I need to demonstrate that Ker(AB)=${\vec{0}}$, just like Ker(A) and Ker(B). I have a not-so-elegant way of showing that (I think): A and B are n * p and p * m, and it has to be the case that n $\geq$ p $\geq$ m, and rref's of A and B are either equal to the identity matrix, or are of form I and row(s) of 0's on the bottom. So, product of rref's of A and B will equal a matrix that is either I or I with row(s) of 0's on the bottom, and is n * m where n $\geq$ m, so it has more equations than variables. Multiplication of matrices of the aforementioned forms can only yield a matrix of the aforementioned form (I or I with row(s) of zeros), so it must also have Ker=${\vec{0}}$. Is there a more simple (though not too advanced, as we are barely mid-way through the course) way of showing that this is true? Also, while I know that Ker(A)=Ker(rref(A)) for any matrix A, I am not sure if I can say that Ker(rref(A) * rref(B))=Ker(AB). Is this statement true?
Thanks!