# Linear maps $f: M_n \longrightarrow \mathbb{R}$ such that $f(AB) = f(BA)$

Let $$M_n$$ be the vector space of $$n \times n$$ matrices over $$\mathbb{R}$$. Find (with proof) all linear maps $$f: M_n \longrightarrow \mathbb{R}$$ such that $$f(AB) = f(BA)$$ for all matrices $$A$$ and $$B$$.

I know that the trace map is one such $$f$$. The determinant map satisfies $$det(AB) = det(BA)$$ -- but isn't a linear map, so it doesn't suffice.

Is there a constructive way to find all of the desired linear maps $$f$$, and show that there are no others ?

Any help would be appreciated. Thanks (=

Note that $$f(AB-BA) = 0$$. The set of commutators is the same as the set of trace zero matrices.
Hence $$f$$ satisfies the condition iff $$\ker f$$ contains all the zero trace matrices.
(Note that this implies that $$f$$ must be a multiple of the trace operator.)
• Basically the kernel of a non zero real valued linear operator must have codimension one. So either $f$ is zero or it has the same kernel as trace (since we have containment). – copper.hat Jan 22 at 4:51