# Proof that a group representation matrix is diagonalizable?

Suppose we have a finite group $G$ and and an $n$-dimensional vector space $V\cong \Bbb C^n$ over the field $\Bbb C$ of complex number. My professor said the other day that for every group element $g$ in $G$ and for every representation $\rho:G\to\text{GL}_n(\Bbb C)$ of $G$ on $V$, the matrix $\rho(g)\in\text{GL}_n(\Bbb C)$ is diagonalizable. I'm having trouble convincing myself that this is true. The text below is from a website that agrees with what my professor said:

$~~~~~$

I believe that all the eigenvalues of $M$ must be $n$-th roots of unity, and that each eigenvalue appears only once as root of $M$'s minimal polynomial. I am having trouble convincing myself, however, that this means $M$ is diagonalizable. How do we know that the eigenvectors of $M$ span the vector space $V$? Couldn't $M$ have a minimal polynomial of degree less than $n$? Any help would be much appreciated.

• I thought that representation matrices are in block-triangular form, not exactly in triangular form, but we do not need that remark for the argument, as the theorem from linear algebra goes: a matrix is diagonalizable iff its minimal polynomial has no multiple roots Sep 18, 2018 at 18:46

Theorem: a matrix over some field $\;F\;$ is diagonalizable iff its minimal polynomial is the product of different linear factors.
Adding a comment to the answer of Timbuc to answer Jasha comment (which is already exhaustive). All you have to know is that the minimal polynomial of the matrix, let's called it $$\mu_M(t) \mid \mu_M(t)$$ (standard argument using $$\mu_M(t) \mid p_M(t)$$, where $$p$$ denotes the characteristical polynomial), and the thesis know is a consequence of Timbuc answer, since if we divide a polynomial with all distinct roots, we must be a finite product of some of them, and so our roots are distinct.
Edit: Actually we don't need the "standard argument", it is sufficient to know that $$x^{n}-1$$ is in the ideal generated by $$\mu_M(t)$$.