How to dynamically fill empty space in a column?

Is there an easy way to get column space from the row space?

• I have a 3*4 matrix 3 6 -6 4 5 10 -10 2 -3 -6 6 5 the question is to find the basis for the column space and basis of the row space. I found the row space by elementary row operations to get it into row echelon form. I got the two vectors of (1,2,-2,6) and (0,0,0,1). I was wondering if there is an easy way to get the basis of the column space besides taking the transpose of my matrix then getting it into row echelon form which is tedious, but not hard. Is there an easier quicker way?

• Answer:

  identify the pivot columns of your matrix in row echelon form (the columns in which the leading 1's occur). these same columns are a basis for your column space. since the reduced row echelon form is: [1 2 -2 6] [0 0 0 1 ] [0 0 0 0 ], a basis for the column space is {(3,5,-3), (4,2,5)}. just by inspection, it is easy to see columns 2 and 3 are scalar multiples of the first column. the rank is 2, so the 4th column has to be LI of the first.