This question is an addition to my [question](https://mathoverflow.net/questions/9541/simultaneous-diagonalization) on simultaneous diagonalization from yesterday and it is probably also obvious but I just don't know this: Let $G$ be a commutative affine algebraic group over an algebraically closed field $k$. Let $G\_s$ be the semisimple part of $G$. Let $\rho:G \rightarrow GL\_n(V)$ be an embedding. Then $\rho(G\_S)$ is a set of commuting diagonalizable endomorphisms and I know from yesterday that I have unique morphisms of algebraic groups $\chi\_i: \rho(G\_s) \rightarrow \mathbb{G}\_m$, $1 \leq i \leq r$, and a decomposition $V = \bigoplus \_{i=1}^r E \_{\chi\_i}$, where $E\_{\chi\_i} = \lbrace v \in V \mid fv = \chi\_i(f)v \ \forall f \in \rho(G\_s) \rbrace$. Now, my question is: are the morphisms $\chi\_i$ independent of $\rho$ so that I get well-defined morphisms $\chi\_i:G\_s \rightarrow \mathbb{G}\_m$? 


If somebody knows what I'm talking about, then please change the title appropriately! :)



"Eigenvalue characters"
Unless I drastically misunderstand your question, of course the characters $\chi\_i$ depend on the representation $\rho$. Try looking at the simplest nontrivial case: $G = \mathbb{G}\_m$ acting on a one-dimensional vector space. In this case, there is exactly one $\chi\_i$ and it is simply a character of $\mathbb{G}\_m$, i.e., is of the form $x \mapsto x^n$ for a unique 
integer $n$. This integer $n$ is determined by (and determines) $\rho$. 


