Thread: need help proving a statment with subspaces dimensions

Hi.

Let be subspaces of .
Suppose , and
I need to prove that

Thanks in advanced!

this is not always TRUE.

if U = W, then dim(U) = dim(W) = n -1, but dim(U∩W) = n-1.

you can't prove things that are not true.

Deveno is right, your problem probably states

sorry, let me rephrase that:

i need to prove that if and , then .

that's not true. see post #2.

Hi, Denovo.
i just noticed it says there that W and U are different.

sorry for the confusion.
the mid exams are next week, so i don't sleep much these days...

then it's a different story.

since W and U are distinct subspaces, say with bases B and B', respectively, then B' must contain a vector not in span(B). for if not, then U ⊆ W, in which case dim(U) = dim(W) forces U = W.

hence dim(U+W) = dim(span(B U B')) = n (glossed over a bit: if we call this vector u_j, with B = {w₁,...,w_n-1}, B' = {u₁,...,u_n-1}, we need to show:

C = BU{u_j} is linearly independent. however, if:

c₁w₁+...+c_n-1w_n-1+c_nu_j = 0, we can distinguish 2 cases:

a) c_n = 0. in this case, the linear independence of the w_i forces c₁ = ... = c_n-1 = 0,

b) c_n ≠ 0. in this case, u_j = (-c₁/c_n)w₁+...+(-c_n-1/c_n)w_n-1, contradicting our choice of u_j.

we also have n-1 = dim(span(B)) < dim(span(B U {u_j}) ≤ dim(B U B') ≤ n).

therefore: dim(U∩W) = dim(U) + dim(W) - dim(U+W) = n-1 + n-1 - n = 2n - 2 - n = n - 2.

thanks again Deveno!

but for the converse one? So far, I only have a result that if K is a set consisting of all nonunits of R, then for any u, v in K, u+v won't be 1R. Does it help? Any suggestion?





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