# Different answers for reduced row echelon form (ROW vector and Column Vector cases)

This is the question I am working on : identify a largest subcollection of linearly independent vectors and express all the remaining vectors as linear combinations of the vectors in the subcollection. (a) [4 points] $v_1 = (5, 2, −3, 1)$, $v_2 = (4, 1, −2, 3)$, $v_3 = (1, 1, −1, −2)$, $v_4 = (3, 4, −1, 2)$, $v_5 = (7, −6, −7, 0)$;

The method I found from one of the stack exchange posts :

1. Put vectors as COLUMNS in a matrix
2. Find RREF
3. Identify non-pivotal columns in RREF matrix and then the corresponding columns in the original matrix are the linearly independent ones.

This works!

But I tried to do this by : 1.Putting vectors as ROWS in a matrix 2.Find RREF 3.Identify non-pivotal rows in RREF and then the corresponding rows in the original matrix are the linearly independent ones.

But I'm getting a wrong set here. What Am I doing wrong? : While doing the RREF, I did permute rows, I am assuming that is the problem here. However, then I should have done Reduced Column Echelon Form right? But I have read that RREF = (RCEF)transpose.

Thoughts: RREF=RCEF transpose works only for the same matrix so if I put vectors as columns, I should do RREF and if as rows then RCEF. Am I right?

Thank you

• Welcome to MSE. Please use MathJax Apr 27, 2018 at 8:06

It depends on what you’re trying to accomplish. Row-reduction produces a basis for the row space of a matrix $M$, but there’s in general no correlation between the resulting non-zero rows and the original rows of the matrix, aside from their spans being equal. This should be fairly obvious since all of the zero rows appear at the bottom of the reduced matrix, but there’s no reason to believe that the first $\operatorname{rank}M$ rows in the original are linearly independent.