Re: I'm puzzled (Q on norms of matrices)


>
> (A1) Ban(A,\sum_i B_i) = \sum_i Ban(A, B_i)
>
> (A2) Ban(\sum_i B_i, A) = \prod_i Ban(B_i, A)
>
> Now, my puzzlement is the following: putting (A1) and (A2) together we
> get
>
> Ban(\sum_i B_i, \sum_j B_k) = \sum_i \prod_k Ban(B_i, B_k)
>
> And also
>
> Ban(\sum_i B_i, \sum_j B_k) = \prod_k \sum_i Ban(B_i,B_k)
>
> Which gives an isometric isomorphism bewteen the right hand sides.
> These can be viewed as matrices of bounded linear operators
> T_ik:B_i->B_k with the respective norms
>
> \sum_i \max_k |T_ik|
>
> \max_k \sum_i |T_ik|
>
> But if we compute this on diagonal matrices, we get the equality
>
> \sum_i |T_ii| = \max_k |T_kk|
>
> At this moment, I have surely screwed up somewhere. Either in the
> "proofs" of (A1) and (A2), which I have not included, or on the
> reasoning following it. Can anyone detect the error? And if it is in
> (A1) or (A2) can you give a counterexample, please?
>

(A1) is incorrect. And you essentially wrote it already.
Counterexample: Let B_1 = B_2 = R,
A = R^2 with the sum norm (l_1 norm, taxicab metric).
Take the operator that corresponds to the identity matrix.
The operator norm is 1, but the sum of the two component norms is 2.

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
.