# Is the product of two independent random matrix full rank?

The elements in matrix $$A$$ and $$B$$ are independent and absolute continous random variable. Is $$AB$$ be full rank with probability one?

First, $$A$$ and $$B$$ must be full rank with probability one.

I try to prove it with the following two ways.

1. I claim $$|ABB^HA^H|$$ is a continuous random variable (I don't know how to prove it.), then $$P( AB \text{ is not full rank})=P(|ABB^HA^H|=0)=0$$. Therefore, $$AB$$ is full rank almost surely.

2. Since $$P(X\in Null(A) \cap Range(B) )=P(X\in Null(A) )\cdot P(X\in Range(B) )=0\cdot \{0,1\}=0$$ (owing to the fact that linear subspace has Lebesgue measure 0), $$AB$$ is full rank almost surely.

New contributor
Hex HE is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.