Let $f: M_n(\mathbb{C})\rightarrow M_n(\mathbb{C})$ be a linear positive map, e.g. $A\geq 0 \implies f(A)\geq 0$. I think I've seen somewhere (although I don't recall where) that it does not have to be the case that when $f$ is a bijection that $f^{-1}$ has to be positive as well. Can somebody give an explicit example of such a map (or a proof that $f^{-1}$ has to be positive as well if I were mistaken).

  • $\begingroup$ By positive do you mean entriwise positive? $\endgroup$ – Zach Boyd Oct 20 '17 at 13:08
  • $\begingroup$ @ZachBoyd: if that were the case, it wouldn't be "operator theory". $\endgroup$ – Martin Argerami Oct 20 '17 at 13:09
  • $\begingroup$ @Zach: Nope, I mean positive semi-definite $\endgroup$ – John Oct 20 '17 at 13:10
  • $\begingroup$ Oops forgot to read the tags. $\endgroup$ – Zach Boyd Oct 20 '17 at 13:11

Let $T:M_n(\mathbb{C})\rightarrow M_n(\mathbb{C})$ be $T(X)=X+\frac{1}{2}X^t$.

This map is clearly a linear positive map. Moreover, $T(X)=\dfrac{3}{2}\dfrac{(X+X^t)}{2}+\dfrac{1}{2}\dfrac{(X-X^t)}{2}$.

Note that, $T(S)=\frac{3}{2}S$ and $T(A)=\frac{1}{2}A$, for every symmetric matrix $S$ and for every anti-symmetric matrix $A$.

Thus, $\frac{3}{2}$ and $\frac{1}{2}$ are the only eigenvalues of $T$.

So $T$ is a bijection and its inverse is $T^{-1}(X)=\dfrac{2}{3}\dfrac{(X+X^t)}{2}+2\dfrac{(X-X^t)}{2}=\frac{4}{3}X-\frac{2}{3}X^t$.

Let $v=(1,i)^t$. So $T^{-1}(v\overline{v}^t)=\frac{4}{3}v\overline{v}^t-\frac{2}{3} \overline{v}v^t$.

Note that $(\frac{4}{3}v\overline{v}^t-\frac{2}{3} \overline{v}v^t)\overline{v}=-\frac{4}{3}\overline{v}$.

So $T^{-1}$ is not a positive map.


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