In order to verify a set of vectors is linearly independent, a way is to form a matrix with the vectors as rows and through row-operations find the echelon matrix and see if there is a zero vector or not.
I was wondering:
can we set such vectors also as columns of a matrix and then reduce such matrix to its row-equivalent echelon one, applying the same procedure as before?
Best Answer
Yes, you can. The result follows from the fact that row rank = column rank.
Or also the (equivalent) fact that $\operatorname{rank} A = \operatorname{rank} A^T.$
As reducing the matrix to its row-reduced echelon form gives you the rank of the matrix, it makes no difference if you start by putting the vectors as columns.