relation between dimension of sums of subspaces and inclusion-exclusion principle

Question

Suppose we have a finite-dimensional vector space $V$ and we have two subspaces $U$ and $W$ of $V$.

It is well known that $\dim(U+W)=\dim(U)+\dim(W)-\dim(U\cap W)$.

Also, suppose $A$ and $B$ are finite sets. It is well known that $|A\cup B|=|A|+|B|-|A\cap B|$ (the inclusion-exclusion principle).

As you can see, the formulae here are similar. Is there a nice way to relate these two facts?

Mike

Does this answer your question? The dimension of the sum of subspaces $(U_1,\ldots,U_n)$ — Golden_Ratio, Jan 11 '22 at 04:53

score 1 · Answer 1 · answered Jan 11 '22 at 04:48

1

As you point out, this parallel holds for two subspaces. However, it does not hold for more than two subspaces in general due to non-distributivity:

$$A\cap (B+C)\neq (A\cap B)+(A\cap C).$$

Check out this MO post for discussion on the same question.

answered Jan 11 '22 at 04:48

Golden_Ratio

1

Also, https://math.stackexchange.com/questions/520910/the-dimension-of-the-sum-of-subspaces-u-1-ldots-u-n – Golden_Ratio Jan 11 '22 at 04:52

1 Answers1