# Cosets and Vector Spaces Question

1. Feb 29, 2012

### Master J

In studying vector spaces, I came across the coset of a vector space.

We have an equivalence relation defined as

u = v $\rightarrow$ u-v $\in$ W

where W is a subspace of V.

the equivalence class that u belongs to is u + W. I can see why u must belong to this equivalence class ( the coset) because of reflexivity, but why W?

Is u - W $\in$ W ?

2. Feb 29, 2012

### Office_Shredder

Staff Emeritus
In general W will not be a subset of u+W. In fact W is contained in u+W if and only if u∈W.

Note that u - W ∈ W doesn't even make sense, because u-W is a set, and you're asking if it's an element of W (on a set theoretic level this does make sense, but in the context of vector spaces, u-W is clearly not a vector)

3. Feb 29, 2012

### micromass

What do you mean??

Do you want to know if $W\in a+W$?? The answer is that this makes no sense. W is a set and can be an element of a+W.

Do you want to now if $W\subseteq a+W$?? In general this is not true unless a is already in W.

4. Feb 29, 2012

### Master J

Perhaps I should have clarified...I meant that W would stand for any element of W.

If the coset of the equivalence relation is u + W, this means as I understand it, that the equivalence relation for u only holds for u itself ( u - u = 0 $\in$ W), and for any element of W. Is that correct?

5. Feb 29, 2012

### Office_Shredder

Staff Emeritus
No. If v is equivalent to u then v-u∈W. Note that if u is NOT contained in W, then u-w is NOT in W for any w∈W - if u-w=v for some v∈W, then u=w+v∈W since v,w∈W.

It probably helps to look at some examples. Let our vector space be R2 and W be the subspace of all points of the form (x,x). Now consider u=(1,0). u+W is the set of all vectors of the form (x+1,x) - literally everything of the form (1,0)+(x,x). So (2,1), (3,2), (4,3) are all equivalent to u. (1,1) is NOT equivalent to u

6. Mar 1, 2012

### Deveno

i tend to think of it like this:

suppose we have a plane, which we will take to be R2.

now suppose we have a proper subspace of R2, which is of the form:

L = {a(x,y) : a in R}.

this is a line going through the origin and the point (x,y).

a coset is thus a set v + L, which is a line parallel to L passing through the point v.

we thus get a 1-dimensional space whose "vectors" (elements) are all parallel lines. to "add 2 lines" u+L and v+L, we take the line parallel to L passing through u+v, or (u+v) + L.

in 3 dimensions, a quotient by a plane, yields a "1-dimensional stack of planes", and a quotient by a line yields a "2-dimensional bundle of parallel lines" (in 3 dimensions we need two vectors to tell us "which line we're on", since one vector just gives us "a line of lines", like pencils aligned to make a fence).

of course, one can't visualize higher dimensions spatially, but the same idea is going on:

if dim(V) = n, and dim(W) = k, then dim(V/W) = n-k (we use k dimensions to create W, and we need the other n-k dimensions to locate which copy of W we're in).

V/W is V, chopped up into "W-sized pieces".

to answer your original question: all the elements of W live in W = 0+W, the 0-vector of V/W (W's "home base").

7. Mar 1, 2012

### Jim Kata

The simple answer is that u-w is not an element of W unless u is an element of W.

8. Mar 1, 2012

### Deveno

indeed, u-w is in the same coset as u+w, since:

(u+w) - (u-w) = 2w ∈ W.

9. Mar 1, 2012

### Master J

Thanks for the input people, you have cleared up a lot!

The notion of cosets is quite confusing, at least to me. They've made their appearance in a chapter on Vector Spaces and I haven't seen them before.

Another minor detail I have come across is this (perhaps my set theory is lacking!):

if 2 cosets are equal, u + W = v + W, where u & v $\in$ U, then u - v $\in$ U $\bigcap$ W.

How is this the case? Is it simply set theory?

10. Mar 2, 2012

### Deveno

suppose u + W = v + W.

then (u - v) + W = (v - v) + W = 0 + W = W (i simply subtracted v + W from "both sides", using the fact that -(v + W) = (-1)(v + W) = (-1)v + W = -v + W).

by definition of a subspace, u - v is also in U, if both u,v are (subspaces are closed under vector addtion and scalar multiplication, so -v = (-1)v is in U when v is, and thus u - v = u + (-v) is in U, since both u and -v are in U).

so u - v is in U, and u - v is in W, and thus is in the set of all elements which are in both sets, which we call U∩W.

thus u - v

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