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!
Best Answer
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.