Suppose $\rho:(\mathbb{C}^{\times})^n \rightarrow GL_m(\mathbb{C})$ is a faithful representation, for some integer $m$.
My professor has told me that the image of $(\mathbb{C}^{\times})^n$ under $\rho$ consists entirely of diagonal matrices.
I have been trying for days to show this, and I can't figure out why it's true. I do know that the image must be an abelian subgroup of $GL_m(\mathbb{C})$, but there are plenty of invertible, non-diagonalizable matrices that commute with other matrices.
Any help would be appreciated!
It is crucial to assume that $\rho$ is a holomorphic map. Then it is uniquely determined by its restriction $\rho_c$ to $(S^1)^n$, which is a compact group. Hence you may choose a hermitian form on $\mathbb C^m$ such that $\rho_c(g)$ is a unitary operator for any $g \in (S^1)^n$. Thus by linear algebra, you may change basis so that all $\rho_c(g)$ are diagonal. Then off-diagonal elements of $\rho(g)$ are holomorphic functions vanishing on $(S^1)^n$, hence vanishing on whole $(\mathbb C^{\times})^n$ (by complex analysis).
Counterexample for the statement if $\rho$ is not taken to be holomorphic: take $m=2$, $n=1$ and $\rho(z) = \frac{z}{|z|} \begin{pmatrix} 1 & \mathrm{log}|z| \\ 0 & 1 \end{pmatrix}$.
By the way, assumption of faithfulness was not important here.