Find a basis for and calculate the dimension of nullA: A = [1 2 7]T, [1 1 2]T, [-2 0 6]T, [0 1 -10]T, [4 -5 -7]T Like most algebra texts mine has pages and pages of proofs with hardly a single example tying it together. Here is what I think I know: If the determinant does not equal 0 than the vectors span R^n and are linearly independent. Therefore, they can makeup a basis for A. In this case since it is not a square matrix I cannot use determinant to find the solution so I opted to find whether it not it spans R^n such as: X1 + 2X2 + 7X3 = 0 X1 + X2 + 2X3 = 0 -2X1 + 6X3 = 0 X2 - 10X3 = 0 4X1 -5X2 -7X3 = 0 The above has a non-trivial solution so the vectors are linearly dependent and do not form a basis. Now I could add a remove vectors until I find a set of vectors with only the trivial solution but it is a pretty time intensive process, especially for a large matrix. Is there a faster way?