Question

Can somebody help me with the following problem:
Inside Rn, suppose dim(A) + dim(B) > n, then show that there must be a non-zero vector in both A & B.
Here is my approach to the problem:
dim(A) + dim(B) = dim(A+B) + dim(A intersection B).
A+B is the coloumn space of the matrix C = [basis(A) basis(B)]. Thus dim(A+B) must be the rank of C. Since dim(A) + dim(B) > n, C will be a short & wide matrix where rank(C) is at most n. Since this is the case, we must have the following to be true to satisfy the inequality:
dim(A intersection B) > 0.
Is it correct?

Answer 1

Sushantpatkar, I’m still thinking about your proof, but in the meantime I have an alternative for you. Let vectors u1, u2, ..., uj be a basis for A, and v1, v2, …, vk be a basis for B, with j+k>n. Let’s start off by assuming that there is no non-zero vector that’s in both A and B, ie, there is no set of nonzero scalar a’s and b’s satisfying a1 u1 + a2 u2 + … + aj uj = b1 v1 + b2 v2 + … + bk vk. Because the u’s and v’s are bases they have to be linearly independent, so there’s no set of nonzero a’s satisfying a1 u1 + a2 u2 + … + aj uj = 0, and no set of nonzero b’s satisfying b1 v1 + b2 v2 + … + bk vk = 0. Now let’s look for sets of a’s and b’s satisfying a1 u1 + a2 u2 + … + aj uj + b1 v1 + b2 v2 + … + bk vk = 0. One possibility is that a1 u1 + a2 u2 + … + aj uj = b1 v1 + b2 v2 + … + bk vk = 0, but this was ruled out by the linear independence of the bases. Another possibility is that a1 u1 + a2 u2 + … + aj uj = -( b1 v1 + b2 v2 + … + bk vk) <> 0; but this is ruled out by the assumption that there’s no nonzero vector in both A and B. So therefore there is no set of of a’s and b’s satisfying a1 u1 + a2 u2 + … + aj uj + b1 v1 + b2 v2 + … + bk vk = 0. However you can’t have j+k>n linearly independent vectors in Rn---that’s the definition of dimension. So we have a contradiction and our initial assumption that there was no vector in both A and B is disproven.

Answer 2

Sushantpatkar, in fact I realise that the result you need comes straight from the result you quote: no need for any complexity! dim(A) + dim(B) = dim(A+B) + dim(A intersection B). We are told dim(A)+dim(B)>n, we know dim(A+B)<=n, so dim(A intersection B)>0: QED. Much neater than my proof!

Answer 3

Thank you JoshDanzinger23.