Only commutators solve this functional equation

Question

I need to prove that the only linear transformations $f\colon M_{n\times n}\rightarrow M_{n\times n}$that solve the functional equation in $\mathcal{L}(M_{n\times n})$ $$f(XY)=f(X)Y+Xf(Y)$$ are the commutators. That is, I have to show that for every linear solution $f$ there exists a matrix $A$ such that $f(X)=[A,X]$ for all $X$. Bearing in mind that every commutator satisfies the equation, I have tried to solve this studying the sets $V_A=\{X\in M_{n \times n}\colon f(X)=[A,X]\}$, where $A\in \ker f$, which happen to be subalgebras that are stable under $f$ and include inverses. Specifically, I have been looking for a maximality argument that leads me to conclude that some of these $V_A$ must be all of $M_{n\times n}$. So far, I have not been able to exploit the finite dimension of $M_{n\times n}$. This problem appears as an exercise at the end of an introductory chapter on matrices and finite dimensional vector spaces in Katsumi Nomizu's Fundamentals of Linear Algebra, so little else beyond the rank-nullity theorem is assumed.

Answer 1

Here is an elementary proof. Denote the underlying field by $\mathbb F$. Pick two vectors $u,v\in\mathbb F^n$ such that $u^Tv=I_1$, the $1\times1$ identity matrix. Since $f$ is linear, so are the mappings $x\mapsto f(xu^T)v\in\mathbb F^n$ and $y\mapsto u^Tf(vy^T)\in\mathbb F^{1\times n}$. Hence there exist two matrices $A$ and $B$ such that $f(xu^T)v=Ax$ and $u^Tf(vy^T)=y^TB$ for all $x,y\in\mathbb F^n$. It follows that \begin{aligned} f(xy^T) &=f\left(x(u^Tv)y^T\right)\\ &=f\left((xu^T)(vy^T)\right)\\ &=f(xu^T)vy^T+xu^Tf(vy^T)\\ &=Axy^T+xy^TB.\\ \end{aligned} Since $M_n(\mathbb F)$ is spanned by the set of all rank-one matrices, the above gives $f(X)=AX+XB$ for all matrices $X$. However, as $f(I)=f(I^2)=f(I)I+If(I)=2f(I)$, we must have $f(I)=0$. Hence $A+B=0$ and $f(X)=AX+XB=[A,X]$.