Discrete Fourier Transform - proof that columns of matrix are orthogonal

Question

The DFT matrix of size $n$ has entries $M_{ij} = \omega^{ij}$ where $\omega$ is the $n$th root of 1. My textbook states that the columns of this matrix are orthogonal because their dot product is 0. The dot product of columns $i$ and $j$ is calculated as $1 + \omega^{j-k} + \omega^{2(j-k)} ... $ but I do not follow. Should it not be $j+k$ in the exponents?

Answer 1

The dot product of vectors $x,y\in\mathbb C^n$ is defined as

$$\sum_{i=1}^nx_i\bar{y_i}\;,$$

where $\bar{y_i}$ is the complex conjugate of $y_i$. Wikipedia gives some motivation for this definition.