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A set $K$ consists of $\mathbb R^{n\times n}$ invertible matrixes such that $\left| K \right| = m$ and its elements are $A_i$, $i=1,2,\cdots,m$. If $\forall {A_i}{\kern 1pt} ,{A_j} \in K$, $A_iA_j$ and ${A_i}^{ - 1}\in K$, and $\sum\limits_{i = 1}^m {tr({A_i}) = 0} $, then show that $\sum\limits_{i = 1}^m {{A_i} = 0} $.

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    @J.M., the question mark has been absorbed in "if..., then" with a missing phrase "show that" or something alike :)2011-10-25

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Let us first note that $A_iA_k=A_jA_k\Leftrightarrow A_i=A_j$ by right multiplication by $A_k^{-1}$ which is an element of $K$.

Therefore we can say that, if $X=\sum_{i=1}^m A_i,$ $XA_j=X,\quad\forall j\in\{1,\dotso,m\},$ therefore $X\left(\sum_{i=1}^m A_i\right)=mX\Rightarrow X^2-mX=0.$ It then follows that the eigenvalues of $X$ are either $0$ or $m>0$, but since $\text{tr}(X)=0,$ then all the eigenvalues are $0$. Therefore $X$ is nilpotent, i.e. there exists $k\in\mathbb N$ such that $X^k=0.$ But then, exploiting the fact (already hinted above) that $X^k=m^{k-1}X,$ we get $m^{k-1}X=0$ an then finally $X=0,$ as we wanted. Hope it is correct. regards.

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    Nice proof, the argument that $XA_j=X$ for all $j$ is particularly clever.2011-10-25