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.



Action of complex torus on a vector space
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.


