We call a $N \times N $ square matrix as a binary-matrix if each of its entries is either 0 and 1. If we pick a $N \times N $ binary matrix at random, is it more likely to be invertible or singular? Propose a method to find the answer and verify it with some level of confident?
My guess is that the chance that the random matrix is singular is more than it is invertible. My guess comes from the observation that we can somehow find a lower bound of the number of singular matrices where entries are 0 or 1.
To show that, we know that there are all $2^{N^2}$ possible binary matrix of size $N$. We need to show that the number of singular matrices is at least $\frac{2^{N^2}}{2}$ then we are done.
There are several ways to obtain singular matrix of 0/1 entries. For example, select a random row (or column) and assign 0 for all of its entries. Or, we assign two random rows (or columns) that all have the same values.
However, I could not find exactly how we can obtain a lower bound $\frac{2^{N^2}}{2}$ to those singular matrices obtained from 0/1. Any advice ?
