Linear Algebra: Finding the basis of a subset Consider $W=\{(v,w,x,y)\ |\ 0=2w-x+y+v\}$ $\subset$ $\mathbb{R}^4$
i) Show that $W$ is a subspace for $\mathbb{R}^4$ 
ii) If $W$ is a subspace, find the basis for $W$
I get how to do problems where we are given a set of vectors then asked to find if they are subspaces, but I'm stumped as to how I approach this question.
To show linear independence, do I have to show that that $(v,w,x,y)$ is not $(0,0,0,0)$?
To show that it spans $\mathbb{R}^4$, do I show that $k*(v,w,x,y) = B$? 
 A: For (i), we know that $W$ is a nonempty subset of a known vector space.  Hence to prove it is a subspace it suffices to prove closure, i.e.


*

*If $(v,w,x,y), (v',w',x',y')$ are both in $W$, then $(v,w,x,y)+(v',w',x',y')$ is in $W$.

*If $(v,w,x,y)$ is in $W$, then for any scalar $k$, $k(v,w,x,y)$ is in $W$.
For (ii) your hint is that $W$ is three-dimensional, so a basis will be a set consisting of three linearly independent vectors from $W$.
A: Ah, so I believe that you might not have the correct definition of a vector subspace.
Start here: A subset $W$ of $\mathbb{R}^n$ is a subspace if:


*

*for every $v, w \in W$, $v+w \in W$.

*for every $v \in W$, $\lambda \cdot v \in W$ for every $\lambda \in \mathbb{R}$.


so start by showing that the $W$ that you have defined above satisfies (1) and (2). This will show that $W$ is a subspace. 
Now to find a basis for $W$, you can do several things. My suggestion would be to think of the following fact:
if $v = (w,x,y,z) \in W$ then it follows that $v\cdot (2,-1,1,1) = 2w-x+y+u = 0$. That is: every $v \in W$ is orthogonal to $(2,-1,1,1) $. Hence, to find a basis for this space, you need to find three linearly independent vectors which are all orthogonal to $(2,-1,1,1) $. I should mention that this doesn't show that the vectors you find actually span $W$. You also need to show this. 
Hopefully this gets you started.
