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EDIT: I decided to ask this question after working on this particular problem. I had the stupidity to think that row-reduced = row reduced echelon form. Brain fart, nothing more to see here...

Wikipedia lists the following two conditions for a matrix to be in reduced row echelon form, in agreement with my Linear Algebra book by Hoffman and Kunze:

  • It is in row echelon form.
  • Every leading coefficient is 1 and is the only nonzero entry in its column

I am told that the reduced row echelon form of a matrix is unique. However, according to the above two conditions only, then is it not possible to have multiple forms of RREF through row interchanges? Ergo, would not an alternative RREF for $I_3$ be the following?

\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}

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  • $\begingroup$ The matrix you show is not in row echelon form. Generally, this requires that the leading coefficient (the first nonzero number from the left, also called the pivot) of a nonzero row is always strictly to the right of the leading coefficient of the row above it. $\endgroup$
    – qaphla
    Jul 30, 2015 at 17:24
  • $\begingroup$ @qaphla I was wondering why an additional condition attesting to such was not given, either on Wikipedia or in my Lin Alg book. As I understand, I am not mistaken in thinking that the above example satisfies the given two conditions, am I? Thanks in advance for the help! $\endgroup$
    – user245273
    Jul 30, 2015 at 17:27
  • $\begingroup$ ??? Read what he wrote again! That matrix does not satisfy those two conditions. It is not in echelon form. The first condition is "is in echelon form". (The "extra condition" you mention is included! It's part of the definition of echelon form.) $\endgroup$
    – David C. Ullrich
    Jul 30, 2015 at 17:30
  • $\begingroup$ @DavidC.Ullrich Well, I made a fool of myself. Turns out I cannot read. Thank you... $\endgroup$
    – user245273
    Jul 30, 2015 at 17:32
  • $\begingroup$ The other day I used the identity $1/N = N^2/N$ in a proof here. These things happen... $\endgroup$
    – David C. Ullrich
    Jul 30, 2015 at 17:35

1 Answer 1

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I don't have Hoffman & Kunze at hand, but the definition of row echelon form (which is embedded in your definition of reduced row echelon form) from wikipediae states:

The leading coefficient (the first nonzero number from the left, also called the pivot) of a nonzero row is always strictly to the right of the leading coefficient of the row above it (some texts add the condition that the leading coefficient must be 1).

So, your example fails to be row echelon since the first nonzero term in the second row is not to the right of the first nonzero entry in the first row.

Since it is not row echelon, it fails the first element of your definition of reduced row echelon, which requires your matrix to be in row echelon form.

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  • $\begingroup$ The first part of your answer is right, but the specifics in the last sentence are not. The second and third rows of the example are in accord with the definition; the definition fails for the second row, because its first 1 is not to the right of the first 1 in the preceding (first) row. $\endgroup$
    – Andreas Blass
    Jul 30, 2015 at 17:28
  • $\begingroup$ @AndreasBlass You are absolutely correct. Fixed now. $\endgroup$
    – Bamboo
    Jul 30, 2015 at 17:29

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