Indecomposable representations of an algebra Let $\rho$ be an indecomposable representation of Algebra A on a finite dimensional vector space V. Let $a \in Z(A)$, how do I show $\rho(a)$ has exactly one eigenvalue. $Z(A)$ represents the centre of algebra $A$. It is easy to see it has one eigenvalue, how do I show it is exactly one?
 A: Let $\Lambda$ be the set of eigenvalues for $\rho(a)$ on $V$, and for $\lambda\in\Lambda$, define
$$V_\lambda=\{v\in V\mid (\rho(a)-\lambda)^Nv=0\mbox{ for }N\gg0\}.$$
Then, we have a vector space decomposition
$$V=\bigoplus_{\lambda\in\Lambda}V_\lambda.$$
[proof: We certainly have $V=\sum_{\lambda\in\Lambda}V_\lambda$, so we only need to show it is direct. To this end, suppose $v\in V_\lambda\cap V_\mu$ is nonzero and $\lambda\neq\mu$. Choose $N$ such that $v':=(\rho(a)-\lambda)^Nv\neq 0$, but $(\rho(a)-\lambda)^{N+1}v=0$. Then, $\rho(a)v'=\lambda v'$. 
Now, take $M> 0$, such that $(\rho(a)-\mu)^Mv=0$. Then,
\begin{align*}
0&=(\rho(a)-\lambda)^N(\rho(a)-\mu)^Mv\\
&=(\rho(a)-\mu)^M(\rho(a)-\lambda)^Nv\\
&=(\rho(a)-\mu)^Mv'\\
&=(\lambda-\mu)^Mv'\neq0
\end{align*}
because $\lambda\neq\mu$. This is a contradiction.]
In fact, we claim that for $\lambda\in\Lambda$, $V_\lambda$ is an $A$-submodule. Indeed, suppose $x\in A$, and $v\in V_\lambda$. Then, there exists $N\gg0$ such that $(\rho(a)-\lambda)^Nv=0$. But, now 
$$
(\rho(a)-\lambda)^Nxv=x(\rho(a)-\lambda)^Nv=0
$$
so $xv\in V_\lambda$ as required.
Finally, we conclude that $|\Lambda|=1$ since $V$ is indecomposable.
