Thank you for your answer! Unfortunately, in my algebra course I always get homeworks which cover such things we haven't learnt before, and will learn the next week. So at this time, I do not have any idea what is a quotient space.
well, the idea is this: suppose we have a subspace of V, called M.
we can define an equivalence relation on V by saying two vectors, u,v are equivalent if u-v is in M.
perhaps we should check that this is a bona-fide equivalence relation:
is u~u? well u-u = 0, and every subspce contains the 0-vector, so this is always true.
suppose u~v. is it also true that v~u? since u~v, we know that u-v is in M. since M is a subspace, this means that am is in M for every a in our field
(which is usually R, the real numbers, or C, the complex numbers, but not always). in particular, we can choose a = -1, so if m is in M, so is -m.
and this means that if u-v is in M, so is -(u-v) = v-u. but then THAT means that v~u. so ~ is reflexive, and symmetric.
finally, we ask if ~ is transitive. that is, if u~v, and v~w, is u~w? so from u~v, we know that u-v is in M. and from v~w, we know that v-w is in M.
since M is a subspace, it is closed under vector addition, so (u-v) + (v-w) = u-w is in M. so ~ is transitive. this means ~ is an equivalence relation on V.
so what does the equivalence class [u] look like? well, it's vectors of the form v, where u-v is in M. but if m is in M, and u-v = m, then u = v+m, for some m.
note that any vector u+m, for any m in M, is in [u], since u - (u+m) = m, which is in M. we define (definitions are important, stop sleeping)
the coset u+M to be the set {u+m: m is in M}. as we saw already, every element of u+M is in [u]. let's prove that every element of [u] is in u+M:
suppose v is in [u], so u-v is in M. then v-u is also in M (as we saw before), so v - u = m, for some m in M, so v = u+m, so v is in u+M.
so u+M is just [u], with the equivalence relation ~ we defined before. now, cosets are just sets, we haven't given then any structure.
but the amazing thing is, we can make them into a vector space! first, we need to decide how to "add" cosets.
since an element of u+M is just u+m, and an element of v+M is just v+m', we might hope that we could just add them together, to get:
[u] + [v] = [u+v], since u+m + v+m' = (u+v) + (m+m'), and m+m' is just "some element of M".
and that's exactly what we're going to do: we will define u+M + v+M to be (u+v)+M. there's just one catch: we want to make sure
that no matter which elements of [u] and [v] we choose, we wind up in the same coset [u+v] = u+v+M.
that is, we need to make sure that if u'+M = u+M, and v'+M=v+M, u'+v'+M = u+v+M (this means that addition is 'well-defined").
now u+M=u'+M means that u' is in u+M, and v'+M=v+M, means v' is in v+M. so u'~u, and v'~v, so u-u' = m, for some m,
and v-v' = m', for some (possibly different) m', so u+v - u'+v' = u-u' + v-v' = m+m', which is, in fact, in M, so u+v~u'+v',
that is, u+v+M = u'+v'+M. so, it works, we have a vector addition. now all we need is a scalar multiplication.
again, the "logical" choice seems to be a(u+M) = au+M. again, we need to check that if u~u', that au~au'.
given that u~u', we know that u-u' is in M, so u-u' = m, for some m. but then: au-au' = a(u-u') = am, and am is in M,
because M is a subspace. so we have a scalar multiplication. it is routine to check that this vector space satisfies the vector space axioms.
so, this space, where our "vectors" are cosets, is called the quotient space of V by M, and written V/M.
(don't think of V/M as being "division" but as grouping together all of V into "translated copies of M"
(M-type clumps) we use u+M to figure out "which copy" of M we're at).
for example, if M is a line, then V/M is a space consisting of a bundle of parallel lines. if M is a plane,
then V/M is a space consisting of a bundle of parallel sheets.