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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
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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
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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
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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
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1 Answer 1

up vote 2 down vote accepted

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 $L$.

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Well choosing L as the set of multiples of the vector (1,1) would show the result to be true. As we get vectors along with the axis on both sides. So obviously that could n't be the choice for our L. Any other hints ? –  Hardy Feb 8 '12 at 18:58
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@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
    
Firstly thank you for your reply. I was interpreting M + N as that too, but Should n't L ∩ M be all the vectors along the x-axis and similar argument for the y-axis one, although i agree i had stuffed up in interpreting M + N as all of R^2 myself. I was interpreting M + N to be just the vectors along the axis which i am unsure about now. Your last reply seems much more feasible. I am a newbie. –  Hardy Feb 8 '12 at 19:59
    
@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
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