Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
4  
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
1  
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 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.