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"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?

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  • $\begingroup$ @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. $\endgroup$ – Hayden Holligan Mar 18 '15 at 23:14
  • $\begingroup$ The product of invertible matrices is invertible. $\endgroup$ – Nishant Mar 18 '15 at 23:14
  • $\begingroup$ @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. $\endgroup$ – Hayden Holligan Mar 18 '15 at 23:18
  • $\begingroup$ 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. $\endgroup$ – Nishant Mar 18 '15 at 23:52
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$AB$ 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.

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  • $\begingroup$ After proving B is invertable, how did you arrive at A=AB? $\endgroup$ – Hayden Holligan Mar 18 '15 at 23:49
  • $\begingroup$ Sorry, see my edit for a better notation. $\endgroup$ – abcdef Mar 19 '15 at 8:57

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