# Basis for a subspace in R^4

1. Oct 21, 2009

### saifatlast

1. The problem statement, all variables and given/known data
Find a basis for $$F=\left\{(x,y,z,w): -x+y+2z-w=0\right\}$$

3. The attempt at a solution

So this looks like a plane to me, but I find 4-d space confusing, so that might be wrong. It does have the form $$\mathbf{x}^T\mathbf{n}=0$$, so that's kind of where I'm getting the idea that it's a plane. Assuming it is a plane (and if not, I'd appreciate some help getting moving in the right direction), my approach is to find 3 vectors in the plane. The plane should be spanned by the linear combinations of these vectors.

This gives

$$F: \left\{ \begin{bmatrix}1\\1\\0\\0\end{bmatrix}, \begin{bmatrix}2\\0\\1\\0\end{bmatrix}, \begin{bmatrix}-1\\0\\0\\1\end{bmatrix} \right\}$$

My questions is, assuming this is correct, how do I prove to myself that it is? I guess if I take the dot product of each of the basis vectors with the normal vector, n, below, I do get 0.

$$\mathbf{n}=\begin{bmatrix}-1\\1\\2\\-1\end{bmatrix}$$

I guess what it comes down to is that I feel like I'm on shaky ground here, and I'm not quite sure how to firm things up.

2. Oct 21, 2009

### Staff: Mentor

Your basis for F is the same one I found, so I would imagine we both used the same technique.

F is not a plane; it is a hyper-plane, a three-dimensional subspace of R4. It is a higher-dimension analog of a plane embedded in R3.

You can convince yourself that the three vectors you found are a basis for F by showing that they are linearly independent, and that they span F. These are actually pretty simple to do, if you have chosen your basis vectors wisely (and you did).

Your idea of dotting the three vectors with the normal to this hyperplane is interesting, and probably reasonable to do. It's a little hard to imagine the normal to a three-D "plane" though.