The dimension of the sum of subspaces $(U_1,\ldots,U_n)$

Question

If $U_1$ and $U_2$ are subspaces of a finite dimensional vector space, then $$\dim(U_1+U_2) = \dim U_1+\dim U_2-\dim(U_1 \cap U_2).$$

How can one generalize this notion to a collection of $n$ subspaces $U_1,\ldots,U_n$?

Or what does $\dim(U_1+\cdots+U_n)$ equal?

What you want is the inclusion-exclusion principle: http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle — Daniel Donnelly, Oct 10 '13 at 02:16
I am not sure that this is true for subspaces. $\dim(U_1+U_2+U_3) \neq \dim U_1+\dim U_2+ \dim U_3 - \dim(U_1 \cap U_2) - \dim(U_1 \cap U_3) -\dim(U_2 \cap U_3) + \dim(U_1 \cap U_2 \cap U_3)$. Which is true for the inclusion-exclusion principle. — St Vincent, Oct 10 '13 at 02:23
No, you don't want inclusion-exclusion. This is one of those really common expectations that fails to be true. — davidlowryduda, Oct 10 '13 at 02:24
After further inspection, this seems to have a more complicated answer than I thought. — St Vincent, Oct 10 '13 at 02:35
The reason that inclusion-exclusion does not apply here is that $+$ and $\cap$ do not obey the distributive law. — Andrew Dudzik, Oct 10 '13 at 03:31
See Axler Linear Algebra Done Right chapter 2 exercise 15. If $U_1={(x,0) | x \in \mathbb{R} }$, $U_2={(0,y) | y \in \mathbb{R} }$, $U_3={(x,x) | x \in \mathbb{R} }$ with $V=\mathbb{R}^2$, it will not work. — Doctor Grizz, Oct 10 '13 at 21:15
Thanks for pointing that out, but I already completed that exercise! — St Vincent, Oct 10 '13 at 21:16
A related post on [mathoverflow.se]: Is there a version of inclusion/exclusion for vector spaces? — Martin Sleziak, Nov 20 '21 at 07:52

score 4 · Accepted Answer · answered Oct 10 '13 at 20:55

We can write $U_1+U_2+\cdots$ as the sum of two subspaces, for example $U_1+(U_2+U_3+\cdots)$, and then: \begin{equation}\dim(U_1+U_2+\cdots)=\dim(U_1)+\dim(U_2+U_3+\cdots)-\dim(U_1\cap(U_2+U_3+\cdots)).\end{equation} So then applying the same argument to the term $\dim(U_2+U_3+\cdots)$ recursively you can then eventually arrive at an equation consisting of the sum of the dimensions of each subspace minus a bunch of nasty terms involving intersections similar to $\dim(U_1\cap(U_2+U_3+\cdots))$.

The dimension of the sum of subspaces $(U_1,\ldots,U_n)$

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