# Interchanging Strong Operator convergent sums

In the book on operator algebras by Stratila and Zsido, they discuss in Ch.2 the idea of taking a Hilbert space $H$ and an index set $I$ and associating to it the Hilbert space that is the direct sum of $card(I)$ copies of $H$ To each bounded linear operator defined on this direct sum, we associate the matrix of operators $x_{ik}$ on $B(H)$ such that:

$x=\sum_{i, k} u_ix_{ik}u_k^*$

where the sum is strongly convergent in the sense that I can sum in i first or in k, and either way the sum converges to the same value. (At least that is what I assume people mean when they write this expression.) I want to verify the usual matrix multiplication rule, where (.) means the operator defined by a matrix:

$(x_{ik})(y_{ik})=(\sum_{j} x_{ij}y_{jk})$

The sum in the right hand member is SO convergent. I can verify that the right hand side inside the parenthesis is SO convergent, but not that it gives rise to a matrix that defines a bounded operator on the direct sum. Furthermore, even if I could verify that, I would then need to interchange two summation symbols to conclude the equality of matrices. Please help me with these two issues.

-
Which norm do you use on the space of matrices. Is it true that $U_i$ defined by $U_j(\oplus_{i\in I}\xi_i)=\xi_j$? –  Norbert Jun 8 '12 at 23:32
Apparently somehow my account seems to be bound to the computer as opposed to the website, and I had a very hard time figuring out how to simply leave a comment. Anyway, yes that is correct on the formula, and there is no norm on the space of matrices. There doesn't have to be. I never speak of convergence of matrices, but if I ever did, I would mean in the operator norm, since by "space of matrices" we mean the space of those arrays that determine a bounded operator on the direct sum. –  Jeff Jun 9 '12 at 1:31
To me the issue here is that if i and j are the indices of the sum for the matrix expression of x, and k, l for y, then j and k are set equal in the product xy by a dirac delta, but after that the j sum is between the i and l sum. On the other hand, for the RHS of my desired equality, the j sum comes innermost. –  Jeff Jun 11 '12 at 18:48
I think, when you define an operator by matrix of operators you need additionally require uniform (by indices) boundedness of the norms of matrix entries. Otherwise matrices can define unbounded operators –  Norbert Jun 11 '12 at 18:53
Norbert that does not address the interchange of the order of summation. Fortunately this argument does, strangely enough without proving that the joint sum in i and j at the same time is the same as the sum in i and j or vice versa. Previously to evaluate the left side of my equality, I wrote out the matrix expansion for x and y, and then multiplied. It hit me only yesterday that what works is to take the matrix expansion of xy, and then insert a partition of 1 as a sum of projections. This makes the sums occur in the correct order to agree with the right side. –  Jeff Jun 14 '12 at 0:43
show 1 more comment