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Let $G=\{M_1, M_2, \ldots ,M_{\ell}\} \subset \mathcal{M}_n(\mathbb{R})$, such that $G$ forms a group for the usual matrix multiplication.

Denote $A= M_1+ \cdots +M_{\ell}$.

Show that $$A=0 \iff \operatorname{tr}(A)=0$$

I am totally stuck here, if someone has any ideas, please share it.

Thank you in advance.


marked as duplicate by darij grinberg, Lord Shark the Unknown, José Carlos Santos linear-algebra Feb 3 at 10:34

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  • $\begingroup$ There must be some typos in your post. $\endgroup$ – Test123 Apr 27 '14 at 0:30
  • $\begingroup$ What do you mean by $\mathcal{M}_n(\mathbb{R})^{\ell}$? What does the $\ell$ mean? $\endgroup$ – Omnomnomnom Apr 27 '14 at 0:33

Here's an idea: using the fact that $G$ forms a group, note that for each $i$, we have $$ M_iG = \{M_iM_1,M_iM_2,\dots,M_iM_\ell\} = G $$ It follows that $$ A^2 = \left(\sum_{i=1}^\ell M_i\right)^2 = \sum_{i=1}^\ell \sum_{j=1}^\ell M_i M_j = \sum_{i=1}^\ell \left(\sum_{M \in G} M \right) = \ell\cdot (M_1 + \cdots + M_\ell) = \ell\cdot A $$ That is, $A^2 = \ell A$, which is to say that $A(A-\ell I) = 0$. What does this allow us to deduce about $A$'s minimal polynomial?

By considering the eigenvalues of $A$ (what can they be?) and noting that $A$ must be diagonalizable (why?), we may conclude that if $A$ has a trace of $0$, it can only be the zero matrix.

  • $\begingroup$ A is diagonalizable (canceled by a polynomial annihilator) , and the eigenvalues are in $\{0,\ell\}$. Using the fact that trace=sum of the eigenvalues I can conclude. Thanks!! $\endgroup$ – user146010 Apr 27 '14 at 11:15