Adam F. answered 09/27/22
Abstract Vector Spaces, Linear Algebra Tutor, Columbia University Grad
Just from glancing at the matrix you can see that the first column [-6, 6] is the same as the third column [-4,4] minus the second [2,-2]. That is to say [-6,6] = [-4,4]-[2,-2]. This shows without much calculation that the vector [1,-1,1,0] must be in the null space. Because this vector effectively takes 1 of the first column, -1 of the second column, 1 of the third column, and 0 of the last. So you get [-6,6] - [-4,4] + [2,-2] = [0,0] which is the zero vector.
We still need to get other vectors in the null space if there are any. But before doing that, a remark:
I just found [1,-1,1,0] through some informed guessing. This of course isn't the official algorithm for finding vectors in the null space. The official procedure is to reduce the matrix into row-echelon form and proceed from there. We could do it that way, it always works, but I just want to demonstrate an alternate solution in case that helps to gain a different perspective.
Anyway, let's keep trying to do some informed guessing. And yeah, we can see that the first column equals the second column plus the fourth column. [-6,6] = [2,-2]+[4,-4]. This tells us the vector [1,-1,0,-1] is in the null space.
Let's find one more. This one's even easier when we see that the 3rd column plus the 4th column is zero. So [0,0,1,1] is in the null space.
We've found three vectors in the null space. The only matrix with four independent vectors in the null space is the zero matrix, and this isn't the zero matrix. So there can't be any more. Therefore a basis for the null space is
[1,-1,1,0], and [1,-1,0,-1], and [0,0,1,1]
These are pretty clearly independent, although if we want to be very rigorous we would have to prove it. I won't do that now since I'm not sure if it's necessary for your assignment. But in any case, hopefully this sheds some light on the subject.