# Prove that if A, B are n x n matrices and AB is a product of elementary matrices, then A is also a product of elementary matrices.

"Prove that if A, B are nxn matrices and AB is a product of elementary matrices, then A is also a product of elementary matrices."

I found an answer that states "AB is a product of elementary matrices if and only if AB is invertible. AB invertible if and only if A and B are invertible. A and B are invertible if and only if A and B are products of elementary matrices."

However, we have not been taught that AB is a product of elementary matrices if and only if AB is invertible. We have only been taught that "If A is an n x n invertible matrix, then A and A^-1 can be written as a product of elementary matrices." So I can't use the IFF statement without proving it first. Any ideas?

• @egreg I know and understand that, but I don't know or haven't seen a proof that a matrix that is a PRODUCT of elementary matrices is invertible. – Hayden Holligan Mar 18 '15 at 23:14
• The product of invertible matrices is invertible. – Nishant Mar 18 '15 at 23:14
• @Nishant can you elaborate? I feel like that would help with the converse of the proof (working backwords from that) but I can't see how it would help here, I may be missing something. – Hayden Holligan Mar 18 '15 at 23:18
• If $A$ and $B$ are invertible, then the inverse of $AB$ is $B^{-1}A^{-1}$. Also, if $A$ and $AB$ are invertible, so is $B$, since $B=A^{-1}(AB)$, and $A^{-1}$ is invertible. – Nishant Mar 18 '15 at 23:52

$AB$ is a product of elementary matrices. So $$AB = {E}_{1}{E}_{2}...{E}_{n}$$ For some elementary matrices ${E}_{i}$. Each ${E}_{i}$ is invertible so $AB$ to.
Now consider the solutions of $BX=0$. Multiplying by A gives: $ABX=0$ and since $AB$ is invertible it follows that $X=0$. So if $BX=0$ then $X=0$. This implies B is invertible. We have $A=A(B{B}^{-1})=(AB){B}^{-1}$ is invertible to, because product of invertible matrices is invertible. So $A,B$ can be written as a product of elementary matrices.