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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.

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No, it is not enough in general: you have to show that the linear combination you formed doesn't belong to $\,W\,$... –  DonAntonio Sep 11 '12 at 16:25
    
Ok. So I would have to show that something like $a[1,1,−2,0]+b[1,0,−1,0]+c[0,0,1,2] = d[0,1,0,2]$. –  tk2 Sep 11 '12 at 16:29
    
Yup, exactly that. –  DonAntonio Sep 11 '12 at 16:31
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2 Answers 2

up vote 2 down vote accepted

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.

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Oops, seems @DonAntonio gave the same hint in the comments before I did. Probably better that this question has an answer though, and it can be edited if the hint is insufficient. –  Matt Pressland Sep 11 '12 at 16:38
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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 $V/W$.

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