Let M, N be the subspaces of the V vector space and $\displaystyle M \subset N$. The quotient space of V by M is V/M and similarly the quotient space of V by N is V/N.

Show that V/M has a subspace which is isomorph to V/N.

Printable View

- Nov 11th 2011, 01:23 PMgotmejerryQuestion about subspaces
Let M, N be the subspaces of the V vector space and $\displaystyle M \subset N$. The quotient space of V by M is V/M and similarly the quotient space of V by N is V/N.

Show that V/M has a subspace which is isomorph to V/N. - Nov 11th 2011, 01:43 PMDrexel28Re: Question about subspaces
I think you might be trying to be "too" clever. You know that since $\displaystyle M\subseteq N$ that ]$\displaystyle \dim M\leqslant \dim N$ and so from first principles $\displaystyle \dim V/N\leqslant \dim V/M$ and so trivially you can find a copy of $\displaystyle V/N$ sitting inside $\displaystyle V/M$. Now, you tell me, what does this subspace look like?

- Nov 11th 2011, 03:57 PMDevenoRe: Question about subspaces
what if V is not finite-dimensional?

- Nov 11th 2011, 04:04 PMDrexel28Re: Question about subspaces
- Nov 12th 2011, 12:45 AMgotmejerryRe: Question about subspaces
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.

- Nov 12th 2011, 02:13 AMDevenoRe: Question about subspaces
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. - Nov 12th 2011, 08:34 AMgotmejerryRe: Question about subspaces
Thanks! I think I understand them now, but how do I solve my original problem:)?

- Nov 12th 2011, 12:56 PMDevenoRe: Question about subspaces
as NonCommAlg pointed out in another thread, consider a basis $\displaystyle \{v_i + N\}$ for V/N. show that $\displaystyle \{v_i + M\}$ is linearly independent in V/M.

two spaces with the same size basis are isomorphic. - Nov 12th 2011, 02:12 PMgotmejerryRe: Question about subspaces
I give this up, I cannot solve it