# 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 $$dim(U + V) \neq 0$$?

## Answers and Replies

Homework Helper
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?

Homework Helper
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.