# Is $\dim\left(U+V\right)=\dim\left(U\cup V\right)$?

If $U$ and $V$ are proper subspaces of a vector space $W$, is $\dim\left(U+V\right)=\dim\left(U\cup V\right)$?

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How do you define the dimension of $U \cup V$ when it's not a subspace? – JSchlather Jan 23 '13 at 22:54
Did you mean $\dim(U)+\dim(V)-\dim(U\cap V)$ on the right? – user7530 Jan 23 '13 at 22:55
I changed \text{dim} to \dim. Writing \dim not only prevents italicization, but also results in proper spacing before and after $\dim$ in expressions like $a\dim b$. – Michael Hardy Jan 23 '13 at 22:58

If $U$ and $V$ are subspaces of $W$, then $U\cup V$ is never a subspace unless $U \subseteq V$ or vice-versa. (This is a simple exercise.)

Nevertheless, if you define the dimension of any subset $X$ of $W$ as the dimension of the smallest subspace that contains $X$, then $\dim(U+V)=\dim(U\cup V)$ because $U+V$ is generated by $U\cup V$.

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Could you define the term on the right? I'm not familiar with this notation (I would interpret it as the expectation value of the union of U and V, which seems non-sensical in this context). – okj Jan 24 '13 at 1:00
@okj, I've edited the answer and removed the notation. – lhf Jan 24 '13 at 1:04

$\dim(U \cup V)$ doesn't make sense if one of $U$ or $V$ isn't a proper subset of the other, as U \cup V isn't a vector space otherwise. A relatively simple result is that $\dim(U+V) = \dim(U) + \dim(V) - \dim(U \cap V)$, which notably doesn't generalize to 3 elements, that is $$\dim(U+V+W) = \dim(U)+\dim(V)+\dim(W)-\dim(U \cap V)-\dim(U \cap W)-\dim(V \cap W)+\dim(U \cap V \cap W)$$ does not neccesarily hold.

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U "union" V is a subspace iff one is contained in the other, In that case how would you define the union and sum? ! Hope that Helps.

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