Can Direct Sums and Subspace Dimensions Determine Vector Space Properties?

[MathematicalPhysicist]

1. let V be a vector space, U1,U2,W subspaces.
prove/disprove: if V=U1#U2 (where # is a direct sum) then:
W=(W^U1)#(W^U2) (^ is intersection).
2. let V be a vector space with dimV=n and U,W be subspaces.
prove that if U doesn't equal W and dimU=dimW=n-1 then U+W=V.

for question two, in oreder to prove this i need to show that dim(U+W)=dimV
which bassically means that: dim(U^W)=n-2, but how do i prove this?

for the first question i think it's correct but i don't know how to prove it, anyone has got any hints for me, thanks in advance.

 

[radou]

I don't have the time to look at it right now, but it may be useful to use the definitions of the sum of vector spaces, as the theorem which states, for a vector space V and two of its subspaces U1 and U2, dim(U1^U2) + dim(U1+U2)= dimU1 + dimU2.

 

[HallsofIvy]

For both problems, start by designating bases for U1 and U2.
For 1, some of those basis vectors will be in V, some not (you will need to show that that if U1 ^ V is not empty, then at least one basis vector of U1 is in V). show that the basis vectors of U1 that are in V, union the basis vectors of U2 that are in V, form a basis for V.

2. let V be a vector space with dimV=n and U,W be subspaces.
prove that if U doesn't equal W and dimU=dimW=n-1 then U+W=V.

Are you assuming that n is not equal to 1? If U and V are not the same, but of the same dimension, each contains a vector not in the other. That vector spans a vector space.

 

[matt grime]

I'm going to take the other view - don't pick bases.

2. U+W is a vector subspace of V. It either has dimension n-1, or n. Which?

1. Let f be any projection onto U1, so that 1-f is a projection onto U2. Thus 1=f+(1-f), and f(1-f)=0=(1-f)f. What can you do now?