# Homework Help: Perpendicular Subspace of R^n

1. Oct 6, 2009

### Shambles

1. The problem statement, all variables and given/known data

3. The attempt at a solution

The terminology in this question confuses me into what I am actually trying to solve. It seems to me that S-perp would naturally be a subspace of real column vectors based on the fact that we specify that S$$\neq$$0. It goes on to mention that S-perp is nonempty which seems obvious in the fact that S is not empty and it asks to show that any scalar multiples of vectors within the subset of S-perp will continue to be elements of S-perp.

So i've reached the thought that either
1) This question is ridiculously simply that intends for me to re-state the obvious or
2) I've missed something completely and it actually requires a long proof. Any insights?

2. Oct 6, 2009

### Staff: Mentor

Here's an example that might give you some insight. Consider R2, and S = {(x, y) | y = 0}. Sperp = {(x, y) | x = 0}. Each element of Sperp in this example is orthogonal to each element of S, and further, it can be shown that Sperp of this example is a subspace of R2.

Your problem is similar. You need to show that your Sperp is a subspace of Rn, which means you need to show the following:
1. Sperp contains the 0 vector.
2. If u and v are in Sperp, then u + v is also in Sperp.
3. If u is in Sperp and c is a scalar, then cu is in Sperp.

3. Oct 6, 2009

### Shambles

Thank you for your assistance. I believe I have constructed a response that is adequate by proving the vectors x,u,v to be elements of S-perp by continually showing that the dot product of the vectors with y (being an element of S) to equal 0 showing orthogonality showing them to be an element of S-perp.

4. Oct 6, 2009

### Staff: Mentor

That's not going to cut it. All you have shown (based on your description) is that the vectors x, u, and v are in Sperp; you haven't shown that Sperp is a subspace of Rn. To do that, you need to show the three things I listed in my previous post.

5. Oct 7, 2009

### Landau

Shambles, do you know what a subspace is, or more precisely, how it is defined? That seems to be the problem.