# Quotient Spaces and Hyperplanes

1. Mar 17, 2014

### Chacabucogod

Hi,

I'm currently reading Shilov's Linear Algebra and he mentions that Hyperplanes are planes that don't pass through the origin. Wouldn't that be a quotient space?

Thank you.

2. Mar 17, 2014

### micromass

Why would a hyperplane be a quotient space?

A hyperplane is for example the line $2x + 3y = 5$ in $\mathbb{R}^2$.

3. Mar 17, 2014

### Chacabucogod

As far as I understand a quotient space is subspace plus a vector that is in the whole space. Isn't that the definition?

4. Mar 17, 2014

### micromass

Not at all.

5. Mar 17, 2014

### Chacabucogod

Well then, the definition I just gave you; is that the definition of the subset v+W?

6. Mar 17, 2014

### Chacabucogod

V mod W is the set of vectors defined by the property that if you substract one from the other the result is in the subspace then?

7. Mar 17, 2014

### micromass

Yes. The quotient space is exactly the set of all such $v+W$ for a given $W$.

So if $W$ is any subspace of $V$, then

$$V/W = \{v+W~\vert~v\in V\}$$

So one could say that the quotient space is the set of all hyperplanes parallel $W$ (I count $W$ itself as a hyperplane, although the OP says it isn't).

8. Mar 17, 2014

### Chacabucogod

Wouldn't V/W make up all the space except for the original subspace?

9. Mar 17, 2014

### micromass

The original subspace is also an element of $V/W$. In fact, $V/W$ is a vector space and $W$ is its zero element.

10. Mar 17, 2014

### Chacabucogod

So in R^3 A quotient space made up by a line would be a plane right?

11. Mar 17, 2014

### micromass

I wouldn't say that. The quotient space would be a $2$-dimensional vector space, and that would isomorphic to a plane. I wouldn't say it actually equals a plane in $\mathbb{R}^3$. In particular, the quotient space won't even be a subset of $\mathbb{R}^3$.

12. Mar 17, 2014

### Chacabucogod

Ok if it isn't a subset of R^3, what is it? What are quotient spaces useful for? What about the quotient space of the plane z=0 in R3; what would that make? Thank you for taking your time to answer my questions by the way.

13. Mar 17, 2014

### micromass

It's just an entirely new vector space. It's the collection of all hyperplanes parallel to a given subspace. It can't be seen as the subspace of something else, it's just something entirely new.

In introductory linear algebra, they are actually quite useless. It is only when you study abstract algebra that quotient spaces become useful. The idea is roughly the following. Consider a vector space $V$ and a subspace $W$. We can form the quotient space $V/W$. The idea is that both the subspace $W$ as $V/W$ are easier to handle than $V$ because they have lower dimension. However, if we "know" both $W$ and $V/W$ then we actually also "know" $V$.

I know this is very vague, but it is quite difficult to give a decent motivation of quotients at this level. Really, you need to see some applications of them before you can really appreciate them.

The intuition behind the quotient space is that you "set $W=0$", meaning that all elements in $W$ become $0$. So you let entire $W$ collapse to $0$.

You are expecting some simple and intuitive answer, but there isn't one. The quotient space is something very abstract, and you need to get used to it. The quotient space of the plane is just the set of all planes parallel to the plane given by the equation $z=0$. So any plane with equation $z=k$ is an element of the quotient space. So we see that the quotient space is isomorphic to $\mathbb{R}$.

14. Mar 18, 2014

### Chacabucogod

Micromass, one more question. What is the the dimension of that new space K/L. For example if we make the K/L space of a line that goes through the x-axis, it would make a 2 dimensional space. Am I right? What would we make out if the K/L space of z=0?What would be its dimension?

15. Mar 18, 2014

### micromass

The dimension of $V/W$ is $\textrm{dim}(V) - \textrm{dim}(W)$. If $V$ is finite-dimensional, at least.

16. Mar 19, 2014

### mathwonk

Although the quotient space is in general not a subspace, nor naturally equivalent to one, it may be of interest that can be viewed as one in the special examples under discussion since post #10, because of the presence of a natural inner product on R^n. I.e. there is a natural way to choose a second subspace orthogonal to the given one, and that second subspace serves as a natural isomorphic model of the quotient space by the given subspace.

E.g. in the case of the quotient of R^3 by the (two dimensional) subspace z=0, the one dimensional subspace consisting of the z axis is a natural model for the quotient space, since it contains exactly one element of each of the hyperplanes in the quotient space.

The elements of the quotient space are equivalence classes of vectors in the original space, and the question of regarding the quotient space as a subspace, is the one of choosing a natural representative of each class. In the abstract setting this is not possible, but in R^n it is.

Nonetheless it may not be wise to do so, since the elements of the quotient space may be more naturally viewed as equivalence classes for understanding the problem in which they arise.