The rank of a matrix of Gaussian random vectors

Question

If I generate a matrix $R\in\mathbb{R^{k\times p}}$, $k<p$, with i.i.d. entries $R_{i,j}\sim\mathcal{N}(0,\sigma^2)$, is there a guarantee that this matrix will have rank $k$, i.e. its columns will be independent?

Answer 1

It is not 'surely' true, but we can say that $$ P(\;\text{rank}(R)=k \;)=1, $$ i.e. $\text{rank}(R)=k$ almost surely. Let $\mathcal{R}=(R_{ij})\in\Bbb R^{kp}$ denote the random vector consisting of $R_{ij}$'s. Note that $\text{rank}(R)<k$ if and only if $RR^T$ is singular and $\det(RR^T)=0$. Hence there is a polynomial $p:\mathbb{R}^{kp}\to \Bbb R$ such that $$ \text{rank}(R)<k\Leftrightarrow p(\mathcal{R})=0. $$ Thus, we have $$ P(\;\text{rank}(R)<k \;)=\int_{p(\mathcal{R})=0}f(\mathcal{R})d\mathcal{R}, $$ where $f(\mathcal{R})$ is the pdf of $\mathcal{R}$. Finally observe that $ \{p(\mathcal{R})=0\} $ has a Lebesgue measure zero, which can be proved by induction on the dimesion and Fubini's theorem. This implies $$ P(\;\text{rank}(R)<k \;)=0 $$ as desired.