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### Finite Dimensional Quotient Space

Date: 11/01/2008 at 10:41:20
From: Debdeep
Subject: quotient space

If V is a finite dimensional vector space and w is a subspace of V
then prove that quotient space v/w is also finite dimensional.

Date: 11/02/2008 at 15:52:24
From: Doctor Jordan
Subject: Re: quotient space

Hi Debdeep,

Let V be a vector space and W be a subspace of V.  For v in V, define
[v] by

[v] = {v + w : w in W}.

In other words, this is the set of co-sets of W in V.  The quotient
space V/W is defined to be the set

{ [v] : v in V }.

Here we define [v_1] + [v_2] = [v_1 + v_2] and a[v]=[av], where a is a
scalar and v_1,v_2 are in V.  You have to verify that the set V/W
defined this way is actually a vector space--in other words that it
satisfies the axioms of vector spaces; you may have already done this

Now we will show that V/W is finite dimensional.  When I was thinking
how to do this, I remembered that we know a formula that involves the
dimension of a vector space and the dimensions of the image and kernel
of a linear transformation: dim(V) = dim(Im(T)) + dim(Ker(T)), if
these are all finite.  My idea is to define some linear transformation
T so that one of the terms in this equation is V/W.

Define T:V->V/W by

T(v) = [v].

You have to verify that this is a linear transformation.  Now, it
should be clear if you look at it for a bit that Im(T) = V/W.  Also,
show that Ker(T) = W.  We know that

dim(V) = dim(Im(T)) + dim(Ker(T)),

hence

dim(V) = dim(V/W) + dim(W),

hence

dim(V/W) = dim(V) - dim(W).

Since dim(V) and dim(W) are finite, this means that dim(V/W) is
finite.  Therefore the quotient space V/W is finite dimensional.

Probably the most important idea to remember from this proof is that
if you want to find the dimension of some vector space, it is often

dim(V) = dim(Im(T)) + dim(Ker(T)),

and then figure out how we need to define T.

- Doctor Jordan, The Math Forum
http://mathforum.org/dr.math/
Associated Topics:
College Modern Algebra

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