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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 (=

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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.)

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    $\begingroup$ See math.stackexchange.com/q/181430/27978 for some proofs of said fact. $\endgroup$ – copper.hat Jan 22 at 4:40
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    $\begingroup$ Marvelous. I didn't know that fact about commutators here. Thanks for your comment. (= $\endgroup$ – michiganbiker898 Jan 22 at 4:45
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    $\begingroup$ @michiganbiker898: I added one more important sentence to the answer. $\endgroup$ – copper.hat Jan 22 at 4:46
  • $\begingroup$ Thanks, I just saw that. I believe I can show why that must be true. (= $\endgroup$ – michiganbiker898 Jan 22 at 4:50
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    $\begingroup$ 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). $\endgroup$ – copper.hat Jan 22 at 4:51

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