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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 …
Nathaniel Johnston's user avatar
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. …
Nathaniel Johnston's user avatar
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 …
Nathaniel Johnston's user avatar
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 …
Nathaniel Johnston's user avatar