0

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 $A$.

postylem
  • 350
Cheburashka
  • 93
  • By singular value decomposition, you may assume that $A$ is a nonnegative diagonal matrix. – user1551 Jul 24 '23 at 13:39
  • Are you sure it is a vector space over $\mathbb{R}$ ? Not on $\mathbb{C}$ ? – Jean Marie Jul 24 '23 at 17:22
  • @JeanMarie Yes, the complex matrix vector space over real numbers. – Cheburashka Jul 25 '23 at 02:08
  • 2
    I'm not seeing how you can do better than get a bound on $\dim \ker F$ here. The fact that you are looking at a vector space over reals combined with the structure of $F$ strongly suggests you write $M_n$ as a direct sum of Hermitian and skew-Hermitian matrices, and try applying $F$ to each subspace – user8675309 Jul 25 '23 at 02:10
  • @user8675309 $F$ maps any matrix to a Hermitian Matrix and I don’t know how to deal with $AX$. I tried to consider $AX$ as a linear map and tried to find its kernel. – Cheburashka Jul 25 '23 at 03:45
  • @user1551 are you sure that SVD is helpful here? Your comment is identical to Ben's here: https://math.stackexchange.com/questions/4653361/matrix-of-operator-on-space-of-symmetrical-matrices/ but that problem is for the much simpler 'product' operator and this is for the 'sum' operator. [If there was less structure I would explicitly reference Kronecker products and Kronecker sums... the former plays very well with SVD and the latter does not.] – user8675309 Jul 25 '23 at 15:51
  • @user8675309 By some changes of bases, I calculate the dimension of the range of $F$ directly without considering nullity. – user1551 Jul 25 '23 at 16:30
  • @user1551 -- thanks. I plan on taking a closer look tomorrow. – user8675309 Jul 25 '23 at 17:18

1 Answers1

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 $$ \operatorname{rank}(F)=\operatorname{rank}(f)=\dim_\mathbb R V=k^2+2k(n-k). $$

user1551
  • 149,263
  • Thank you very much. At last, I think the dim of $H$ is $\frac{1}{2} k(k+1)$ and the dim of $W$ is $k(n-k)$. Since if we decide the $W$ we also decide the $W^*$, we should only count the dim once. – Cheburashka Jul 26 '23 at 06:58
  • 1
    @Cheburashka No. The underlying field here is $\mathbb R$, not $\mathbb C$. The $\mathbb R$-vector space of all $k\times k$ Hermitian matrices is $k^2$ and the $\mathbb R$-vector space of all $k\times(n-k)$ complex matrices is $2k(n-k)$. – user1551 Jul 26 '23 at 07:16
  • Thank you so much! – Cheburashka Jul 26 '23 at 07:43