positive linear bijection whose inverse isn't positive

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).

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