I want to prove that:
iid Gaussian random matrix $A\in M_n$(I mean whose elements are iid Gaussian) has full rank with probability 1
Below is my consideration:
$$1-P(\text{full rank})=P(\text{columns are linear dependent})\le \sum_{i=1}^n P(a_i \,\text{is a linear combination of other columns}),\,\text{by the union bound}$$
Denote:
$$P_i:=P(a_i \,\text{is a linear combination of other columns})$$
and now I try to prove $P_i=0$
$$P_i=P(a_i=\lambda_1a_1+\dots+\lambda_{i-1}a_{i-1}+\lambda_{i+1}a_{i+1}+\dots+\lambda_na_n,\, \text{for some} \, \Lambda_i:=(\lambda_1,\dots,\lambda_n)^T\in \mathbb{R}^{n-1})$$
Now for fixed $\Lambda_i$
$$P_i=0$$ by direct calculation.
But now $\Lambda_i$ can be any value in $\mathbb{R}^{n-1}$, I am not sure whether
uncountable infinite zeros add up to still zero?
Best Answer
Hint: What is the probability that the determinant of that matrix is zero. Note that the determinant is a function of its entries and is thus a continuous random variable. (laplace expansion)