If $M=\sum_{g\in G}\rho(g)$ then $\operatorname{tr}(M)=0$ implies $M=0$

Question

Let G be a finite group and $\rho:G\to GL_n(\mathbb C)$ a representation.

a) If $M=\sum_{g\in G} \rho (g) \neq 0$ then prove that there is a non-zero vector $v$ such that $\rho(g)v=v $ for every $g\in G$.

b) If $\sum_{g\in G} \chi(g)=0$ then prove that $M=0$

The first part is easy, as if $M\neq 0$, then there s a vector $w$ such that $Mw=v \neq 0$. However, then $\rho(g)v= \rho(g) M w= Mw=v $ for every g Any hint would be welcome.

For b) however I am a bit stuck. $\sum_{g\in G} \chi(g)=\operatorname{tr}(M)$. Suppose that $M\neq 0$, and consider v be the vector found in (a), then if I complete $v$ to a basis of $\mathbb C^n$ then the first column of every matrix $\rho(g)$ would be the vector $[1,0,...,0]$. However, on the rest of the diagonal I could have negative values so I dont see why trace being $0$ implies $M=0$

Answer 1

Notice that $$ \sum_{g\in G} \chi(g)$$ is the inner product of the representation with the trivial representation, hence if it equals $0$ it means that the decomposition in irreducible representation does not contain the trivial representation. But if $M\neq 0$ then $\rho$ has a one dimensional invariant subspace, i.e. a trivial subrepresentation, thus $M=0$.