It is known that for a random matrix $A\in \mathbb{C}^{M\times K}$ with $M>K$, if each element is i.i.d. Gaussian generated, then $A$ has full column rank with probability one.
Now, assume that each element of $A$ is in the form of $e^{j\theta}$, where $\theta$'s are i.i.d. uniformly distributed in $(0,2\pi)$. Can I prove that $A$ has full column rank, with probability one? Thanks.
Best Answer
Yes, a result of this form is true very generally. It's equivalent to prove that $k$ vectors $v_1, \dots v_k \in \mathbb{C}^m$ chosen iid wrt the distribution given by $e^{i \theta}$ in each coordinate are linearly independent with probability $1$.
This is true with a much more general distribution on random vectors: it suffices for the vectors to be chosen iid wrt any distribution on $\mathbb{C}^m$ such that the probability of the vector lying in any particular hyperplane through the origin is zero. A large class of examples is any distribution with a continuous pdf such that the induced distribution on the unit sphere given by sampling a random vector and scaling it to unit length also has a continuous pdf which is nowhere zero; in any case your distribution qualifies.
To prove this we sample the vectors $v_1, v_2, \dots$ one at a time, observing that at each step $\text{span}(v_1, \dots v_i)$ is at most an $i$-dimensional subspace of $\mathbb{C}^m$, and so by hypothesis $v_{i+1}$ avoids it with probability $1$.