Question of Linear Independence of V/W So here is the question I have:
Let $V=\mathbb{R}^4$ and $W=\operatorname{span}\{[0,1,0,2]\}$.  Determine whether the set $S=\{[1,1,-2,0]+W,[1,0,-1,0]+W,[0,0,1,2]+W\}$ is linearly independent in $V/W$ and prove that your answer is correct.
I have taken the set S and multiplied each vector by a scalar, i.e.
$a[1,1,-2,0]+b[1,0,-1,0]+c[0,0,1,2]=[0,0,0,0]$
and have determined that $a=b=c=0$, so it is linearly independent. I'm not sure whether this is enough though and the $+W$ in $S$ is troubling me.  Thanks in advance.
 A: That proves that $[1,1,-2,0]$, $[1,0,-1,0]$ and $[0,0,1,2]$ are linearly independent in $\mathbb{R}^4$, but not necessarily in $V/W$.
To prove independence in $V/W$ you need to remember that the $0$ element of $V/W$ is $[0,0,0,0]+W$, which is really an equivalence class, consisting of all the vectors in $W$. So independence fails if there are $a,b,c$ such that:$$a([1,1,-2,0]+W)+b([1,0,-1,0]+W)+c([0,0,1,2]+W)=[0,0,0,0]+W$$
or equivalently, such that $a[1,1,-2,0]+b[1,0,-1,0]+c[0,0,1,2]$ is in $W$, so is some multiple of $[0,1,0,2]$.
You should try again with this in mind, and I can expand this answer if you still get stuck.
A: Let $v_1 = [1,1,-2,0]$, $v_2 = [1,0,-1,0]$, $v_3 = [0,0,1,2]$, $w = [0,1,0,2]$.
Suppose $a_1v_1 + a_2v_2 + a_3v_3 \in W$.
Then $a_1v_1 + a_2v_2 + a_3v_3 = bw$ for some $b$.
We claim that $a_1 = a_2 = a_3 = b = 0$.
Let $A = \left( \begin{array}{ccc}
1 & 1 & -2 & 0\\
1 & 0 & -1 & 0\\
0 & 0 & 1 & 2\\
0 & 1 & 0 & 2\\
\end{array} \right)$
It's easy to see that det $A = 1$.
Hence $v_1, v_2, v_3, w$ are linearly independent.
Hence $a_1 = a_2 = a_3 = b = 0$ as claimed.
Hence $S$ is linearly independent in $V/W$.
