Dimension of Homomorphisms of representations I'm looking for a counterexample of two representations $\rho$ and $\sigma$ of a group $G$ (also infinite) on a (better finite-dimensional) vector space over the fields $\mathbb{R}$ and $\mathbb{C}$, with the characteristic that $$\dim(\operatorname{Hom}(\rho,\sigma)) \neq \dim(\operatorname{Hom}(\sigma,\rho)).$$
Thank you to everybody.
 A: Hint: Consider $G={\mathbb Z}$. In this case, $G$-modules are the same as modules over the ring of Laurent polynomials $R := k[X^{\pm 1}]$ (the action of $n\in {\mathbb Z}$ corresponding to the action of $X^n$). Now you might search for examples where, say, $\rho$ has torsion but $\sigma$ does not.
Phrased differently, try to look for a representation of ${\mathbb Z}$ where all $n\in{\mathbb Z}$ act linearly independently (e.g., try some shifting operation), and compare it to some representation where there are dependencies (e.g., the trivial representation).
For a finite-dimensional example, e.g. pick as $\sigma$ the action of ${\mathbb Z}$ on $k^2$ given by $1$ acting as $\begin{pmatrix} 1 & 1 \\ 0 & 2\end{pmatrix}$, and as $\rho$ again the trivial representation. 
A: For a simple example, take $G=\mathbb{Z}^2$ and define $\rho(a,b) = \begin{pmatrix} 1 & a & b \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}\in GL(\mathbb{C}^3)$.
Then consider the trivial representation $\sigma:G\to GL(\mathbb{C})$.
Now, $\dim Hom(\rho, \sigma) = 2$ while $\dim Hom(\sigma, \rho) = 1$.
