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Determine the trace and determinant of the linear operator (on the space $\mathbb{F^{n\times n}}$) that sends the matrix $M\to AMB$ where $A$ and $B$ are $n\times n$ matricies

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Let $L_A$ be left multiplication by $A$. Then you can split $F^{n\times n}$ into the direct sum of $L_A$-invariant $n$ copies of $F^n$ (the columns of $M$), on each of which $L_A$ acts by the usual multiplication of a matrix by a column vector. Hence $\det(L_A)=[\det(A)]^n$. Similarly, $\det(R_B)=[\det(B)]^n$. Hence $\det(L_AR_B)=\det(L_A)\det(R_B)=[\det(A)]^n[\det(B)]^n.$ To compute $tr(L_AL_B)$, refine the previous argument to $L_AR_B=A\otimes B^*$ (thinking of $F^{n\times n}=Hom(F^n, F^n)=F^n\otimes (F^n)^*$), hence $tr(L_AR_B)=tr(A)tr(B)$.

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    $\begingroup$ Very nice. It also follows that nothing is gained from restricting to the square matrix case. If acting on $M_{n,m}(F)$ on the left by $A\in M_{n,n}(F)$ and on the right by $B\in M_{m,m}(F)$, one gets $\det(L_AR_B)=\det(A)^m\det(B)^n$, and as before $\def\tr{\operatorname{tr}}\tr(L_AR_B)=\tr(A)\tr(B)$. $\endgroup$ – Marc van Leeuwen Mar 9 '14 at 8:43
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    $\begingroup$ I can not understand this part "you can split $F^{n\times n}$ into the direct sum of $L_A$-invariant $n$ copies of $F^n$ ". Could you please add some details? $\endgroup$ – Majid May 20 '18 at 21:29
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    $\begingroup$ @Gil Bor, I am still waiting for your favor to explain your solution in more details. $\endgroup$ – Majid May 21 '18 at 18:48
  • $\begingroup$ @Gil Bor could you explain the part you said "split $F^{n\times n} $ into the direct sum of $L_A$-invariant $n$ copies of $F^n$ (the columns of M) " $\endgroup$ – J. Kyei Oct 17 '18 at 1:20
  • $\begingroup$ Sorry for the delay. Let $V_j$ be the subspace of $Mat_{n,n}(F)$ consisting of matrices whose only non-zero elements are in their $j$-th column. Then (1) $Mat_{n,n}(F)=V_1\oplus\ldots\oplus V_n$, (2) each $V_j$ is $L_A$ invariant, (3) each $V_j$ is isomorphic to $Mat_{n,1}(F)$ in an obvious way, such that the action of $L_A$ on $M_j$ goes over to multiplying a column vector on the left by $A$. It follows that the determinant of the action of $L_A$ on $M_j$ is $\det(A)$ and that $\det(L_A)=[\det(A)]^n.$ $\endgroup$ – Gil Bor Jan 2 at 23:06

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