Reduced Echelon Form - Which is correct?

[L = K - U]

Hi everyone,

I am teaching myself Linear Algebra and I am confused with the terminology used in the subject.

I am studying Linear Algebra based on Anton's. In the textbook, an augmented matrix in REF needs to have the first nonzero number in a given row to be 1. However, in other textbooks, the first nonzero number in a given row can be any number.

Which is right? It is based solely on preference?

Thanks :)

 

[Mark44]

Hi everyone,

I am teaching myself Linear Algebra and I am confused with the terminology used in the subject.

I am studying Linear Algebra based on Anton's. In the textbook, an augmented matrix in REF needs to have the first nonzero number in a given row to be 1. However, in other textbooks, the first nonzero number in a given row can be any number.

Which is right? It is based solely on preference?

Thanks :)

According to this wiki article, it varies. See the last part of the 2nd bullet.
From https://en.wikipedia.org/wiki/Row_echelon_form:

Specifically, a matrix is in row echelon form if

• all nonzero rows (rows with at least one nonzero element) are above any rows of all zeroes (all zero rows, if any, belong at the bottom of the matrix), and
• 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).

 

[S.G. Janssens]

There is a difference between the row echelon form (REF) and the reduced row echelon form (RREF). In his wiki quote, Mark44 gives the definition of the REF. To obtain the RREF one additionally requires that

• The leading entry in each nonzero row is 1
• Each leading 1 is the only nonzero entry in its column

One can show that each matrix is row equivalent to exactly one matrix in RREF. However, a (nonzero) matrix is always equivalent to more than one matrix in REF. This last statement follows clearly from the first statement: just multiply any nonzero row in the RREF with a nonzero scalar.

In conclusion: The condition that the leading entry in each nonzero row equals 1 is a normalization condition that helps to ensure uniqueness of the RREF.

 

[mathwonk]

to see that a matrix determines a unique RREF note that the row space of a matrix is well determined by the matrix, namely the span of the rows. If the dimension of the row space is r, assume for simplicity that the projection of the row space onto the r dimensional subspace of R^n spanned by the first r standard basis vectors is an isomorphism. Then viewing R^n as R^r x R^(n-r), we can think of the row space as the graph of a linear map from R^r-->R^(n-r). Hence the row space detrmines this graph and hence this linear map. Now just look at the values this map takes on the r basis vectors of R^r, i.e. the first r basis vectiors of R^n. These values, when added onto the basis vectors themselves, are exactly the rows of the reduced echelon form. Hence the reduced echelon form is uniquely determined by the matrix.

E.g. if r = 2, and n = 4, the row space is the graph of a map from R^2-->R^2, and if the value of this map on (1,0) is say (3,5), the first row of the reduced echelon form is (1 0 3 5).

From this point of view the fact the pivot entries are all 1, corresponds to the fact that the standard basis vectors have a single 1 in them.

So to recap, you are looking for a nice basis of the row space. If the row space projects isomorphically onto the subspace R^r spanned by e1,...,er, then just take the r vectors in the row space that project isomorphically to these standard basis of R^r, and those are the rows of the RREF.

This proves it (both existence and uniqueness of RREF) in the most common case where the first r columns are all pivots, and can be adapted to the general case when this fails, by examining more closely the projections of the row space onto the various subspaces spanned by initial sets of standard vectors.