7
$\begingroup$

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:

$~~~~~$Picture

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.

$\endgroup$
1
  • $\begingroup$ 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 $\endgroup$
    – AgentSmith
    Sep 18, 2018 at 18:46

2 Answers 2

4
$\begingroup$

The theorem that solves this is (indeed, from linear algebra):

Theorem: a matrix over some field $\;F\;$ is diagonalizable iff its minimal polynomial is the product of different linear factors.

Now, we know that over an algebraic closed field any polynomial decomposes in a product of linear factors, and thus this product will be of different linear factors iff the poloynomial has no repeated roots (i.e., if all its roots are simple).

$\endgroup$
1
  • 1
    $\begingroup$ Thanks! Now I just have to figure out why that theorem is true... $\endgroup$
    – Jasha
    Oct 12, 2014 at 14:46
0
$\begingroup$

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 $\mu_M(t)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.