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Suppose the collection $\{A_1, A_2,... A_k\}$ forms a group under matrix multiplication, where each $A_i$ is an $n×n$ real matrix. Let $A=\sum_{i=1}^kA_i$.

  1. Show that $A^2=kA$
  2. If the trace of $A$ is zero, then show that $A$ is the zero matrix

My try: Since $G=\{A_1, A_2,... A_k\}$ forms a group under matrix multiplication, there exists $A_p=I$ for some $ p\in \{1,2,...k\}$. WLOG let $A_1=I$. Also being group, for each $A_i \in G$ there exist $A_j \in G$ for some $j\in\{1, 2,...,k\}$ I cannot proceed further how to solve this problem. Please help me. Thank you in advance

Debprasad Kundu
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  • If $G$ is a group and $g\in G$ some fixed element then the set ${gh: h\in G}$ is a permutation of the elements of $G$. So expand the product $A^2$ into a sum of products and then partition the sum by grouping together elements with the same multiplier on the left. –  Jun 05 '21 at 09:29

1 Answers1

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  1. Notice multiplying by $A_i$ permutes the group, $A^2=\sum_{i=1}^k \left(\sum_{j=1}^k A_iA_j\right)=\sum_{i=1}^k \left(\sum_{j=1}^k A_j\right)=kA.$

  2. I'm resisting the intuition and temptation to use representation theory here. Although I think that would be a really interesting (and easy) approach. That said, I'm not completely confident that the following is the best approach to this.

If $\operatorname{tr}(A)=0$, then from 1 we know $\operatorname{tr}(A^n)=0$ for all $n\geq 1$. One can see this as $A^n=k^{n-1}A$. See this page (or simply think about eigenvalues over $\mathbb C$), we know $A$ is nilpotent so $A^n=0$ for some $n$. Therefore $A=0$ using $A^n=k^{n-1}A$ again.

August Liu
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