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1 Context

Fix natural $N > 0$ for $\mathbb{C}^N$. Let

$$ \omega := e^{2 \pi i / N} $$

for

$$ e_n(k) := {1 \over \sqrt{N}} \omega^{kn} \text{ for } k \in \{0,1 \ldots, N-1\} $$

2 Problem

From a book on Harmonic Analysis:

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3 Attempt

Let $l,j \in \{0,1, \ldots, N-1\}$ s.t. $l \ne j$.

$$ \langle e_l, e_j \rangle = \sum_{k=0}^{N-1} {1 \over \sqrt{N}}\omega^{kl} \overline{{1 \over \sqrt{N}}\omega^{kj}} = \sum_{k=0}^{N-1} {1 \over N}e^{2 \pi kl / N} e^{-2 \pi kj / N} = \ldots ? \ldots = 0 $$

user1770201
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    Note that $e^{2\pi k i/N} e^{-2\pi j i/N} = e^{2\pi (k - j) i/N}$, another root of unity. If $k - j \neq 0$, then this root of unity is not $1$. – Theo Bendit Nov 09 '17 at 12:44
  • @TheoBendit: Does this mean that we could replace the constant ${1 \over \sqrt{N}}$ with any arbitrary constant $c$ in the definition of $e_n$, and still have that $\langle e_l, e_j \rangle = 0$ for $l \ne j$? – user1770201 Nov 10 '17 at 13:35
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    Yeah. You'd get an orthogonal basis, not an orthonormal one. – Theo Bendit Nov 10 '17 at 13:56

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