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I've been having difficulty working on a proof the intersection of two subspaces. This is for linear algebra. Let me give the question and then I'll talk about my attempt:

"The intersection $$M\cap N = \{v\in V: v\in M\,, and, v\in N\}$$ is a vector subspace of $V$ if and only if $M\subset N$ or $M\subset N$."

I do not want to go into the extreme details of the proof, unless needed, since I believe that I understand these concepts well enough. But I believe that this theorem is not true.

In order for this theorem to be true, the statement given above must be biconditional: $$M \cap N \leftrightarrow (M\subset N) \vee (N\subset M)$$ Based on the way I'm looking at it, I believe that if we start with $M \cap N \to (M\subset N) \vee (N\subset M)$, this way is true. If you have an intersection of a subspace, M must contain N and N must contain M since they would be the same where there is an intersection, by the definition of intersection. Like at (0,0).

The other way, however, I don't see as true. Here is the other way: $(M\subset N) \vee (N\subset M) \to M \cap N $. Because the left side of that has an OR, I think that it is not necessarily true. To be an intersection, both of those must be true. Obviously the $\vee$/OR can be true if only one of it's values i.e. $(M\subset N)$ and $(N\subset M)$ is true. Because both parts of the OR part of the statement are not necessarily both True and both need to be true for $M \cap N $ to be a subspace, I do not believe that this direction for the biconditional is true.

Also, if I am correct, I cannot think of a counter example. What 2 subspaces M and N could fulfill the conditions I outlined above? I could not find any, indicating that my logic could be incorrect. Am I completely wrong and the biconditional is true both ways?

Thanks so much!

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    $\begingroup$ Intersection of subspaces is always another subspace. But union of subspaces is a subspace iff one includes another. $\endgroup$ – lEm Oct 30 '16 at 3:27
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    $\begingroup$ The first implication is not correct. Take $V=\mathbb {R^2}$, $M$ the x-axis and $N$ the y-axis. Their intersection is the origin, so it is a subspace. But neither M nor N contain the other subspace. $\endgroup$ – lEm Oct 30 '16 at 3:30
  • $\begingroup$ @John: I would like to point out one thing: Other than the title, the first sentence of your post says intersection of subspace; this is meaningless. One talks of intersection of two subspaces. So this clarity is needed before you can go further. $\endgroup$ – P Vanchinathan Oct 30 '16 at 3:50
  • $\begingroup$ @Bubububu Thanks for the reply. I can definitely see how what I said was inaccurate now. But the only other related question is, if the intersection of two subspaces always contain another subspace, what relevance does M⊂N or N⊂M from the question have? Trying to understand the idea of containment and applying it to the idea of the intersection of subspaces is definitely stumping me. Is it only containment along the area/line of intersection? Based on how I read it at first it didn't seem to be, but I definitely could be wrong. $\endgroup$ – John Oct 30 '16 at 4:40
  • $\begingroup$ @PVanchinathan Apologies for the poor title/confusing wording. Yes it is an intersection of 2 subspaces, M and N, and I fixed the title/wording to reflect that. $\endgroup$ – John Oct 30 '16 at 4:40
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The intersection of two subspaces is always a subspace so your reasoning is wrong. Specifically, if $M \cap N$ is a subspace, then what you can deduce is that $M \cap N$ is a subspace of both $M$ and $N$ but I don't see why you argue that $M$ must contain $N$ or $N$ must contain $N$. For a concrete example, take $V = \mathbb{R}^3$ and $M = \operatorname{span} \{ e_1, e_2 \}$ and $N = \operatorname{span} \{ e_2, e_3 \}$. The set $M \cap N$ is the intersection of the $xy$-plane with the $yz$-plane which is the $y$-axis.

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  • $\begingroup$ That is really insightful and extremely helpful. I can't believe I didn't see that before. My additional question is: how does the "subspace of V if and only if M⊂N or N⊂M" play into your response? I definitely see now that an intersection of the plane indicates what I said is not correct--but does the containment of N and M even play a role, especially if the intersection of two subspaces is always a subspace? $\endgroup$ – John Oct 30 '16 at 4:33
  • $\begingroup$ @John: It this context, the only role it plays is to confuse you :) Namely, you can prove that the intersection of two subspaces is always a subspace. Given that, the statement "The intersection of two subspaces is a subspace if and only there is some containment" is false. The containment plays no role in the question. $\endgroup$ – levap Oct 30 '16 at 4:39
  • $\begingroup$ Wow, that's devious trick! Okay just to make sure I understand: It should be enough to prove that in any case M∩N is a subspace, and then show that because that is a subspace, the subspaces M and N can be given where neither subspace contain each other (such as the example you gave above). Thus, although the intersection of two subspaces is a subspace, because M does not contain N and vice versa (also from Bubububu above, my idea of containment was definitely not accurate before!), the entire "theorem" given must be false. And containment isn't really related to the question. $\endgroup$ – John Oct 30 '16 at 4:54
  • $\begingroup$ @John: Yep, that pretty much sums it up. BTW, if, instead of intersection, you were asked the question about the union of subspaces then it would be a completely different question and in that case, the containment relations would have been relevant to the question. $\endgroup$ – levap Oct 30 '16 at 4:57
  • $\begingroup$ I'll definitely keep that in mind! Thank you so much for the help. $\endgroup$ – John Oct 30 '16 at 5:01

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