# Dimension of vector spaces

1. Nov 21, 2008

### Marin

Hi all!

I´m trying to prove following two inequalities but I somehow got stuck:

U, W are subspaces of V with dimV = n

1) dimV >= dim(U+W)

2) dim(U+W)>=dimU and dim(U+W)>=dimW

Could you give me some hints?

2. Nov 21, 2008

### Office_Shredder

Staff Emeritus
Just notice that if U and W are subspaces of V, then U+W is too. So pick a basis, and see what you can do with it. The same principle works for the second part also

3. Nov 21, 2008

### Marin

ok, let u_1,.....,u_n be a basis of U and w_1,....,w_m be a basis of W.

Every vector x of U+W can be therefore expressed as a lin combination of some u_1,..u_n,w_1,..,w_n , which implies that u_1,..u_n,w_1,..,w_n is a generating set of U+W. But we know that the basis of U+W must be a minimal generating set, so letting p be its dimension this yields: p<=m+n, or dim(U+W)<= dimU + dimV. Now we consider w element of W, which we also find in W+U, but cannot be spanned by u_1,...,u_n only, so dimW<=dim(W+U), where W is a subspace of W+U. but W+U is also a subspace of V, so in the end, we have

dim(U+W)<= dimV

is this a regular proof, or I did something wrong?

and another question: dimU + dimW should not necesserily be smaller or equal to dimV, should it?

4. Nov 21, 2008