# Counter example for a result of intersection of subspaces

I am struggling with this question from Halmos's text, please ignore the imperative language.

"Suppose that $L, M$ and $N$ are subspaces of a vector space. Show that the equation

$$L \cap (M + N) = (L \cap M) + (L \cap N)$$

is not necessarily true.

Since each of these subspaces has origin in them, clearly there intersections could not be empty. I wasn't able to formulate an example where this result did n't hold. Any help would be highly appreciated.

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What kind of examples have you tried? –  Qiaochu Yuan Feb 8 '12 at 18:31
As a first step, try to show that one side is a subspace of the other. What goes wrong when you try to prove the reverse inclusion? –  Aaron Feb 8 '12 at 18:32
Hint: If $M$ is the whole space, this equation becomes $L = L + (L \cap N)$. Can you finish it from here? –  student Feb 8 '12 at 18:35
This was discussed here: mathoverflow.net/questions/17740/… –  mt_ Feb 8 '12 at 18:36
@Leandro Could i do subtract L from both sides and come to the conclusion (L∩N) = a null set which is obviously not true as the origin belongs to all subspaces and since intersection of two subspaces is also a subspace. Hence we have a contradiction if this result was true. –  Hardy Feb 8 '12 at 18:49
Work for example in $\mathbb{R}^2$.
Let $M$ be the set of multiples of the vector $(1,0)$, let $N$ be the set of multiples of $(0,1)$. It's your turn, you can choose $L$.
@Hardy: Maybe I am misinterpreting $+$. My interpretation is that it is the space generated by $M$ and $N$. With that interpretation, and $M$ and $N$ as in my answer, and $L$ as you mention, we have that $M+N$ is all of $\mathbb{R}^2$. So the left-hand side is just $L$. Look now at the right-hand side. We have $L\cap M$ contains only the $0$ vector, and the same is true of $L\cap N$. So the sum of the two also only contains the $0$ vector. Thus left in this case is $L$, right is $\{0\}$, not equal. If I am misinterpreting the meaning of $+$, please tell me. –  André Nicolas Feb 8 '12 at 19:46
@Hardy: Definitely $L\cap M$ just contains the $0$ vector. Think of it geometrically. You can think of $L$ as all points on the line $y=x$, and $M$ as the points on the $x$-axis. The only point they have in common is the origin. And I checked the standard usage of $U+V$, when both are subspaces $W$. It is the set of all points of $W$ of the shape $u+v$, where $u\in U$ and $v\in V$. –  André Nicolas Feb 8 '12 at 21:03