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Could intersection of a subspace with its complement be non empty.

Is it possible for a finite dimensional vector space to have 2 disjoint subspaces of the same dimension ? Any help would be much appreciated.

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marked as duplicate by Arturo Magidin, Nate Eldredge, Norbert, Hagen von Eitzen, mixedmath Oct 9 '12 at 17:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

What do you mean by disjoint? Of course it's not as disjoint in the usual sense (for two sets) since a subspace should at least contain $0$. –  Davide Giraudo Feb 8 '12 at 20:16
@Hardy: although I understand that when one starts studying a subject many questions come to mind, I believe it's best that you first put some amount of effort to them before asking everything here. I say that because some of your problems seem to be almost immediate consequences of the definitions themselves. Don't take this personally, though; it's just a small advice. –  student Feb 8 '12 at 20:20
Concerning your question, it seems you've already asked this very same thing in a previous question: math.stackexchange.com/questions/107184/… –  student Feb 8 '12 at 20:29
When it was suggested that you separate the first paragraph of your previous question because only one topic per post is appropriate, it was not to suggest you should take a 2-question post and turn it into three posts, leaving the two questions in a single post, and then repeating each question separately. The suggestion was to remove one of the questions from the first post, and make a second post containing only that question. Voting to close this one. –  Arturo Magidin Feb 8 '12 at 20:33
Hi guys i think the title of the other post is not quite correct any more. I apologise if i duplicated questions somewhat to me they still seemed different enough. For the record, I do try to put in effort to find answers to my questions. –  Hardy Feb 8 '12 at 22:20
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1 Answer

up vote 4 down vote accepted

As David Giraudo points out, any subspace $U\subseteq V$ is going to contain the zero vector (it has to: the subspace is a vector space so is closed under multiplication by elements of the underlying scalar field, and zero is in the scalar field, so multiplying by it tells us the zero vector is in the subspace). In this way, no two vector subspaces are disjoint as sets. In fact, the smallest subspace is trivial, $\{0\}$.

There is a linear notion of "disjoint" here: orthogonal. For example, both a line and a plane through the origin are subspaces, and if they intersect at a right angle they are orthogonal. More generally, two vector subspaces $U$ and $W$ are orthogonal if for every $u\in U, w\in W$, the vectors $u,w$ are perpendicular. The orthogonal complement of a subspace is the maximal orthogonal subspace to it (see Wikipedia). So saying $U,W$ are orthogonal is equivalent both $U\subseteq W^\perp$ and $W\subseteq U^\perp$.

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