For example. Find the dimension of the subspace of R

^{4}, <u, v, w, x>, satisfying 3u- 2v+ 3w- x= 0. (Yes, we can consider it the null space of the operator that maps any <u, v, w, x> to 3u- 2v+ 3w- x in

.) Since that is a single equation I know immediately that the dimension of this subspace is 4- 1= 3. In fact, I can solve for x, x= 3u- 2v+ 3w, so I can write any member of that subspace as <u, v, w, 3u- 2v+3w>= u(1, 0, 0, 3>+ v<0, 1, 0, -2>+ w<0, 0, 1, 3> showing that {<1, 0, 0, 3>, <0, 1, 0, -2>, <0, 0, 1, 3>} is a basis.

If I were talking about the subspace satifying both that and the equation u+ v- w+ x= 0 (Which we can think of as the null space for the linear transformation that takes <u, v, w, x> to <3u-2v+3w-x. u+v-w+x> in

. Any subspace is the null space of

**some** linear transformation!), I can see that there are two equations and, once I have checked that they are independent, know that this subspace will have dimension 4- 2= 2. If I add the two equations, I eliminate x and get 4u- v+ 2w= 0. I can solve that for v: v= 4u+ 2w. Putting that into the first equation, 3u-2(4u+2w)+ 3w- x= -5u- w- x= 0 so x= -5u- w. That means that any member of this subspace can be written as <u, 4u+2w, w, -5u- w>= u<1, 4, 0, -5>+ v<0, 2, 1, -1> which means that {<1, 4, 0, -5>,<0, 2, 1, -1>} is a basis for the subspace.