Action of complex torus on a vector space Consider a torus $T$ over $\mathbb{C}$. Let  $\rho: T\rightarrow \operatorname{GL}_{n}(\mathbb C)$ be a finite dimensional complex representation.
Is there an elementary way (undergrad level) to see that $\rho(T)$ is diagonalizable?
The standard way uses algebra of functions on $k[T]$, tensor products and symmetric algebras, but it is too involved, can we see it directly?
We already know by Lie–Kolchin that the image is trigonalizable.
 A: You need to put some niceness hypothesis on the function $\rho$, such as "algebraic" or "analytic". Otherwise, the map $\mathbb{C}^{\ast} \to \text{GL}_2(\mathbb{C})$ by $z \mapsto \begin{bmatrix} 1 & \log |z| \\ 0 & 1 \end{bmatrix}$ is a representation and not diagonalizable. Since you didn't tell me which hypothesis to use, I will take $\rho$ to be algebraic, meaning that each of the $n^2$ entries in the matrix $\rho(z_1, z_2, \dots, z_d)$ is a Laurent polynomial in $z$. (You used $n$ for both the dimension of $T$ and the dimension of the representation, but I assume you didn't want that; I'll put $d = \dim T$.)
So we can write
$$\rho(z_1, z_2, \dots, z_d) = \sum_{(k_1, k_2, \ldots, k_d)} P_{(k_1, k_2, \ldots, k_d)} z_1^{k_1} z_2^{k_2} \cdots z_d^{k_d}$$
where each $P_k$ is an $n \times n$ matrix.
From the equation $\rho(xy) = \rho(x) \rho(y)$, we deduce that
$$\sum_{(k_1, \ldots, k_d)} P_{(k_1, \ldots, k_d)} x_1^{k_1} \cdots x_d^{k_d} y_1^{k_1} \cdots y_d^{k_d} = \sum_{(i_1, \ldots, i_d),\ (j_1, \ldots, j_d)} P_{(i_1, \ldots, i_d)} P_{(j_1, \ldots, j_d)} x_1^{i_1} \cdots x_d^{i_d} y_1^{j_1} \cdots y_d^{j_d}.$$
Comparing the coefficient of $x_1^{i_1} \cdots x_d^{i_d} y_1^{j_1} \cdots y_d^{j_d}$ on each side, we deduce that
$$P_{(i_1, \ldots, i_d)} P_{(j_1, \ldots, j_d)} = \begin{cases} P_{(i_1, \ldots, i_d)} & (i_1, \ldots, i_d) = (j_1, \ldots, j_d) \\ 0 &  (i_1, \ldots, i_d) \neq (j_1, \ldots, j_d)\\ \end{cases}.$$
In other words, the $P$'s are mutually orthogonal idempotents.
Moreover, from the equation $\rho(1) = \text{Id}_n$, we deduce that $\sum P_{(k_1, \ldots, k_d)} = \text{Id}_n$. So the $P$'s are a complete set of mutually orthogonal idempotents.
Now use your favorite proof that mutually orthogonal idempotents are simultaneously diagonalizable.
