Consider $n\times n$ matrices in $M_n(\mathbb{C})$ and consider the transposition map:
$$T: M_n(\mathbb{C}) \to M_n(\mathbb{C})\\A\mapsto A^T$$
This map is positive since it maps positive definite matrices to positive definite matrices.
On the other hand, the eigenvalues of $T$ are $\pm 1$ (see here)
, so they are not all positive.
Now, if $T$ were Hermitian/symmetric then this would contradict the fact that a Hermitian matrix is pos. def. iff all eigenvalues are positive.
So, am I correct to assume that this means that transposition is not Hermitian (or symmetric in the real case)? Probably, a stupid question but: To check if transposition is not symmetric, I want to take the tranpose. Is the transpose of $T$ simply the identity map?
Best Answer
It's clear what taking the transpose of a linear map over $\Bbb C^n$ means, but it's not clear what the "transpose" of a map over $M_n(\Bbb C)$ should refer to.
To answer this question, consider the following definition of the "transpose" of a map over $\Bbb C^n$. Let $\langle \cdot, \cdot \rangle$ denote the usual "dot product" over $\Bbb C^n$, which is to say that $\langle x, y \rangle = x^\top y = \sum_i x_iy_i$. For a map $A$ over $\Bbb C^n$, the transpose of $A$ is the unique map $B$ such that for all $x,y \in \Bbb C^n$, we have $$ \langle Ax,y \rangle = \langle x, By \rangle. $$
With this as motivation, one answer to this question is as follows. Define the bilinear map $(\cdot,\cdot)$ over $M_n(\Bbb C)$ by $ (A,B) = \operatorname{tr}(AB^\top) = \sum_{i,j} a_{ij}b_{ij}. $ We can say that the "transpose" of $T$ is the unique map $S$ over $M_n(\Bbb C)$ for such that for all $A,B \in M_n(\Bbb C)$, $$ (T(A),B) = (A,S(B)) $$ With this definition in mind, we note that if $T$ is the transpose map, $$ (T(A),B) = \operatorname{tr}(A^\top B) = \operatorname{tr}[(A^\top B)^\top] = \operatorname{tr}(B^\top A) = \operatorname{tr}(AB^\top) = (A,T(B)). $$ In other words, relative to this definition of the "transpose", the transpose map is "symmetric".
Along similar lines, if we transfer the definition of "positive definite" to this context, we find that $T$ fails to be positive definite even though it is a "positive" map in the sense you described. For a map $A$ over $\Bbb C^n$, if $A$ is positive definite, then every non-zero real vector $x \in \Bbb R^n$ should satisfy $\langle Ax,x \rangle > 0$.
Analogously, for a map $T$ over $\Bbb C^n$, if $T$ were "positive definite" by a reasonable definition, then it should hold that $(T(X),X) > 0$ holds for all non-zero $X \in M_n(\Bbb R)$. However, if we take $X$ to be a non-zero, real, skew-symmetric matrix, then it's easy to see that $$ (T(X),X) = (-X,X) = -(X,X) < 0. $$