Given a represenation $\rho:G \to Gl(V)$ and a subrepresentation $W \subset V$, is $\rho_V(g) = \rho_W(g)?$

I am asking for clarifications of the basic definitions in the representation theory of a subrepresentation and a character of a subrepresentation.

Given a represenation $\rho:G \to Gl(V)$ and a subrepresentation $W \subset V$, is $\rho_V(g) = \rho_W(g)?$

I think it should be true because a subrepresentation of $(V,\rho)$ is a pair $(W, \rho)$ such that $W \subseteq V \wedge \forall g \in G, \forall x \in W, \rho(g)(x) \in W$. So $G$ is mapped to the same group of matrices. If $W$ a proper subset of $V$, then only the basis changes and $\rho_V = \rho_W$ always but $V \neq W$.

One corollary says: if $V$, $W$ are representations of $G$, then $V \cong W \iff \chi_V = \chi_W$.

By definition the character is $\chi(g) = Tr(\rho(g))$. So the subrepresentations of some representation should have the same characters. But it could be that $V \neq W$ and therefore $V \ncong W$ although $\rho_V = \rho_W$.

So where is the problem in my understanding and the definitions given?

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Your problem is that even though the same matrix can represent an element in both $GL(W)$ and $GL(V)$, it's trace depends on which way you want to think of it. For example, consider the representation of $\mathbb R$ on $V=\mathbb R^2$ given by $$t \mapsto \operatorname{diag}(e^t, e^{2t}).$$ Then a subrepresentation is given by $W = span(1,0)$. Then $\chi_V(t) = e^t + e^{2t}$ but $\chi_W(t) = e^t$.

EDITED for more details

There is a slight abuse of notation in that really you want to think of the subrepresentation as the composition $G \to GL(V) \to GL(W)$. Now write what the matrix representation looks like when $\rho(g)$ is thought of as an elemento f $GL(W)$.

Also, maybe it will help to think more abstractly about the trace. If $V$ is an $n$-dimensional vector space and $T: V \to V$ is a linear map, then you can define it's trace as follows. Let $e_1, \ldots, e_n$ be a basis for $T$. If $T e_j = \sum_k T_{jk} e_k$ then, by definition, $$\operatorname{tr} T = \sum_j e_{jj}.$$ Now you can think of $T$ as the $n\times n$ matrix whose entries are $T_{ij}$ and then the trace is the sum of the diagonal elements.

In the example I gave, when you consider the representation on $V = \mathbb R^2$ you can take the basis (1,0) and (0,1) and the matrix for $\rho(t)$ is just $\operatorname{diag}(e^t, e^{2t})$. Now consider this matrix as a map from $\operatorname{span}(1,0) = W \to W$. We can choose $(1,0)$ as a basis for $W$. Then $\rho(t) (1,0) = (e^t, 0) = e^t (1,0)$ so the trace is $e^t$. In other words, the $1 \times 1$ matrix that represents $\rho(t) \in GL(W) = GL(1) \simeq \mathbb R$ is just the number $e^t$.

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I know that $W$ is spanned by one vector only. But how did you get $\chi_W(t) = e^t$? You are not changing the matrix $\rho(g)$ and still $\rho_V(g) = \rho_W(g)$ So the trace should be the same. –  David Toth Mar 7 '13 at 14:03
Let me know if my edit helps. –  Eric O. Korman Mar 7 '13 at 14:27
Your great answer reveals my weakness in ability to compute the trace of the matrices under different bases. How would we compute the trace of a matrix representing $(1 2 3)$ of $S_3$ in a permutation representation in a subrepresentation spanned by a) $W_a=span((1,1,1))$, $W_b=span((1,0,-1),(-1,1,0))$? Thanks a lot. –  David Toth Mar 7 '13 at 16:06
@DavidToth : For $W_a$, $(123)(1,1,1) = (1,1,1)$ the trace is 1 (the coefficient on $(1,1,1)$). For $W_b$ we have $(123)(1,0,-1) = (-1,1,0)$ and $(123)(-1,1,0) = (0,-1,1) = -(1,0,-1) - (-1,1,0)$ so the trace is 0 + -1 = -1, with the 0 coming from the coefficient of (123)(1,0,-1) on (1,0,-1) and the -1 coming from the cofficient of (123)(-1,1,0) on (-1,1,0). –  Eric O. Korman Mar 7 '13 at 16:32