Problem and Proof. Rank of a Random matrix is full rank with probability 1? I am wondering if a random matrix $ A\in \mathbb{C}^{M\times N}$ where its $(m,n)$th element is from a complex Gaussian distribution $\mathcal{CN}(0,\sigma_{mn}^2)$ has a full rank with probability $1$. It seems that the conclusion holds when all $\sigma_{mn}^2\ (m=1,\cdots,M,n=1,\cdots,N)$ are the same. Is this conclusion still holding for these different $\sigma_{mn}^2$ (at least two of them are different)? If yes, how to prove it? If not, what's the reason?

Work by induction on $m$, so long as $m\lt n$.
Assume an $m×n$ matrix is rank $m$. Pick a nonpivot column.  Removing that column, the remainder is still rank $m$.
Add a random  $m+1th$ row.  Since the first $m$ rows are linearly independent, there is at most one linear combination of the first $m$ rows that equals the $m+1th$.  If no linear combination will do it, the matrix is already rank $m+1$.
Now replace the nonpivot column, and consider its $m+1th$ entry.  There is at most one value that leaves the $m+1th$ row still a linear combination of the first $m$ rows.  So the probability it is still rank $m$ is zero.
In summary, if a matrix is full rank and there are any nonpivot columns left, then adding a random new row almost certainly increases the rank. ( If there are no nonpivot columns, the matrix is already full rank.)
