Looks to me like it is just a matter ofdoingthe calculation.

The null space is the set of all (x, y, z) such that

That means we must have 2x+ z= 0, -4x- 2z= 0, and -4x- 2z= 0, all of which reduce to the single equation z= -2x. Of course, y does not appear in that equation and can be anything. That tells us that any vector in the nullity is of the form <x, y, -2x>= x<1, 0, -2>+ y<0, 1, 0> so {<1, 0, -2>, <0, 1, 0>} is a basis for the null space.

Of course, any vector in the range is of the form

or <2x+ z, -4x- 2z, -4x- 2z>= (2x+z)<1, -2, -2> for any numbers x, z so <1, -2, -2> is a basis for the range, as you say.

But <1, -2, -2>= <1, 0, -2>- 2<0, 1, 0> and so is itself in the null space. That means that the range is contained in the null space: the intersection of range and null space is just the range itself. A basis for the intersection is a basis for the range, <1, -2, -2>.

And, Since Ax is in the range which is in the null space of A, for all x: .