Equivariant map, how do I see that addition of an invertible and noninvertible intertwiner is invertible? Let $A$ be a $k$-algebra and $\rho$ a finite dimensional representation of $A$ that does not admit a nontrivial decomposition $\rho = \rho_1 \oplus \rho_2$ as a direct sum of two nonzero subrepresentations, i.e. $\rho$ is indecomposable. Let $x$, $y \in \text{Hom}_A(\rho, \rho)$ be a pair of intertwiners such that $x$ is invertible and $y$ is not invertible. How do I see that the map $x + y$ is invertible?
According to Wikipedia here, the definition of intertwiner is as follows.

In representation theory, a vector space equipped with a group that acts by linear transformations of the space is called a linear representation of the group. A linear map that commutes with the action is called an intertwiner. That is, an intertwiner is just an equivariant linear map between two linear representations. Alternatively, an intertwiner for representations of a group $G$ over a field $K$ is the same thing as a module homomorphism of $K[G]$-modules, where $K[G]$ is the group ring of $G$.

 A: We suppose that $\rho$ is a representation of $A$ in the finite dimensional $k$-vector space $V$. We are going to show that $y$ is nilpotent. Consider the sequence $y^n, n$ an integer. Since $dim(V)<\infty$, there exists $N$, such that for $n>N, Im(y^n)=Im(y^N), ker(y^n)=ker(y^N)$. Let $n>N$, $ker(y^n)\cap Im(y^n)=\{0\}$. To see this consider $u\in ker(y^n)\cap Im(y^n)$, there exists $v\in V$ such that $y^n(v)=u, y^{2n}(v)=y^n(u)=0$, this implies that $v\in ker(y^{2n})=ker(y^n)$. This implies that $y^n(v)=u=0$.
Let $a\in A$, and $u\in ker(y^n), y^n(\rho(a)u)=\rho(a)(y^n(u))=0$ let $v\in Im(y^n)$, write $v=y^n(u), \rho(a)(v)=\rho(a)(y^n(u))=y^n(\rho(a)u)$. This implies that $Im(y^n)$ and $ker(y^n)$ are stable by $\rho$, since $Im(y^n)\cap ker(y^n)=0$ and $dim(ker(y^n))+dim(Im(y^n))=dim(V)$, we deduce that $V=ker(y^n)\oplus Im(y^n))$ since $V$ is indecomposable, we deduce that $ker(y^n)=0$ or $Im(y^n)=0$. The fact that $ker(y^n)=0$ is equivalent to saying that $y^n$ is invertible and henceforth $y$ is invertible, since we have assumed that $y$ is not invertible, we deduce that $Im(y^n)=0$ and $y$ is nilpotent.
Remark that $ker(y)\subset ker(x^{-1}y)$ since $x^{-1}y$ is interwinable, we deduce that it is also nilpotent since its kernel is not zero. Suppose that $(x^{-1}v)^n=0$
We have $(Id_V+x^{-1}v)(Id_V+(-x^{-1}y)+(-x^{-1}y)^2+..+(-x^{-1}y)^{n-1})=Id_V+(-1)^{n-1}(x^{-1}y)^n=Id_V$. We deduce that $Id_V+x^{-1}y$ is invertible, this implies that $x(Id_V+x^{-1}y)=x+y$ is invertible since it is the composition of invertible morphisms.
