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If that is possible could you please correct my understanding about complement of a subspace.

From what i recall from set theory. A complement of a set B is the set U - B where U is the universal set. So B and U - B are disjoint, but in case of subspaces of a finite dimensional vector space all subspaces of the same dimension have origin in them, hence the intersection of such a subspace with it's complement is n't disjoint. Is n't this contradictory ?

Any help would be much appreciated.

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By a complement of a subspace $B \subseteq U$ one usually means another subspace $A \subseteq U$ such that $A \oplus B = U$. There are many such complements in general for a given $B$. –  student Feb 8 '12 at 19:59
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What you say isn't incorrect - it just means that the set-theoretic complement of a subspace is never a subspace. So set-theoretic complements in linear algebra are uninteresting, and instead we have the definition mentioned by Leandro above. –  Matt Feb 8 '12 at 20:05
    
Thanks for that guys, any answers for the other 2 questions would also be much appreciated, it would help clear my understanding. –  Hardy Feb 8 '12 at 20:08
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Most of the question seems to revolve around the misunderstanding of "complement" in the set-theoretic instead of the vector-space sense, which Leandro cleared up; but what does the part about "vector spaces over different fields" have to do with that? Or is that a separate question? If so, please ask it separately; the site works best if there's one question per post. –  joriki Feb 8 '12 at 20:10
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Since you have separated the first paragraph into its own separate question, perhaps you should remove it from this one. –  Arturo Magidin Feb 8 '12 at 20:27

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The word "complement" has different meanings depending on the context. In Set Theory, we often talk about "[relative] complement", which is the concept you describe. As you note, a set and its (relative) complement are disjoint. Moreover, the relative complement is unique: there is one and only one set that qualifies as the relative complement of $A$.

In linear algebra, though, "complement of a subspace" has a different meaning. We say that a subspace $\mathbf{Z}$ of the vector space $\mathbf{V}$ is a complement of the subspace $\mathbf{W}$ if and only if (i) $\mathbf{V}=\mathbf{W}+\mathbf{Z}$; and (ii) $\mathbf{W}\cap\mathbf{Z}=\{\mathbf{0}\}$. In general, there are many different possible complements, and none are disjoint from $\mathbf{W}$ (however, since $\{\mathbf{0}\}$ is the smallest that the intersection of two subspaces can be, we usually say the intersection is "trivial"). For example, if $\mathbf{V}=\mathbb{R}^2$ (as a real vector space) and $\mathbf{W}$ is the $x$-axis, then any line through the origin except the $x$-axis is a complement of the $x$-axis.

There is even a further concept of "complement" in linear algebra, when we have a notion of "inner product"; then we talk about the orthogonal complement of a set/subspace: if $\mathbf{V}$ is a vector space with inner product $\langle \cdot,\cdot\rangle$, then the orthogonal complement of a set $S$ is the set of all vectors that are orthogonal to $S$, and is denoted $S^{\perp}$; in general, $S\cap S^{\perp}\subseteq \{\mathbf{0}\}$.

So, in summary: the same word ("complement") has different meanings depending on context; you are trying to apply one meaning (set-theoretic) in the wrong context (linear algebra). You should look for the linear meaning instead. If you think it's bad to have the same word mean different things... well, you are right, but since the contexts are so different, it usually doesn't matter. (If you think that's bad, just wait until you run into the term "normal"...)

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Thanks sorry i just created 2 new questions splitting this one up as i was suggested. –  Hardy Feb 8 '12 at 20:15
    
@Hardy: Please delete the first paragraph from this question, then. Thank you. –  Arturo Magidin Feb 8 '12 at 20:17
    
@Hardy: You should not have repeated this question in a new one; the suggestion was to create a total of two posts, to cover your two questions. You've now created a total of three posts to cover two questions: the two new posts, and then this post with the two questions again. –  Arturo Magidin Feb 8 '12 at 20:34
    
Sorry for not responding earlier i was commuting. I asked 3 questions in total now as per suggestion i have split them into 3 seperate questions. –  Hardy Feb 8 '12 at 21:47
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@Hardy: No, not at all. For example, in $\mathbb{R}^2$, subspaces of dimension 1 correspond to lines through the origin. There are uncountably many different subspaces of dimension $1$, and any two have only the origin in common; there are uncountably many subspaces of dimension $2$, and any two distinct ones have only a line in common. The only dimensions for which there is only one subspace of that dimension are dimension $0$, and dimension $n=\dim(\mathbf{V})$. Two subspaces of the same dimension are isomorphic, but they need not be the same. –  Arturo Magidin Feb 9 '12 at 2:18

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