1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Yes thanks a lot for adding more details to your answer. I understand now perfectly. $\endgroup$ – Yan Zhuang Feb 23 at 17:22
2
$\begingroup$
  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.

$\endgroup$
1
  • $\begingroup$ Thanks a lot as well. :) $\endgroup$ – Yan Zhuang Feb 23 at 17:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.