Is there something called the Reduced Column echleon form?

Question

I recently asked a question where I couldn't find the rank of a matrix. The question is :

Problem on Finding the rank from a Matrix which has a variable

At the time I believed in the answer, and now when I went back to that problem I realized something weird. How did this answer work.

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The answer was a continuation from where I have reduced the matrix by reduce row echleon form. And the answer continues as adding individual columns to another column and with scalar multiples of a column added to another (like how we do for rows in echleon form). I have never learnt anything other than row reduction.

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I don't know what this is called, but how is this valid?

Common sense suggest me that it's not valid. I even tried a Google search on "Reduced Column Echleon form" and nothing came up. Can someone please tell what's going on ?

And also what is the name given to this process?

Answer 1

I suppose you can define the reduced column echelon form as the reduced row echelon form of the transpose, but it doesn't see much (if any) usage.

The fact of the matter here is that the rank of a matrix is equal to the rank of its transpose. Since elementary row operations do not change rank, neither will elementary column operations, since you may simply view them as row operations done on the transpose. This means that, for the purposes of finding rank at least, you are free to use any combination of elementary row or column operations at your leisure.

If you really don't like looking at this process in terms of operations on columns, you can think of it as taking the transpose of the matrix (which preserves rank), performing the needed elementary row operations, and then transposing back.

Answer 2

I asked this exact question to my Linear Algebra professor (whose area of study is abstract and linear), and she said that the processes works almost identically, and it is just convention to do ROW reduced echelon, instead of column. You could, although it is not necessary to view a $N \times M$ matrix as a $M \times N$ and to proceed doing "column" operations.

Honestly, when I was first learning LinAlg, and was told to find something reading columns, using row reduction, I just reduced the columns.

Answer 3

It depends on what you want to do.

If you strictly do row operations, then you preserve the Kernel (Null Space) and the Row Space (and thus the Rank).

If you strictly do column operations, then you preserve the Cokernel (Left Null Space) and the Column Space (and thus the Rank).

If you mix and match the types of operations (row or column) then you still preserve the rank, because either type of operation preserves the rank, but the actual Column/Row Spaces will probably be lost.

If you want to solve an equation of the form Ax=b (where A is a matrix and x,b are column vectors), then you should do row operations on A. You can also swap the columns, but you have to swap the related variables in x, and it's not really a major benefit.

If you want to solve an equation of the form xA=b (where A is a matrix and x,b are row vectors), then you should do column operations on A. You can also swap the rows, but you have to swap the related variables in x, and it's not really a major benefit.

"Reduced column echelon form" is a good enough name for it, though since it's not commonly taught, I can't really say it's official. I think the reason that rref is taught instead of rcef is a consequence of Ax=b being the more common notation, and teachers not wanting to waste time or cause confusion by teaching the exact same techniques on a technicality.