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The calculation of $\dim(U + V + W)$

Given a linear space $V$ and subspaces $A_i \subseteq V$ such that $1\leq i \leq n.$

To find $\dim(A_1 + A_2 +\cdots + A_n)$ it seems we can use inclusion exclusion. Is there any other way of finding it?

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marked as duplicate by lhf, Martin Sleziak, Qiaochu Yuan May 22 '12 at 13:12

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Do you have any extra assumptions e.g. that you know $dim(A_{i})$ for all $i's$? –  m.woj May 22 '12 at 10:38
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How are you using inclusion-exclusion? The obvious thing is false. –  Chris Eagle May 22 '12 at 11:20
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You say that "To find $\dim (A_1+A_2+⋯+A_n)$ it seems we can use inclusion exclusion", so I want to point out that while for $n=2$ the inclusion-exclusion formula $$ \dim (A+B) = \dim A +\dim B - \dim A\cap B$$ is true, it fails for $n=3$: in general $$\dim (A+B+C) \neq \dim A+\dim B + \dim C - \dim(A\cap B) - \dim(B\cap C) - \dim (A \cap C) + \dim(A\cap B \cap C).$$ Look at the example of three distinct lines in $\mathbb{R}^2$.

The problem is that subspaces of a vector space don't form a distributive lattice, in other words $A \cap (B + C) \neq A \cap B + A \cap C$. Repeatedly using the $n=2$ formula will give you an expresion for $\dim (A+B+C)$, but not just in terms of the terms appearing on the right of the displayed equation above.

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Are there a set of circumstances / restrictions in which it generally holds for n>2? –  Robert S. Barnes May 22 '12 at 12:55
    
I would guess that, for a collection of subspaces $S_1,\ldots,S_n$ every proper subcollection of which is distributive (i.e. intersection distributes over sum), $S_1,\ldots,S_n$ obeys inclusion-exculsion iff it distributive. –  mt_ May 22 '12 at 13:08
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It depends on what you already know about the $A_i$.

For example, you can write down the list of all basis vectors of all $A_i$ and determine the rank by row reduction.

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