# Representation with 1 as eigenvalue cannot be irreducible

I am trying to solve an exercise which states that "if $\rho:G\to GL_2(\mathbb{C})$ is a representation with 1 as an eigenvalue of every $\rho(g)$ then $\rho$ cannot be irreducible $(|G|<\infty)$. At another point it asks to prove that $\rho$ is the sum of two representations.

My try: Since the degree of $\rho(g)$ is 2, the character of $\rho$ will be $x(g)=1+\lambda_g$ where $\lambda_g$ is the other eigenvalue of $\rho(g)$. Then, the inner product of $x$ and the trivial character of $G$ will be $$1+\frac{\sum_g\lambda_g}{|G|}$$ If $\rho$ is irreducible then the last integer must be zero ($\rho$ cannot be the trivial representation due to its dimension). Hence $$\sum_g\lambda_g=-|G|$$ must hold.

Intuitively that cannot happen since every $\lambda_g$ is some root of unity but I can't find an elegant way to prove that their sum cannot be $-|G|$

So I am asking for any thoughts/suggestions about that. Also the second question seems wrong to me so if anyone can verify I will be grateful.

Excuse me for the long post.

• I take it you are assuming $G$ is finite? Commented May 1, 2017 at 12:17
• @ancientmathematician yes! Commented May 1, 2017 at 12:18
• Better edit, then because the second bit is, I reckon, false in the case $G=\mathbb{Z}$. Commented May 1, 2017 at 12:19
• Just did, thanks! Commented May 1, 2017 at 12:21
• Now what is your definition of irreducible? Commented May 1, 2017 at 12:22

For elements $\lambda_1, \dots, \lambda_n \in S^1$ we always have $| \sum \lambda_i| \leq n$, with equality if and only if $\lambda_1 = \lambda_2 = \dots = \lambda_n$. This is clear for $n=2$ (you can use Cauchy-Schwartz inequality) and follows by induction for the general case. Applying this to your exercise you have $\lambda_g = -1$ for all $g \in G$ which contradicts the hypothesis that the representation was irreducible.