Quotient Spaces

This text is about the construction of quotient spaces. Mainly to test Blogger in conjunction with MathJax.

Given a vector space \(V\) over a field \(K\) and a subspace \(U\subseteq V\) one can define the relation 

\[\forall v,w\in V: v\sim w\Longleftrightarrow v-w\in U\]

which actually is an equivalence relation since it is

reflexive:

\[v\in V:v\sim v\Leftrightarrow v-v=0\in U\]

symmetric:

\[v,w\in V:v\sim w \Leftrightarrow \exists! u\in U:v-w=u \Leftrightarrow w-v=-u\in U\Leftrightarrow w\sim v\]

transitive:

\[v,w,q\in V: v\sim q\wedge q\sim w \Leftrightarrow \exists!u\in U:v-q=u \wedge \exists!u'\in U:w-q=u'\]

\[\Rightarrow u-u'=v-q-w+q=v-w\in U\Leftrightarrow v\sim w\]

The equivalence class \([v]\) of \(v\in V\) is given by 

\[[v]=\{w\in V|v\sim w\}\]

\(w\in[v]:v\sim w\Leftrightarrow \exists!u\in U:v-w=u\Rightarrow w = v-u\) which leads us to assume that

\[v+U:=\{v+u|u\in U\}=[v]\]

First we check if \(v+U\subseteq [v]\), so let \(w\in v+U\) then there exists \(u\in U\) such that \(w=v+u\). The last equation leads to \(w-v=u\Leftrightarrow w\sim v\Leftrightarrow w\in[v]\). Next we take a look at \([v]\subseteq v+U\). For every \(w\in[v]\) there is \(v\sim w\Leftrightarrow \exists! u\in U:v-w=u\Rightarrow w=v-u \in v+U\). So the assumption

\[v+U=[v]\]

is valid.

Quotient Space

The equivalence classes \(v+U\) of \(V\) define a vector space \(V/U\), a so called quotient space, if we define the vector addition

\[[v]+[w]:=[v+w]\forall v,w\in V\]

and the scalar multiplication 

\[\lambda [v] := [\lambda v] \forall \lambda \in K;\forall v \in V\]

Since an equivalence class can be represented by several vectors, the first step should be to show that the above operations are actually well defined. So we take vectors \(v,v',w,w'\in V:[v]=[v']\wedge [w]=[w']\) which should yield \([v+w]=[v'+w']\) or in other words

\[v+w\sim v'+w'\Leftrightarrow v-v'+w-w' \in U\] 

Because there are \(u,u'\in U\) such that \(v-v' = u\) and \(w-w'=u'\) we get

\[ v-v'+w-w'\Leftrightarrow u+u'\in U\]

which just shows that \(v+w\) and \(v'+w'\) describe the same equivalence class. The same idea we can apply to show that the scalar multiplication is well defined as well. What is left is the proof that the addition and scalar multiplication satisfy a vector space which is just long winding and technical, but quite easy.

What about the dimension of a quotient space? Lets assume the vector space \(V\) is of finite dimension \(n\) then \(dimU=:m\leq n\). So we can take a basis sequence \((b_i)_{i=1,2,...,m}\) of \(U\) and extend it to be sequence of \(V\) with another vector sequence \((b_i)_{i=m+1,m+2,...,n}\).  The equivalence class of any vector taken from \(U\) is actually the subspace \(U\) itself. We can partition the vector space \(V-U\) with vector subspaces \(V',W'\). Then we get for any non zero \(v\in V',w\in W'\)

\[v+U \neq w+U\]

since \(v\sim w\) does not exist. We can extend this idea and realize that the equivalence classes \(b_i+U\) for \(i=m+1,m+2,...,n\) are pairwise disjoint and for \(i=1,2,...,m\) they are equal \(0+U\). We can assume that 

\[B:=(b_i + U)_{i=m+1,m+2,...,n}\]

is a basis sequence of the quotient space. So this set should be linearly independent as well as a generating system of the quotient space. When we take a look at the following equation

\[0+U=\sum_{i=m+1}^{n}\lambda_ib_i+U\]

whereby the \(\lambda_i\) are field elements of \(K\), then we can almost directly see that the presumed basis of the quotient space is indeed linearly independent, just note that \(\sum_{i=m+1}^n \lambda_ib_i = 0\Rightarrow \lambda_i = 0\forall i\). Further we can't express any element of \(U\) as a linear combination of the vectors \((b_i)_{i=m+1,m+2,...,n}\). What's left is to show that \(B\) spans \(V/U\). So we take a vector \(v+U\in V/U\). To avoid any triviality we let \(v\notin U\). So there should exist field elements \(\lambda_i\) such that

\[v+U=\sum_{i=m+1}^n\lambda_ib_i+U\]

Such elements do exist since \(v\notin U\), so the same field elements we used above also satisfy 

\[v=\sum_{i=0}^n\lambda_ib_i\]

\(B\) is shown to be linearly independent as well as a generating system of \(V/U\) and therefore a basis system. As a direct result we also get the dimension formula of quotient spaces

\[dim V= dim (V/U) -dim U\]