T a,b = λ if λ 1/ and µ 1 < λ 1 µ 1 λ (1 µ1 )(1 λ ) otherwise (9) Summarising the calculation in a basis independent way, we can state the following. Proposition 11 Suppose that a, b B(C ) are commuting normal operators and that a / a b / b. If a, b attain their norms at a common unit vector, then T a,b = a b. If not T a,b = b a a if a 3/ a and b < 3 b ( a b )/( a a ) a b + a b a b ( a a )( b b ) otherwise (10) Proof. Note that in a suitable orthonormal basis of C, a, b will both be represented by diagonal matrices. 4 A formula for self-adjoint operators Our aim here is to present a proof of a formula from [1] that follows a similar approach to the one used in section. For a linear operator T: B(H) B(H) we denote by T the associated operator defined by T (x) = T(x ). We call T self-adjoint if T = T. Lemma 1 ([18]) For T: B(H) B(H) a self-adjoint elementary operator, there is a representation Tx = l i=0 ε ic i xc i with c i B(H), ε i { 1, 1} for each i and l T cb = c i c i. 1

13 Lemma 13 ([18]) Let T = T : B(H) B(H) be an elementary operator, Tx = k c ixc i l i=k+1 c ixc i with 0 k l and (c i ) l linearly independent. (We include k = 0 for the case where the first summand is absent and when k = l the second summand is absent.) Then the ordered pair (k, l k) (which we could call the signature ) is the same for all such representations of T. Example 14 ([1]) For T: B(H) B(H) given by Tx = axb + bxa with a, b linearly independent, we have T cb = inf { raa + sbb + ti(ab ) : r > 0, s > 0, t R, rs t = 1 } (where I(ab ) = (ab ba )/(i) is the imaginary part). Proof. We can rewrite Tx = c 1 xc 1 c xc if we take c 1 = (a + b)/ and c = (a b)/. Note for later use that we can undo this change by a = (c 1 + c )/, b = (c 1 c )/. According to Lemma 1 and Lemma 13 we can find T cb as the infimum of c 1 (c 1 ) + c (c ) where [c 1, c ] = [c 1, c ]α and α is an invertible matrix with the property that [ ] [ ] α α = As unitary diagonal α have no effect on the estimate c 1 (c 1 ) +c (c ) we can work modulo these unitaries and then elementary analysis of the possibilities shows that we need only consider the cases [ ] p p 1e α = iθ p 1e iθ p (with p 1, θ R). This leads us to consider only Hence [c 1, c ] = [pc 1 + p 1e iθ c, p 1e iθ c 1 + pc ]. T cb = inf p 1,θ R c 1(c 1) + c (c ) = inf (p 1)(c 1 c 1 + c c ) + 4p p 1R(e iθ c 1 c ) 13

14 = inf (p 1)(aa + bb ) + p p 1cos θ(aa bb ) +4p p 1sin θi(ab ) = inf (p 1 + p p 1cos θ)aa p 1,θ R +(p 1 p p 1 cosθ)bb +4p p 1sin θi(ab ) The claimed formula follows by taking r = p 1 + p p 1 cosθ, s = p 1 p p 1 cosθ and t = p p 1sin θ, noting that rs t = 1. We can recover p and cosθ from r, s (with r > 0, s > 0, rs 1) using r + s = (p 1), r s = 4p p 1cosθ. From the sign of t = ± rs 1 we get sin θ and so θ modulo π. Remark 15 In [1] it is also shown that, for T as in the example above, T cb = T. A more general result can be found in [18]. References [1] Pere Ara and Martin Mathieu, Local multipliers of C*-algebras, Springer Verlag (003). [] Mohamed Barraa and Mohamed Boumazgour, A lower bound of the norm of the operator X AXB + BXA, Extracta Math. 16 (001) 3 7. [3] Rajendra Bhatia, Matrix Analalyis, Graduate Texts in Mathematics 169, Springer-Verlag, New York (1997). [4] Ariel Blanco, M. Boumazgour and T. Ransford, On the norm of elementary operators, preprint (003). [5] L. J. Bunce, C.-H. Chu, L. L. Stachó and B. Zalar, On prime JB -triples, Quart. J. Math. Oxford () 49 (1998) [6] M. Cabrera and A. Rodriguez, Nondegenerately ultraprime Jordan Banach algebras, Proc. London Math. Soc. 69 (1994) [7] Erik Christensen and Allan M. Sinclair, A survey of completely bounded operators, Bull. London Math. Soc. 1 (1989)

15 [8] Edward G. Effros and Zhong-Jin Ruan, Operator Spaces, London Mathematical Society Monographs 3, Oxford Science Publications (000). [9] U. Haagerup, The α-tensor product of C*-algebras, unpublished manuscript, Univ. of Odense (1980). [10] Roger A. Horn and Charles R. Johnson, Matrix Analysis, Cambridge University Press (1985). [11] Bojan Magajna and Aleksej Turnšek, On the norm of symmetrised twosided multiplications, Bull. Australian Math. Soc. 67 (003) [1] Bojan Magajna, The norm of a symmetric elementary operator, Proc. Amer. Math. Soc. (to appear). [13] Martin Mathieu, Properties of the product of two derivations of a C*- algebra, Canad. Math. Bull. 3 (1989) [14] Martin Mathieu, More properties of the product of two derivations of a C*-algebra, Bull. Australian Math. Soc. 4 (1990) [15] V. I. Paulsen, Completely Bounded Maps and Dilations, Pitman Notes in Mathematics Series Vol. 146, Pitman, New York, [16] R. R. Smith, Completely bounded module maps and the Haagerup tensor product, J. Funct. Anal. 10 (1991) [17] L. L. Stachó and B. Zalar, On the norm of Jordan elementary operators in standard operator algebras, Publ. Math. Debrecen 49 (1996) [18] Richard M. Timoney, Computing the norms of elementary operators, Ilinois J. Math (to appear). 000 Mathematics Subject Classification: 47B47, 46L07, 47L5 School of Mathematics Trinity College Dublin Ireland richardt@maths.tcd.ie 15