Completely bounded operators

Continuing with the linear algebra theme, let’s consider the space \mathcal M of all 2\times 2 matrices with real entries. This is a 4-dimensional vector space; we can treat a matrix (a_{ij}) as a vector (a_{11},a_{12},a_{21},a_{22}). So a general linear map T\colon \mathcal M\to \mathcal M can be described by a 4\times 4 matrix. That is, the space of linear maps L(\mathcal M) is 16-dimensional.

However, some linear maps T\colon \mathcal M\to \mathcal M arise more naturally than others; they are somehow more “matricial”. For example, left multiplication by a fixed matrix A\in\mathcal M. Or right multiplication by a fixed B\in\mathcal M. Or, to hit both with one stone, the two-sided multiplication map T_{A,B}\in L(\mathcal M) defined by X\mapsto AXB. The matrix of this linear map is the Kronecker product A\otimes B^T:

\displaystyle A\otimes B^T = \begin{pmatrix}  a_{11}b_{11} & a_{11}b_{21} & a_{12}b_{11} & a_{12}b_{21} \\  a_{11}b_{12} & a_{11}b_{22} & a_{12}b_{12} & a_{12}b_{22} \\  a_{21}b_{11} & a_{21}b_{21} & a_{22}b_{11} & a_{22}b_{21} \\  a_{21}b_{12} & a_{21}b_{22} & a_{22}b_{12} & a_{22}b_{22} \end{pmatrix}

Observe that the set of Kronecker products is not closed under addition. This is somehow unsatisfactory. It would be nice if “matricial” transformations, whatever they are, formed a linear subspace of L(\mathcal M). This can be fixed by generalizing the two-sided multiplication map: any representation \pi of \mathcal M on some vector space V and any pair of operators A\colon V\to \mathcal M, B\colon \mathcal M\to B, give rise to a linear map X\mapsto A\pi(X)B. Given another triple (\tilde \pi, \tilde A, \tilde B), we can take the direct product of representations and arrange the operators accordingly:

\displaystyle \begin{pmatrix} A & \tilde A\end{pmatrix} \begin{pmatrix} \pi(X) & 0 \\ 0 & \tilde \pi(X) \end{pmatrix} \begin{pmatrix} B \\ \tilde B\end{pmatrix} = A\pi(X)B+\tilde A\tilde\pi(X)\tilde B

Hurray, we have a linear subspace of L(\mathcal M)! Now, what is this subspace? Maybe all L(\mathcal M)? Hm.

Let \mathcal M_{2n} be the space of square matrices of size 2n. We can think of them as block matrices with n\times n blocks of size 2\times 2. A linear map T\in L(\mathcal M) induces T_n\in L(\mathcal M_{2n}): just apply T to every block. If T comes from a triple (\pi,A,B) as above, then T_n can be obtained if we let \mathcal M_{2n} act on \bigoplus^n V blockwise and multiply on left and right by A\otimes I_n and B\otimes I_n. In this case the norm of T_n is bounded by \|A\|\|B\|the bound is independent of n. Note that the norm of T_n is taken with respect to the matrix norm on \mathcal M_{2n}

An operator T\in L(\mathcal M) such that \sup_n \|T_n\|<\infty is called completely bounded. We just saw that any operator of the form X\mapsto A\pi(X)B is completely bounded. The converse is also true; it was proved independently by Haagerup, Paulsen and Wittstock (and not just for 2\times 2 matrices, of course).

This certainly helps. Still, which operators on \mathcal M are completely bounded? So far we saw that they form a linear space of dimension between 8 and 16…

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