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I have a question which is pretty basic about basis and verctor spaces.

Generally, if I have a basis K of vector space V, Why it is not a basis of a subspace W of V (real subspace)?

The vectors in basis K are linearily independent and for every vector of W I can find a linear combination of the vectors in the basis K that equals to the vector in W.

The only problem I can see is that when i make some linear combinations of the vectors in basis K , I get vectors that belong to V but does not belong to the space W, is this the problem here? When I get vectors beyond my space that means it is not the proper basis , because it is spanning more vectors than my space? does it have to span exactly the group and not beyond? Thank you very much.

Sorry for my english and I don't know how to edit the question to be pleasant to read.

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marked as duplicate by Asaf Karagila, Julian Kuelshammer, Start wearing purple, Lord_Farin, user1729 Jul 1 '13 at 11:41

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 definition of "basis" are you using? – Chris Eagle Jul 1 '13 at 10:21
I just can't understand this question: if a set $\,K\,$ is a basis for vector space $\,V\,$, why would it be a basis for a subspace $\,W\le V\;$ ?! – DonAntonio Jul 1 '13 at 10:22
basis of vector space : vectors should be linearily independent and to span the vector space – user84585 Jul 1 '13 at 10:23
Yes, it has to span exactly the space $W$. It also has to consist entirely of vectors that are in $W$. (There’s no need to apologize for your English: the question is quite clear.) – Brian M. Scott Jul 1 '13 at 10:24
That’s correct, though it can be said more simply. $B$ is a basis for $W$ if (1) $B$ is linearly independent, and (2) $W$ is precisely the set of all linear combinations of elements of $B$. Condition (2) implies that $B\subseteq W$. (2) in turn breaks down as you said: each $w\in W$ is a linear combination of the vectors in $B$, and each linear combination of the vectors in $B$ is in $W$. – Brian M. Scott Jul 1 '13 at 10:35
up vote 1 down vote accepted

Yes, that's exactly right. Some set of vectors is a "basis" for V if those vectors are linearly independent and span V. Informally, "spanning" means that V is the smallest vector space that contains all of those vectors; "linearly independent" means that there are no redundant vectors (i.e. if you take one out, the new set of vectors spans a strictly smaller space).

Of course, linearly independent vectors will stay linearly independent regardless of the ambient space you consider them in. But vectors that span a space W will not necessarily span a space V. The ambient space matters. (If they span W, then they cannot span V, unless V = W.)

It might help you to try to work out which subspace of $\mathbb{R}^3$ the vectors $\begin{pmatrix}1\\0\\1\end{pmatrix}, \begin{pmatrix}2\\0\\3\end{pmatrix}$ span. Then show that they are linearly independent, to show that they are a basis of this subspace. Once you've worked that out, try to work out why can't they be a basis of anything smaller, bigger, or different.

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I think I understood. As I mentioned in the comments above, does it mean that: 1. for every vector in W i can find parameters from the field that will set a linear comination with the vectors from the basis, and this combination is equal to the vector from W. 2. every linear combination is a vector, and this vector must be from W? – user84585 Jul 1 '13 at 10:36
That's what "spanning" (Brian's condition (2)) means, yes. A basis must span and be linearly independent. – Billy Jul 1 '13 at 10:37
(For example, $\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix}, \begin{pmatrix}5\\5\end{pmatrix}, \begin{pmatrix}-9\\ \pi\end{pmatrix}$ clearly spans $\mathbb{R}^2$, but it's not a basis, because there are some redundant vectors in there.) – Billy Jul 1 '13 at 10:38

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