The rank of a linear transformation into the space of $n \times n$ matrices

Question

Denote the $n\times n$ complex matrix space as $M_n$ and it is a vector space over $\mathbb R$.
Matrix $A\in M_n$ is of rank $k<n$, and we consider the following linear map $$F:M_n \to M_n$$ $$F(X)=\frac{1}{2}(\overline{(AX)^T}+AX)$$ $A^T$means the transpose of $A$. And I need to find the rank of $F$.
I am trying to find the kernel of the map $F$, which is to find the dimension of $\overline{(AX)^T}=-AX$. But I don't know how to use the rank of $A$.

Answer 1

Let $A=P(D\oplus0)Q$ where $P$ and $Q$ are invertible and $D\in GL_k(\mathbb C)$. One may express $A$ this way by using singular value decomposition, Smith normal form or rank decomposition.

Define $g(X)=Q^{-1}XP^\ast$ and $h(X)=2P^{-1}X(P^\ast)^{-1}$. Since $g$ and $h$ are isomorphisms on $(M_n(\mathbb C),\mathbb R)$, the rank of $F$ is identical to the rank of $$ f=h\circ F\circ g: X\mapsto X^\ast(D^\ast\oplus0)+(D\oplus0)X. $$ Partition $X$ as $\pmatrix{Y&Z\\ \ast&\ast}$, where $Y$ is $k\times k$. Then $$ f(X)=\pmatrix{Y^\ast D^\ast+DY&DZ\\ Z^\ast D^\ast&0}. $$ Hence $$ \operatorname{range}(f)\subseteq V:=\left\{\pmatrix{H&W\\ W^\ast&0}:H=H^\ast\in M_k(\mathbb C),\ W\in M_{k,n-k}(\mathbb C)\right\}. $$ We actually have $\operatorname{range}(f)=V$, because the equation $$ f\pmatrix{Y&Z\\ \ast&\ast}=\pmatrix{H&W\\ W^\ast&0} $$ when $H$ is Hermitian is solvable by putting $Y=\frac12D^{-1}H$ and $Z=D^{-1}W$. It follows that $$ \operatorname{rank}(F)=\operatorname{rank}(f)=\dim_\mathbb R V=k^2+2k(n-k). $$