Orthonormal basis of $\mathbb{C}^{r}$

Question

I have to prove that $\{f_1,f_2,..,f_r\}$ is an orthonormal basis of $\mathbb{C}^{r}$ where:

$f_j=\frac{1}{\sqrt{r}}(1,e^{2i\pi\frac{j-1}{r}},e^{4i\pi\frac{j-1}{r}},...,e^{2(r-1)i\pi\frac{j-1}{r}})$

for $j$=1,2,...,r

I have no ideia how to even begin

Answer 1

First of all, we have $$ \langle f_j, f_j\rangle = \|f_j\|^2 = \frac{1}{r} \left(1+\left|e^{2i\pi\frac{j-1}{r}}\right|^2+\dots+ \left|e^{2(r-1)i\pi\frac{j-1}{r}}\right|^2\right) = \frac 1r r = 1 $$ Now, for $j \neq k$, define $\alpha = e^{2 \pi i(j-k)/r}$ and show that $$ \langle f_j, f_k \rangle = \frac 1r \left(1 + \alpha + \alpha^2 + \cdots + \alpha^{r-1} \right) = \frac 1r \frac{1-\alpha^r}{1-\alpha} $$ Why is this equal to zero?

Answer 2

Remember that the dot/inner product on the complexes is defined in a special way. Why not try it with $r=2$ or $3$ first? That gives you $1/\sqrt{2}(1,1)$ and $1/\sqrt{2}(1,-1)$, which I think you'll agree are unit vectors. Then try it with the cube roots of unity and remember that $e^{2i\pi/3}\overline{e^{2i\pi/3}}=1$ and you should be in business...