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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?

tianhang
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  • Intuitively, the probability should already be $1$ , if all the standard deviations are positive, but I have no idea how to formally prove it. – Peter Sep 15 '20 at 09:30
  • Thanks. I found the proof for the case where all $\sigma_{mn}$ are the same, i.e., link. Is there any possible approach to prove analogically. – tianhang Sep 15 '20 at 09:39

1 Answers1

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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.)

Empy2
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