Hi.

Let $\displaystyle U,W$ be subspaces of $\displaystyle V$.

Suppose $\displaystyle dim(V)=n$, and $\displaystyle dim(U)=dim(W)=n-1$

I need to prove that $\displaystyle dim(U\cap W)=n-2$

Thanks in advanced!

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- Jan 11th 2013, 09:47 AMStormeyneed help proving a statment with subspaces dimensions
Hi.

Let $\displaystyle U,W$ be subspaces of $\displaystyle V$.

Suppose $\displaystyle dim(V)=n$, and $\displaystyle dim(U)=dim(W)=n-1$

I need to prove that $\displaystyle dim(U\cap W)=n-2$

Thanks in advanced! - Jan 11th 2013, 10:32 AMDevenoRe: need help proving a statment with subspaces dimensions
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. - Jan 11th 2013, 10:59 AMvincisonfireRe: need help proving a statment with subspaces dimensions
Deveno is right, your problem probably states $\displaystyle dim(U\cap W)\geq n-2$

- Jan 11th 2013, 11:33 AMStormeyRe: need help proving a statment with subspaces dimensions
sorry, let me rephrase that:

i need to prove that**if**$\displaystyle dim(V)=n$**and**$\displaystyle dim(U)=dim(W)=n-1$,**then**$\displaystyle dim(U\cap W)=n-2$. - Jan 11th 2013, 12:35 PMDevenoRe: need help proving a statment with subspaces dimensions
that's not true. see post #2.

- Jan 11th 2013, 01:03 PMStormeyRe: need help proving a statment with subspaces dimensions
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... (Doh) - Jan 11th 2013, 03:24 PMDevenoRe: need help proving a statment with subspaces dimensions
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 = {w_{1},...,w_{n-1}}, B' = {u_{1},...,u_{n-1}}, we need to show:

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

c_{1}w_{1}+...+c_{n-1}w_{n-1}+c_{n}u_{j}= 0, we can distinguish 2 cases:

a) c_{n}= 0. in this case, the linear independence of the w_{i}forces c_{1}= ... = c_{n-1}= 0,

b) c_{n}≠ 0. in this case, u_{j}= (-c_{1}/c_{n})w_{1}+...+(-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. - Jan 12th 2013, 12:05 AMStormeyRe: need help proving a statment with subspaces dimensions
thanks again Deveno!

- Jan 14th 2013, 06:39 PMLoidaWardRe: need help proving a statment with subspaces dimensions
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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