# vector space and its subspace

The definition of subspace from the Friedberg book :

A subset $W$ of a vector space $V$ over field $F$ is called a subspace of $V$ if $W$ is a vector space over $F$ under the operations of addition and scalar multiplication defined on $V$.

say our Field is $\Re$, and let $V$ consists of vectors "$a_n(i)$"(a vector with only one dimension is considered for simple explanation ) where $a, n \in N$ , if $W$ has to be a subspace of $V$, then $W$ has to be a subset of $a_n(i)$, i.e. $(i, 2i, 3i,...)$, now if consider $i$ and $2i$ to be forming $W$, then because of the addition property $3i$ has to be there in the set of $W$, if $3i$ is there then $4i$ has to be there in $W$ because of addition property, this goes on and we have to exhaust the original vector space $V$, so what is the subspace $W$? and also the example that I have considered here, does $V$ satisfy the addition and scalar multiplication property to be a vector space? so I have two questions:

1-Is the example considered here is a valid vector space and if yes, then

2- what can be its subspace?

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It's really hard to understand your notation. Subscripts are completely unnecessary for describing a generic element of a vector space, unless you are taking pains to describe an element with respect to a basis. –  rschwieb Aug 3 '12 at 12:04
I don't understand what your set $V$ looks like, but it sounds like it is not a vector space. –  Jyrki Lahtonen Aug 3 '12 at 12:11
It is not clear enough to be sure, but it looks as if you are saying that a $1$-dimensional space $V$ will not have many subspaces. And that is true. A $1$-dimensional space has two subspaces: (i) the subspace consisting of the $0$-vector, and that's all and (ii) $V$ itself. But any space (over the reals) of dimension $\ge 2$ has infinitely many subspaces. –  André Nicolas Aug 3 '12 at 12:21

## 1 Answer

I hope to add more after the question is clarified a little, but here is a start:

If you have a vector space $V$, there will often be lots of subspaces.

For instance, if $v\in V$ is any nonzero element, then $\langle v \rangle=\{\alpha v\mid f \in \mathbb{F}\}$ will be a subspace of $V$, since it's clearly closed under addition and scalar multiplication. It's a one dimensional subspace, so any nonzero vector will generate this little space. If your field is of characteristic $0$, then it will also contain $2v,3v,4v,\dots nv$ because $2,3,\dots n$ are elements of the field $\mathbb{F}$. However there are many more scalars than the integers...

If you pick $w\notin \langle v\rangle$, then $\langle w \rangle$ will produce another subspace of its own, but it won't share any elements with $\langle v \rangle$ except for $0$.

If you take the $v$ and $w$ we have chosen, you can also find another subspace $\langle v,w \rangle=\{\alpha v +\beta w\mid \alpha,\beta\in \mathbb{F}\}$, which is a two dimensional subspace.

It also may be possible to find another $z\in V$ such that $\langle v,z\rangle$, but it is not equal to $\langle v,w\rangle$.

Hopefully you can see how this works for even larger collections of vectors.

If this topic is very new to you, I would highly recommend getting your head around the concept of linear independence as early as possible.

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i need to get these things cleared, but thanx for the explanation –  Vikram Aug 4 '12 at 11:31
Community user is aware of the crusade and bumped this post...+1 –  Shuhao Cao Jul 27 '13 at 4:47