# How to prove that an invertible matrix is a product of elementary matrices?

Let $$A = (a_{ij})$$ be an invertible $$n\times n$$ matrix. I wonder how to prove that $$A$$ is a product of elementary matrices. I suspect that we need to transform it into the identity matrix by using elementary row operations, but how to do it exactly?

P.S. I've checked questions which could be considered similar and neither of them deals with this exact (general) situation. Please don't mark this question as a duplicate unless you find a precise answer.

• You want to prove it is a product of elementary matrices. The RREF of an invertible matrix is the identity, because all columns become pivot columns. – egreg Jun 15 '19 at 23:31
• @egreg Thanks for noticing a mistake, I will edit. However, but why all column become pivot column? How to prove it exactly? – Jxt921 Jun 15 '19 at 23:33
• @egreg Oh, I've found a source regarding this. Apparently, the only invertible RREG if the identity. – Jxt921 Jun 15 '19 at 23:36
• An invertible $n\times n$ matrix has rank $n$, so the elimination must find $n$ pivots. – egreg Jun 15 '19 at 23:41
• @egreg Oh, that's clever. Thanks. – Jxt921 Jun 15 '19 at 23:48

If A is an invertible nxn matrix, $$A\bar x = \bar 0$$ has only the trivial solution of $$\bar 0$$ because $$A^{-1}A\bar x = \bar x = A^{-1}\bar0 =\bar0 \tag{1}\label{1}$$ This implies A is a product of elementary matrices because it implies A is row equivalent to I
Let's assume the A is in REF, if one of the diagonal elements were $$0$$ we would have a row filled with $$0$$'s and would have more unknowns than equations (proof n>m implies nontrivial solution omitted) and as a consequence has a nontrivial solution. Therefore, since we have no nontrivial solution $$\ref{1}$$, I is the RREF of A and A is composed of elementary row operations of I