# Since $U$ and $W$ are distinct, shouldn't $U \cap W = 0$?

Suppose $U$ and $W$ are distinct four-dimensional subspaces of a vector space $V$, where $\dim V = 6$. Find the possible dimension of $U \cap W$.

Since $U$ and $W$ are distinct, shouldn't $U \cap W = 0$?

• Two or three. Any other questions? :) – heptagon Aug 19 '16 at 18:35
• I would have said that "distinct" just meant that they were not the same subspaces. Doesn't preclude non-trivial intersection. – lulu Aug 19 '16 at 18:35
• Think of the planes $z=0$ and $x=0$ in $\mathbb{R}^3$. Each is a two dimensional subspace of $\mathbb{R}^3$ yet their intersection is the $y$-axis. – John Wayland Bales Aug 19 '16 at 18:37

## 2 Answers

I think you're mistaking the word "distinct" for the word "disjoint".

The word "disjoint" means $U\cap W=\varnothing$, i.e. there are no elements in common between $U$ and $W$. Actually, it is impossible for two subspaces of a vector space to be disjoint, since they always must both contain the $0$ element. Thus $0\in U\cap W$ for any two subspaces $U$ and $W$. However, if the intersection of $U$ and $W$ is only the zero element and nothing else, this may sometimes be expressed in symbols as $U\cap W=0$.

The word "distinct" just means $U\neq W$, i.e. they aren't identical, but they can still have elements in common.

• I do not see anything that tells user362453 uses the word 'distinct' mistakenly. His problem seems to be well posed and makes sense. Anyway, I do not see how does your answer actually answer the question. – heptagon Aug 20 '16 at 15:33
• I think this answer is sensible. – user228113 Aug 20 '16 at 15:53

By Grassmann formula for two subspaces $U$ and $W$ of a vector space $V$: $$\dim (U\cap W) = \dim U+\dim W-\dim(U+W)$$ Here, $\dim U=\dim W = 4$ and $4\leq \dim(U+W)\leq 6$ since $W\subseteq U+W\subseteq V$. Thus $-6\leq -\dim(U+W)\leq -4$ and we obtain $$2 = 8-6 \leq \dim (U\cap W) = 8-\dim(U+W) \leq 8-4 = 4.$$

Finally, $\dim(U\cap W)=4$ means that $U\cap W=U=W$ (for dimension reasons), hence $\dim(U\cap W) = 2$ or $3$. Thus, $U\cap V$ can't be $0$.

In fact, distinct subspaces are simply different (i.e. there's a vector belonging to only one of them), but they can have a non-zero intersection.