# Find elements of Quotient space.

Find two distinct elements of $\mathbb R^3 / W_3$ where $W_3$ = {(-1, 1, 0), (1, 0, 1)}.

Solution: First, span of $W_3$ is (-a+b, a, b) and any element of $\mathbb R^3 / W_3$ is of the form $$(a, b, c) + (-a+b, a, b)$$ $$= (b, a+b, c+b) = a(0, 1, 0)+b(1, 1, 1)+c(0, 0, 1)$$ so, (0, 1, 0), (1, 1, 1) and (0, 0, 1) are the distinct elements of $\mathbb R^3 / W_3$

Have I done it correctly? Or there is some other method to find the elements.

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I think that you misunderstand what the quotient space is. Let $V=\operatorname{span}W_3$; as you said, $V$ is the set of all vectors in $\Bbb R^3$ of the form $\langle b-a,a,b\rangle$, where $a,b\in\Bbb R$. The elements of $\Bbb R^3/V$ are sets of vectors in $\Bbb R^3$: specifically, they are the sets $$\langle x,y,z\rangle+V\;,$$ where $\langle x,y,z\rangle$ ranges over all vectors in $\Bbb R^3$. For example, $\langle 0,1,2\rangle+V$ is one of the vectors in $\Bbb R^3/V$, and $\langle 0,3,4\rangle+V$ is another. But despite appearance, they are not distinct elements of $\Bbb R^3/V$: $$\langle 0,1,2\rangle+V=\langle 0,3,4\rangle+V\;.$$

How can I tell? $\langle 0,1,2\rangle+V$ is the set of all vectors in $\Bbb R^3$ of the form $\langle 0,1,2\rangle+\langle x,y,z\rangle$, where $\langle x,y,z\rangle\in V$. In other words, $\langle 0,1,2\rangle+V$ is the set of all vectors in $\Bbb R^3$ of the form $$\langle 0,1,2\rangle+\langle b-a,a,b\rangle=\langle b-1,a+1,b+2\rangle\tag{1}$$ for $a,b\in\Bbb R$. Is $\langle 0,3,4\rangle$ of the form $(1)$? Yes, with $a=b=2$. Thus, $$\langle 0,3,4\rangle\in\langle 0,1,2\rangle+V\;,$$ and it follows from basic properties of quotients that $$\langle 0,1,2\rangle+V=\langle 0,3,4\rangle+V\;.\tag{2}$$

A basic result that is useful here:

$\qquad\qquad\langle u,v,w\rangle+V=\langle x,y,z\rangle+V$ if and only if $\langle x,y,z\rangle-\langle u,v,w\rangle\in V$.

In my example, for instance, $\langle 0,3,4\rangle-\langle 0,1,2\rangle=\langle 0,2,2\rangle\in V$, so I knew right away that $(2)$ was true.

The problem requires you to find two distinct elements of $\Bbb R^3/V$. To do this, you must find $\langle u,v,w\rangle,\langle x,y,z\rangle\in\Bbb R^3$ such that $\langle u,v,w\rangle+V\ne\langle x,y,z\rangle+V$. This means ensuring that $\langle x,y,z\rangle-\langle u,v,w\rangle\notin V$. One easy way to start is to let $\langle u,v,w\rangle=\langle 0,0,0\rangle$, so that $\langle u,v,w\rangle+V=V$. Now just pick a suitable $\langle x,y,z\rangle$ and let your other element be $\langle x,y,z\rangle+V$.

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Certainly, I misunderstood quotient space. But now, after your explanation I understand the mystery of quotient space. So, Thank you very much. – Faisal Apr 8 '12 at 20:55

I'm not sure what you mean by "the distinct elements of $\mathbb R^3 / W_3$" since there are $\mathbb R$ many elements in that quotient space (it has dimension one).

I'm also not sure what you're doing there. The span of $W_3$ are all vectors that are a linear combination of the two vectors in $W_3$. Quotienting $\mathbb R^3$ by $W_3$ means that you "identify" all vectors in $W_3$ with the zero vector. What this means is that $[v] \in \mathbb R^3 / \operatorname{span}{ W_3}$ is identical to zero if $v \in \mathbb R^3$ is in $W_3$ which means if you can write $v$ as a linear combination of $(-1,1,0)$ and $(1,0,1)$.

To get two distinct vectors $[v_1], [v_2] \in \mathbb R^3 / \operatorname{span}{ W_3}$ you pick two (different) vectors that you can't write as such a linear combination.

For example, let's say $(1,0,0)$ and $(2,0,0)$. Let's verify our claim:

Assume $(1,0,0) = a(-1,1,0) + b(1,0,1)$. It immediately follows that $a=b=0$ from the second and third equation/component. But then $a + b \neq 1$.

Hope this helps.

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No. $(0,1,0)$ and $(1,1,1)$ represent the same element in the quotient, because they differ by an element of $W_3$. Look, $\mathbb R^3$ is 3 dimensional, and $W_3$ is spanned by two linearly independent vectors, so it is 2 dimensional. The quotient will be 3-2=1 dimensional. An element of the quotient is of the form $(c,a,b)$ where $c \neq b-a$. For example $(0,1,0)$ is an element of the quotient. The only distinct elements in the quotient are multiples of your original vector $(0,1,0)$, since the quotient is 1 dimensional. So the elements $(0,c,0)$ are all distinct in the quotient. Any two values of $c$ will do.

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