# Extending the Intersection of Subspace

For two subspace, one can express the dimension of the sum as $$\dim(U_1 + U_2) = \dim U_1 + \dim U_2 - \dim (U_1 \cap U_2).$$

However, the obvious extension to three subspacess fails, in the sense that $$\dim (U_1 + U_2 + U_3) \neq \dim U_1 + \dim U_2 + \dim U_3 - \dim (U_1 \cap U_2) - \dim (U_1 \cap U_3) \\ - \dim (U_2 \cap U_3) + \dim (U_1 \cap U_2 \cap U_3).$$

I showed the above with a counter example but can't seem to understand it intuitively. Why does it fail? In general, is the "obvious extension" true for $\dim(U_1+....+U_n)$ when $n$ is even and false when $n$ is odd? Is it just that two a special case?

• @user123429: I suspect it may be true if the number of subspaces is less-than or equal-to the dimension of the ambient space. – user99680 May 9 '14 at 2:45
• okay deleting my answer since the original poster is more interested in the small dimensional cases. Can you post the counter example if you're still interested in this problem. – N8tron May 9 '14 at 17:29
• Let $U_1 ={(x, 0)}, U_2 = {(0, y)}, U_3 = {(z,z)}$. They are all obviousy 1 dimensional because they are spanned by (1, 0), (0, 1) and (1, 1) respectively. $U_1 + U_2 + U_3$ has dimension 2 since any $vector (x, y) \in R$ can be written Span {(x,0), (0, y)}. However note the intersection of any two is only the zero subspace, as is the intersection of all three. Thus using the formula given in the question you would get 2 = 3. – user123429 May 9 '14 at 20:02

Using the original formula we get this: $$\dim(U_1 + U_2 + U_3) = \dim(U_1+U_2) + \dim(U_3) - \dim((U_1+U_2)\cap U_3)\\ = \dim(U_1) + \dim(U_2) - \dim(U_1 \cap U_2) + \dim(U_3) - \dim( (U_1 + U_2) \cap U_3)\\ = \dim(U_1) + \dim(U_2) + \dim(U_3) - \dim(U_1 \cap U_2) - \dim( (U_1+U_2) \cap U_3)$$
The issue is how the "+" operation interacts with the "$\cap$" operation for subspaces, which is not very obvious.
The "obvious extension" more or less assumes that $(U_1 + U_2)\cap U_3 = (U_1 \cap U_3) + (U_2 \cap U_3)$, and this is not true in general. Now it is true that, $(U_1 + U_2)\cap U_3$ contains $(U_1 \cap U_3) + (U_2 \cap U_3)$, so if your formula doesn't work it will overestimate the dimension.
2 is a special case. The counterexample you have for 3 works for any $n\ge3$; take $n$ distinct one-dimensional subspaces of ${\bf R}^2$. Their sum is ${\bf R}^2$, which has dimension 2; the sum of the dimensions is $n$; the intersections all have dimension zero, so the "equation" gives $2=n$.