Show that $A=0 \iff \operatorname{tr}(A)=0$ where $A= M_1+ \cdots +M_{\ell}$. [duplicate]

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.