# basis of space and subspace [duplicate]

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.

-

## marked as duplicate by Asaf Karagila, Julian Kuelshammer, L.G., Lord_Farin, user1729Jul 1 '13 at 11:41

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

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.
(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