1
$\begingroup$

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.
$\endgroup$

0

Your Answer

Hex HE is a new contributor. Be nice, and check out our Code of Conduct.

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.