# 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. May 9, 2014 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. May 9, 2014 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. May 9, 2014 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 $2=n$.