Showing that $U_{A,B}$ is a subspace

Question

Show that $$ U_{A,B} = \{X \in \mathbb{K}^{n \times n}, f(X^T) = f(X)^T \} $$ is a subspace of $V = \mathbb{K}^{n \times n}$ with $f: V \to V, X \mapsto AXB$ and $A,B \in V$.

I'm not sure what to show. Showing $f((v+w)^T) = f(v+w)^T $, with $v,w \in V$ does not get me anywhere and it feels not right. But this could just be me lacking intuition.

Answer 1

Note that the kernel of any linear map is a subspace. With that said, it suffices to verify that $U_{A,B}$ is indeed the kernel of the linear map $g:V \to V$ defined by $g(X) = f(X^T) - f(X)^T$.

Answer 2

To show that $U_{A,B}$ is a subspace you need to show two things:

1. $X+Y\in U_{A,B} \forall X, Y \in U_{A,B}.$ This is straightforward since, $$f((X+Y)^T) = A(X+Y)^TB = A(X^T + Y^T)B \\= AX^TB + AY^TB \\= f(X^T)+f(Y^T) \\= f(X)^T + f(Y)^T \\= (f(X)+f(Y))^T \\= f(X+Y)^T$$ since $f$ is linear.

2. Next $aX\in U_{A,B}$ for scalar $a\in \mathbb{K}$ and matrix $X\in U_{A,B}.$ This is again trivial since $$f(aX^T) \\= aAX^TB \\= af(X)^T \\ =(af(X))^T \\= (f(aX))^T$$ where I used the fact that $f$ is linear.