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I apologize if this is a silly question( which may have been asked before), I was wondering after seeing a post on this list on math-overflow

When is it true that $\dim(U \cap (V+W))=\dim(U \cap V + U \cap W$)?

This is not true in general, since we can set $U,V,W$ to be three distinct lines in $\mathbb{R}^{2}$. For example, if if I set $U=\langle (1,0)\rangle$, $V=\langle (0,1)\rangle$, and $W=\langle (1,1) \rangle$. We then have that

$\dim(U \cap (V+W))=\dim(U)=1$

$\dim(U \cap V + U \cap W)=0$

I was just curious more than anything.

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Since $U\cap (V+W)$ and $(U\cap V)+(U\cap W)$ are both vector spaces, with the former containing the latter, they have the same dimension exactly when they are equal. Hence, the original question is true precisely when $$U\cap ((V+W)\setminus V\setminus W)=\emptyset$$

In the example given, $(1,0)\in U\cap ((V+W)\setminus V\setminus W)$.

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  • $\begingroup$ Thanks, this looks to be what I am looking for. In the last line, did you mean $(1,0)$? $\endgroup$ – user135520 Jun 15 '15 at 20:45
  • $\begingroup$ @user135520, quite right, thanks. Fixed. $\endgroup$ – vadim123 Jun 15 '15 at 22:46

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