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

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.

marked as duplicate by darij grinberg, Lord Shark the Unknown, José Carlos Santos linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 3 at 10:34

• 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!! – user146010 Apr 27 '14 at 11:15