Consider the linear operator $f_A$ on $K^{n×n}$ defined by $f_A(X)=AX-XA$ where $K$ is field. Is $f_A$ is surjective? If no, then under what conditions $f_A$ is surjective?
My attempt: I can see easily that $f_A$ is not surjective if I take $K=\mathbb{R}$. Hence in general $f_A$ is not surjective.
But I am not able to determine the condition under which $f_A$ is surjective.
It is given in a hint that, if $char(K)$ does not divide $n,$ then $f_A$ is not surjective. So does that mean, if I consider $K$ to be a finite field having characteristic which divides $n,$ then $f_A$ is surjective? How? Please help!
Hint: Examine the trace of $f_A(X)$.