Let $V=\mathbb{R}^3$ and $W=\{ (x,y,z): x+y+z=0\}$ Describe $V/W$ geometrically and contrsuct an explicit isomorphism $W^\perp \cong V/W$

Question

For an isomorphism I let $\phi:W^\perp\to V/W$ be defined as $\phi(x)=[x]$

To show $\phi$ is injective suppose $x\neq y$ for $x,y \in W^\perp$.

Since $W=\{ (x,y,z): x+y+z=0\}$, $(1,1,1)$ is the vector normal to the plane $W$.

And so $x=a(1,1,1)$ and $y=b(1,1,1)$ where $a\neq b$.

Then $x-y=(a-b,a-b,a-b)$ and $a-b+a-b+a-b=3(a-b)\neq 0$

so $x-y \not\in W$ and $[x]\neq [y]$. Thus $\phi$ is injective.

I'm not sure how to show it's onto.

To show $\phi$ is onto, let $[y] \in V/W$

then maybe I have that $y=w+v$ for some vector $w\in W$ and $v\in V$. So that $\phi (w+v)=[w+v]=[y]$?

I think describing V/W geometrically would mean showing $V/W=[span(1,1,1)]$

Answer 1

$W^\perp$ is one dimensional since it's isomorphic to a one dimensional space, namely, $span\{(1,1,1)\}$.

Since $dim(W) = 2$, we have that $dim(V/W) = 1$

You have an injective map, $\phi$, between two one dimensional spaces, $W^\perp $ and $V/W$

Rank nullity says the the dimension of the image of $\phi$ is one and hence is all of $V/W$

Answer 2

As Jaime pointed out, surjectivity is free via rank-nullity. If you want a more elementary proof, here it is: Let $[x]\in\mathbb{R}^3/W$. If $y\in W$ then $[x+y]=[x]$. Hence, we need only show that $W^\perp+W=\mathbb{R}^3$. Indeed, suppose $(a,b,c)\in\mathbb{R}^3$. Set $d:=\frac{a+b+c}{3}$, and observe that $(d,d,d)+(a-d,b-d,c-d)=(a,b,c)$ with $a-d+b-d+c-d=0$.

As for the geometry, well, there is no natural geometric description of $\mathbb{R}^3/W$, as it does not sit in any natural geometric space. However since $\mathbb{R}^3/W$ is isomorphic to $W^\perp$, and $W^\perp$ is just $\mathbb{R}(1,1,1)$, that is as good as a geometric description.

Answer 3

Geometrically, you can think of W as the plane through the origin with normal vector (1,1,1). You can think of the quotient space as translates of this plane along the normal vector. So you have this infinite family of parallel planes indexed by how far it's translated from W, which will be a scalar multiple of (1,1,1). Thats why it's 1-dimensional.