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31
votes
Square root of doubly positive symmetric matrices
Robert Bryant already showed via an example that the answer is "no". To come up with lots of counterexamples, recall that (under some mild assumptions) if $A$ has maximal eigenvalue $\lambda_{\text{ma …
18
votes
Accepted
positive not completely positive maps
A couple of side notes:
This map shows that Choi's theorem on complete positivity is optimal in some sense: if $k \geq n$ then $k$-positivity implies complete positivity, but if $k < n$ then $k$-positivity … and $(k+1)$-positivity are indeed different sets. …
7
votes
Accepted
Tensor product of positive linear maps is positive
No. A standard example is given by $A_1 = A_2 = B_1 = B_2 = M_2(\mathbb{C})$, where we choose $\pi_1$ to be the identity map and $\pi_2$ to be the transpose map. These maps are positive, but $\pi_1 \o …
4
votes
Accepted
Questions about hermitian positive semidefinite matrices
You noted in your "Edit 2" that these $n \times n$ matrices $A$ are exactly those that can be written in the form
$$
A = \sum_j \mathbf{v_j}\mathbf{v}_{\mathbf{j}}^*,
$$
where each $\mathbf{v_j}$ has …