Dimension of the sum of three subspaces

Question

We know that $$\dim(U_1 + U_2) = \dim U_1 + \dim U_2 - \dim(U_1 \cap U_2)$$ if $U_1$ and $U_2$ are finite dimensional subspaces.

For three finite dimensional subspaces prove or give a counterexample for the following:

$$ \begin{align} \dim(U_1 + U_2 + U_3) &= \dim U_1 + \dim U_2 + \dim U_3 \\ &- \dim(U_1 \cap U_2) - \dim(U_2 \cap U_3) - \dim(U_1 \cap U_2)\\ &+ \dim(U_1 \cap U_2 \cap U_3) \end{align}$$

Its basically the formula for the union of three sets. I think that this is false but the only reason I can think it would be is because the union of subspaces is rarely a subspace itself, even though it applies to sets, so it seems to me like this formula which works for sets shouldn't work for subsets because it would disrupt overlapping elements of $U_1 + U_2 + U_3$ and run into complications with the sum being closed under addition or something like that. Help much appreciated thank you!

For two subspaces $U_1$ and $U_2$, the subset $U_1+U_2$ is defined to be the subspace generated by sums $u_1+u_2$ for $u_1\in U_1$ and $u_2\in U_2$. So, $U_1+U_2$ is not the union. — J126, Jul 23 '14 at 02:31
You might want to take a look at this popular mathoverflow post: Examples of Common False Beliefs in Mathematics. — Peter Woolfitt, Jul 23 '14 at 02:32
Nice question. I would try to adapt the proof for two subspaces using basis. While at it, it might be nice to try to generalize it. — Ivo Terek, Jul 23 '14 at 02:37
Ah thank you for the link, if it's a common question would it be good to just delete this question if its redundant? — Soaps, Jul 23 '14 at 02:51
@Soaps According to this post on the Math.SE meta it is acceptable to post answers from mathoverflow to the same question on Math.SE - therefore this is what I have done. — Peter Woolfitt, Jul 23 '14 at 03:19

score 8 · Answer 1 · edited Apr 13 '17 at 12:58

8

This is the counterexample given by Willie Wong on a mathoverflow post with this question:

Consider $3$ distinct lines through the origin in $\mathbb{R^2}$ as $U_1$, $U_2$, and $U_3$.

Then the left hand side of the equation becomes $$\mathrm{Dim}(U_1+U_2+U_3)=\mathrm{Dim}(\mathbb{R^2})=2,$$ while the right hand side of the equation becomes

$$\mathrm{Dim}(U_1)+\mathrm{Dim}(U_2)+\mathrm{Dim}(U_3)-\mathrm{Dim}(U_1\cap U_2)-\mathrm{Dim}(U_1\cap U_3)-\mathrm{Dim}(U_2\cap U_3)+\mathrm{Dim}(U_1\cap U_2\cap U_3)\\=3\mathrm{Dim}(\mathbb{R})-3\mathrm{Dim}(\{0\})+\mathrm{Dim}(\{0\})=3,$$

and as $2\ne3$ this equation does not hold.

edited Apr 13 '17 at 12:58

Community

1

answered Jul 23 '14 at 03:15

Peter Woolfitt

21,114

i am not getting how $Dim(U_{1}+U_{2}+U_{3})=2$? – TheStudent Jun 15 '21 at 04:47
2

@honeykumar $U_1+U_2+U_3$ is the subspace consisting of all elements $u_1+u_2+u_3$ where $u_1\in U_1$, $u_2\in U_2$, and $u_3\in U_3$. Since $U_1$, and $U_2$ are distinct lines in $\mathbb{R}^2$, any element $\mathbb{R}^2$ can be expressed as a sum of $2$ elements, one from each line (think projection). Hence $U_1+U_2=\mathbb{R}^2$, and since $U_3$ is also in $\mathbb{R}^2$, we have $U_3+\mathbb{R}^2=\mathbb{R}^2$, so in conclusion $U_1+U_2+U_3=\mathbb{R}^2$ which has dimension 2. – Peter Woolfitt Jun 19 '21 at 01:33

score 3 · Answer 2 · edited Dec 27 '22 at 15:08

3

That is not true. But, for 3 linearly independent spaces, it is true,and can be generalized by the inclusion–exclusion principle, to any finite number of spaces.

edited Dec 27 '22 at 15:08

J. W. Tanner

60,406

answered Jul 23 '14 at 02:58

user160823

690

Dimension of the sum of three subspaces

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