Given $N$ distinct real numbers $x_1,\ldots, x_N$, how can I show that there exist real numbers $a_1,\ldots, a_N$ so that the following matrix is invertible?
$$\begin{bmatrix} \exp(ia_1 x_1) & \exp(i a_2 x_1) & \cdots & \exp(i a_N x_1)\\ \exp(ia_1 x_2) & \exp(i a_2 x_2) & \cdots & \exp(i a_N x_2)\\ \vdots & \vdots & \ddots & \vdots\\ \exp(ia_1 x_N) & \exp(i a_2 x_N) & \cdots & \exp(i a_N x_N)\\ \end{bmatrix}$$
Coming up with the $a_n$ explicitly seems hard. Would a proof by contradiction work? (Suppose it is not invertible for any choice of the $a_n$. Then we must have some $x_i=x_j$ for $i \ne j$?)
a) First note that if we have a sum $$f(a)=\sum_{k=1}^N \lambda_k \exp(iax_k)$$ with the $x_k\in \mathbb{R}$ distincts, then this function is zero on $\mathbb{R}$ if and only if all $\lambda_k$ are zero. To see this, use induction on $N$, and if $f(a)=0$ for all $a$, compute the derivative of $g(a)=f(a)\exp(-iax_N)$.
b) Now prove your assertion by induction on $N$. I leave to you the case $N=1,N=2$. Now to show that the assertion is true for $N$ if it is true for $N-1$, you have by the induction hypothesis that there exists $a_1,\cdots,a_{N-1}$ such that the $(N-1)\times (N-1)$ determinant is non zero, that we fix. If we put $a=a_N$ in the $N\times N$ determinant, and developping it with respect to the last column, we get an expression of the form $f(a)$ of a). In addition, the coefficient of $\exp(iax_N)$ is the non zero determinant $(N-1)\times (N-1)$. Hence by a) $f(a)$ is not the zero function, and there exists $a=a_N$ such that$f(a_N)$ is not zero, and we are done.