Linear algebra - dimension and intersection

[zwingtip]

Homework Statement

let $$V$$ be a finite dimensional vector space of dimension n. For $$W \leq V$$ define the codimension of $$W$$ in $$V$$ to be $$codim(W) = dim(V) - dim(W)$$. Let $$W_i, 1 \leq i \leq r$$ be subspaces of $$V$$ and $$S = \cap_{i=1}^{r}W_i$$. Prove:

$$codim(S) \leq \sum_{i=1}^{r} codim(W_i)$$

Homework Equations

$$dim(U + V) = dim(U) + dim(V) - dim(U \cap V) \sum_{i=1}^{r} codim(W_i) = \sum_{i=1}^{r} (n - dim(W_i)$$

The Attempt at a Solution

I'm completely lost here. I know I need to prove this by induction. Any tips to point me in the right direction? Can I use the fact that $$codim(S) = n - dim(\cap_{i=1}^{r} W_i$$ and $$dim(U \cap V) < dim(U) + dim(V)$$ assuming $$dim(U + V) \neq 0$$?

 

[lanedance]

only had a quick look, but not convinced this is true..

take the case when say dim(W1) = dim(W2) = 1, in Rn, then codim(W1) = codim(W2) = n-1

now say the intersection S is only the zero vector then dim(S) = 0 and codim(S) = n >n-1> Can up with oher higherdimensional arguments similarly

 

[zwingtip]

Still doesn't make sense to me. What if the intersection contains more than just the zero vector? Sorry for being stupid. Help?

 

[lanedance]

actually i think my exampel was wrong as
codim(S) = n < codim(W1) + codim(W1) = 2n-2, which works for for n>=3

now take R3, say you have 2 planes P1 & P2 that intersect in a line L, then
dim(R3) = 3
dim(P1) = dim(P2) = 3, then codim(P1) = codim(P2) = 2
dim(L) = 1, codim(L) = 2

so
codim(L) = 2 <= codim(P1) + codim(P2) = 1 + 1 = 2

 

[zwingtip]

I think I've got it with an induction proof from the base cases r=1, r=2 and the inductive step being that it holds for r>2 then manipulating it algebraically for an entire page of 5mm graph paper. THanks.