Let $M$ be a $m$ rows and $n$ columns matrix over GF(2). And Let $M^{'}$ be a $r$ rows and $n$ columns submatrix of $M$ ($r <= m$). Note that rows of $M^{'}$ is randomly selected from $M$. Is it possible that the $M^{'}$ is full row-rank?


If it is, why?


If it is not, how about the case where $r < m$?


Edit:


1) I forgot to mention $m > n$.


2) I have run some experiments on Matlab. In experiments, I first randomly generated $10^{6}$ matrices over GF(2). Each matrix has $m = 288$ rows and $n = 216$ columns. Turn out they all have the rank of $216$. Then, I randomly generated $10^{6}$ matrices over GF(2). Each matrix has $m = 84$ rows and $n = 216$ columns. Turn out they all have the rank of $84$. So I am very confused now.



Full row-rank submatrix of full column-rank matrix over GF(2)
There's a formula to compute the probablity of an $m×n$ $(m≥n)$ matrix to be of full rank: if the matrix has i.i.d uniformly distributed elements drawn from $\{0,1\}$, the probability for it to be of full rank (i.e. rank n) is $∏^{n−1}\_{k=0}\frac{2^m−2^k}{2^m}$. 


The formula can be proved by induction: Take the columns of a full-rank matrix as linearly-independent vectors. For $n=0$, the matrix is full-rank iff not all elements are $0$, so the formula holds.


Assume the formula holds for $n=k$. The $(k+1)$th vector should not lie in the linear space spanned by the first $k$ vectors, so the number of possibilities for the $(k+1)$th vector to be independent with the previous vectors is $2^m-2^k$. As there're $2^m$ $m$-dimensional vectors in $\text{GF(2)}$, we should append a term $\frac{2^m-2^k}{2^m}$, so the formula holds for $n=k+1$.


Hope this answers your question.


