# Is there only a unique way to decompose a matrix as a series of multiplication of elementary matrix ? (RREF and invertible matrix)

We know that for each given matrix, there exists only one unique matrix in row reduced echelon form.

We also know that the transformations of a matrix to another can be decomposed to a series of elementary matrices: $$A=E_nE_{n-1}E_{n-2}\dots E_1 (P) \text{ (where P is the original matrix)}$$

As a result, I would like to ask:

1. Does it exist only one way to transform a matrix to the RREF ?
2. Is the matrix resulted from $$E_nE_{n-1}E_{n-2}\dots E_1$$ that transforms a matrix to its RREF unique?

We also know that to find the inverse of an invertible matrix, we can also apply a series of multiplication of elementary matrices.

In that case, is $$E_nE_{n-1}E_{n-2}\dots E_1$$ unique in finding the inverse matrix?

Try row reducing $$\begin{pmatrix} 1 & 0 \\ -1 & 0 \end{pmatrix}$$ in two ways: first, by just adding row 1 to row 2, and second, by swapping the rows, adding (the new) row 1 to (the new) row 2, and then scaling row 1 by -1. Pretty sure you get different products of $$E$$'s.

In more detail, consider the matrix $$\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$$ (implementing the first RREF) and the matrix product $$\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$ (implementing the second). Are they the same?

As a metamathematical note, I came up with this example from a general intuition that the product of $$E$$s may depend on which rows you choose to use when clearing out entries from a given column. Often you have some choice in that, and apparently it does make a difference.

• Yes thanks a lot for adding more details to your answer. I understand now perfectly. – Yan Zhuang Feb 23 at 17:22
1. The sequence of transformations is not unique. For example you can insert an operation, then undo it by applying the inverse of the operation.

2. The matrix $$E_nE_{n-1}\cdots E_1$$ is not unique unless the original matrix $$P$$ is of full rank. For example, consider $$P:=\begin{pmatrix}1&0\\1&0\end{pmatrix}$$, whose RREF is $$A:=\begin{pmatrix}1&0\\0&0\end{pmatrix}$$. Try writing $$A$$ as the product $$EP$$, for $$E:= \begin{pmatrix}a&b\\c&d\end{pmatrix}$$. After expanding $$EP$$, you will find that the only constraints are $$a+b=1$$ and $$c+d=0$$. This means you have some freedom in selecting $$a, b, c, d$$.

3. If $$P$$ has full rank, then the RREF $$A$$ is the identity matrix. So if $$E$$ satisfies the equation $$A=EP$$, then $$E$$ must be the inverse of $$P$$ and is therefore unique.

• Thanks a lot as well. :) – Yan Zhuang Feb 23 at 17:32