Dimension proof of the intersection of 3 subspaces

harvesl
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Homework Statement



Assume V = \mathbb{R}^n where n \geq 3. Suppose that U,W,X are three distinct subspaces of dimension n-1; is it true then that dim(U \cap W \cap X) = n-3? Either give a proof, or find a counterexample.

The Attempt at a Solution



The question previous to this was showing that for subspaces U,W of dimension n-1 \longrightarrow dim(U \cap W) = n - 2 which I was able to prove fine. Now, I'm thinking this is true, because W \cap X will be a subspace of dimension n-2 so if we set W \cap X = Z, where Z is a subspace we turn this problem into dim(U \cap Z) and since all three were distinct we should have dim(U \cap Z) = n-3 but I'm not entirely sure.

Thanks!
 
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You seem to be on the right track.
 
voko said:
You seem to be on the right track.

Thanks.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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