# Subspaces and Orthogonal Complements

• Mar 17th 2010, 03:18 PM
uberbandgeek6
Subspaces and Orthogonal Complements
If a subspace S is contained in a subspace V, prove that the orthogonal complement of S contains the orthogonal complement of V.

I'm really not sure how to go about this. I think that if V contains S, then S must have a smaller dimension than V, and since one orthogonal complement should also be contained by the other, it would make sense that S(perp) contains V(perp) else there would be too many dimensions in them, but I don't know how to explicitly prove this (sorry if that explanation was confusing, I'm not really even sure of it).
• Mar 17th 2010, 03:29 PM
Drexel28
Quote:

Originally Posted by uberbandgeek6
If a subspace S is contained in a subspace V, prove that the orthogonal complement of S contains the orthogonal complement of V.

I'm really not sure how to go about this. I think that if V contains S, then S must have a smaller dimension than V, and since one orthogonal complement should also be contained by the other, it would make sense that S(perp) contains V(perp) else there would be too many dimensions in them, but I don't know how to explicitly prove this (sorry if that explanation was confusing, I'm not really even sure of it).

Let $\displaystyle v\in V^{\perp}$ then given any $\displaystyle v'\in V$ we have that $\displaystyle \langle v,v'\rangle =0$. But, since $\displaystyle S\subseteq V$ we see in particular that $\displaystyle \langle s,v\rangle =0$ for every $\displaystyle s\in S$ . Thus $\displaystyle V^{\perp}\subseteq S^{\perp}$