Blog > The Other Superoperator Isomorphism

The Other Superoperator Isomorphism

November 20th, 2009

A few months ago, I spent two posts describing the Choi-Jamiolkowski isomorphism between linear operators from Mn to Mm (often referred to as “superoperators“) and linear operators living in the space Mn ⊗ Mm. However, there is another isomorphism between superoperators and regular operators — one that I’m not sure of any name for but which has just as many interesting properties.

Recall from Section 1 of this post that any superoperator Φ can be written as

\Phi(X)=\sum_iA_iXB_i.for some operators {Ai} and {Bi}. The isomorphism that I am going to focus on in this post is the one given by associating Φ with the operator

M_\Phi:=\sum_iA_i\otimes B_i^{T}.

The main reason that MΦ can be so useful is that it retains the operator structure of Φ. In particular, if you define vec(X) to be the vectorization of the operator X, then

{\rm vec}(\Phi(X))=M_\Phi{\rm vec}(X).

In other words, if you treat X as a vector, then MΦ is the operator describing the action of Φ on X. From this it becomes simple to compute some basic quantities describing Φ. For example, the induced Frobenius norm,


is equal to the standard operator norm of MΦ. If n = m then we can define the eigenvalues {λ} and the eigenmatrices {V} of Φ in the obvious way via

\Phi(V)=\lambda V.

Then the eigenvalues of Φ are exactly the eigenvalues of MΦ, and the corresponding eigenvectors of MΦ are the vectorizations of the eigenmatrices of Φ. It is similarly easy to check whether Φ is invertible (by checking whether or not det(MΦ) = 0), find the inverse if it exists, or find the nullspace (and a pseudoinverse) if it doesn’t.

Finally, here’s a question for the interested reader to think about: why is the transpose required on the Bi operators for this isomorphism to make sense? That is, why can we not define an isomorphism between Φ and the operator

\sum_iA_i\otimes B_i?

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