# Show that $A=0 \iff \mathrm{tr}(A)=0$ where $A= M_1+ \cdots +M_{\ell}$.

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

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

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

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

What do you mean by $\mathcal{M}_n(\mathbb{R})^{\ell}$? What does the $\ell$ mean? – 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.
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!! – Edwin Apr 27 '14 at 11:15