# Comparing probability and dimension forumulas, concidence?

So we know that if given two finite vector spaces $V,W$ then the $\dim(V\cup W)=\dim(V)+\dim(W)-\dim(V\cap W)$ This curiously corresponds with the Probability formula of, given two probabilities $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Question, is this a mere coincidence or is there something deeper going on? Probabilities are always numbers between 0 and 1, the dimension of a vector space is, I assume, always a natural number. So anything special going on?

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I wonder what it is that you call the dimension of $V\cup W$. –  Did Oct 14 '12 at 22:29
Perhaps what's going on is that for any measure $\mu$ you get this equation, and dimension is a sort of counting measure on vector spaces. –  Gerry Myerson Oct 14 '12 at 22:29
@did, remind me to look for hidden meanings in your comments in the future. –  Gerry Myerson Oct 14 '12 at 22:51

## 3 Answers

WARNING: This does not work for sums of more than two subspaces.

$$\dim(U+V+W)$$ may differ from $$\dim U+\dim V+\dim W-\dim(U\cap V)-\dim(U\cap W)-\dim(V\cap W)+\dim(U\cap V\cap W).$$ For example, it doesn't work if $U,V,W$ are distinct one-dimensional subspaces of a two-dimensional space.

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This is a consequence of the additivity property, both of the dimension of a subspace and of probability, as Gerry commented, they both are some kind of measures.

If $U,V$ are 'disjoint' subspaces (i.e. $U\cap V=\{0\}$), then $\dim(U+V)=\dim U+\dim V$.

If $A,B$ are disjoint 'events' (subsets of the probability space), then $P(A\cup B)=P(A)+P(B)$.

This is the basic rule for area/volume, and any measure. There are also Projection valued measures, which can be used for some kind of probability using (projections to) subspaces of a Hilbert space..

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I think this point of view may mislead people into thinking that this works for more than two subspaces, i.e. that in general the dimension of the sum of three subspaces is $\dim U+\dim V+\dim W-\dim(U\cap V)-\dim(U\cap W)-\dim(V\cap W)+\dim(U\cap V\cap W)$. But that is false. –  Michael Hardy Oct 15 '12 at 13:05

I guess you are looking for the Inclusion–exclusion principle.

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