# Finite dimensional real vector space

1. Oct 5, 2005

### Benny

Hi can someone assist me with the following question?

Q. Let V be a finite dimensional real vector space with inner product < , > and let W be a subspace of V. Then the orthogonal complement of W is defined as follows.

$$W^o = \{ v \in V: < v,w > = 0,w \in W\}$$

Prove the following:

a) $$W^o$$ is a subspace of V.
b) $$W \cap W^o = \left\{ {\mathop 0\limits^ \to } \right\}$$
c) $$\dim W + \dim W^o = \dim V$$

My working:

I can do the first part but the others are a problem for me.

b) W and W^o are both subspaces of V and so they both contain the zero vector. Then their intersection also contains the zero vector. Suppose the intersection contains some non-zero vector say f. Then we must have <f,f> = 0 for some non-zero vector f. But this contradicts some inner product property which says <f,f> = 0 iff f = zero vector. So from that I conclude that $$W \cap W^o = \left\{ {\mathop 0\limits^ \to } \right\}$$.

c) I can't think of a way to do this one. I know that dim(V) >= dim(W), dim(W_0) because any linearly independent set in V has most k elements where k is the number of vectors in a basis for V.

Can someone help me with part c or check my answer for part b? Any help appreciated.

2. Oct 5, 2005

### HallsofIvy

Staff Emeritus
Your answer to (b) is completely correct.

To do (c), choose a basis for $$W$$ and a basis for $$W^0$$.
Show that their union is a basis for $$V$$.

3. Oct 5, 2005

### Benny

Ok thanks for your help HallsofIvy, I'll give that a go.