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Cosets and Vector Spaces Question

  1. Feb 29, 2012 #1
    In studying vector spaces, I came across the coset of a vector space.

    We have an equivalence relation defined as

    u = v [itex]\rightarrow[/itex] u-v [itex]\in[/itex] 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 [itex]\in[/itex] W ?
  2. jcsd
  3. Feb 29, 2012 #2


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    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)
  4. Feb 29, 2012 #3
    What do you mean??

    Do you want to know if [itex]W\in a+W[/itex]?? 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 [itex]W\subseteq a+W[/itex]?? In general this is not true unless a is already in W.
  5. Feb 29, 2012 #4
    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 [itex]\in[/itex] W), and for any element of W. Is that correct?
  6. Feb 29, 2012 #5


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    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
  7. Mar 1, 2012 #6


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    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").
  8. Mar 1, 2012 #7
    The simple answer is that u-w is not an element of W unless u is an element of W.
  9. Mar 1, 2012 #8


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

    (u+w) - (u-w) = 2w ∈ W.
  10. Mar 1, 2012 #9
    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 [itex]\in[/itex] U, then u - v [itex]\in[/itex] U [itex]\bigcap[/itex] W.

    How is this the case? Is it simply set theory?
  11. Mar 2, 2012 #10


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    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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