Find a basis for the column space and the rank of the matrix

Find a basis for the column space and the rank of the matrix.

[(7,-3,-6,14);(2,-6,-3,4);(-2,-2,1,-4);(2,-2,-2,4)]

(a) a basis for the column spac

e

Solution

First reduce the matirx to row echelon form.

Divide row1 by 7

Add (-2 * row1) to row2

Add (2 * row1) to row3

Add (-2 * row1) to row4

Divide row2 by -36/7

Add (20/7 * row2) to row3

Add (8/7 * row2) to row4

Add (3/7 * row2) to row1

The matrix has 2 pivot columns (hilighted in yellow) and 2 free columns; because the matrix has 2 pivots, the rank of the matrix is 2.

Let\'s take the \'free\' part of the reduced row echelon form matrix (hilighted below in yellow)...

and turn it into its own matrix:

Let\'s multiply this matrix by -1:

Now, we add the Identity Matrix to the rows in our new matrix which correspond to the \'free\' columns in the original matrix, making sure our number of rows equals the number of columns in the original matrix (otherwise, we couldn\'t multiply the original matrix against our new matrix):

Finally, the Null Space of our matrix is defined by scalar multiples of these column vectors: