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I need to show that the Fourier matrix is invertible using the fact that the determinant of the Vandermonde matrix is given by

$$ det(V_n) = \prod_{s,t=0 \\ s<t}^n (z_t - z_s). $$

It is not clear how I should do that but is it possible to prove it using the fact that it is a unitary matrix and so it is invertible?

wrong_path
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1 Answers1

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Hint

The matrix $V$ is invertible if and only if $\det(V) \ne 0$. Take a look at the polynomial $$ \det(V_n) = \prod_{s,t=0 \\ s<t}^n (z_t - z_s). $$ and try to guess why $z_t - z_s$ is never zero.

Andrei Kulunchakov
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  • Thanks. Well, since we are considering distinct complex numbers there is no way that product is zero. But how does it relate to $F_n$? – wrong_path Feb 25 '17 at 09:21
  • @LorenzoFabbri, $F_n$ is the Vandermonde matrix for ${\displaystyle \omega =e^{-2\pi i/n}} $ – Andrei Kulunchakov Feb 25 '17 at 09:24
  • Yes, I know that. But is this sufficient as a proof. I am basically saying that $F_n$ is $V_n$ with a particular choice of $w$ and since $det(V_n)$ is not zero then also $F_n$ is invertible. Thanks a lot. – wrong_path Feb 25 '17 at 09:26
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    @LorenzoFabbri, Yes, it is sufficient. As soon as you express your matrix in the Vandermonde form, you can use any implication known for Vandermonde matrices. So, the invertibility comes straightforwardly. Basically, I should correct myself and say that $F_n$ is the Vandermonde matrix for the powers $\omega^k$ of ${\displaystyle \omega =e^{-2\pi i/n}}$ from 0 to $n-1$. As soon as they are different, we have invertibility. – Andrei Kulunchakov Feb 25 '17 at 09:30