Let \(U\leq V\). We construct a new vector space from \(U\) and \(V\) which is an “abstract complement” to \(U\). The elements of this vector space are equivalence classes for the following equivalence relation:
Definition. Let \(U\leq V\). Say that \(v,w\in V\) are congruent modulo \(U\) if \(v-w\in U\). In this case, we write \(v\equiv w\mod U\).
Warning. This is emphatically not the relation of congruence modulo an integer \(n\) that you studied in Algebra 1A: here the relation is between vectors in a vector space. However, both notions of congruence are examples of a general construction in
group theory.
Lemma2.10. Congruence modulo \(U\) is an equivalence relation.
Proof. Exercise3! □
3 This is question 3 on exercise sheet 2.
Thus each \(v\in V\) lies in exactly one equivalence class \([v]\sub V\).
What do these equivalence classes look like? Note that \(w\equiv v\mod U\) if and only if \(w-v\in U\) or, equivalently, \(w=v+u\), for some \(u\in U\).
Definition. For \(v\in V\), \(U\leq V\), the set \(v+U:=\set {v+u\st u\in U}\sub V\) is called a coset of \(U\) and \(v\) is called a coset representative of \(v+U\).
We conclude that the equivalence class of \(v\) modulo \(U\) is the coset \(v+U\).
Figure 2.4: A subspace \(U\leq \R ^{2}\) and a coset \(v+U\).
Remark. In geometry, cosets of vector subspaces are called affine subspaces. Examples include lines in \(\R ^2\) and lines and planes in \(\R ^3\) irrespective of whether they contain zero (as vector subspaces must).
Example. Fibres of a linear map: let \(\phi :V\to W\) be a linear map and let \(w\in \im \phi \). Then the fibre of \(\phi \) over \(w\) is defined by:
Unless \(w=0\), this is not a linear subspace but notice that \(v,v'\) are in the same fibre if and only if \(\phi (v)=\phi (v')\), or, equivalently, \(\phi (v-v')=0\) or \(v-v'\in \ker \phi \). We conclude that the fibres of \(\phi \) are exactly
the cosets of \(\ker \phi \):
This is a subset of the power set4 \(\mathcal {P}(V)\) of \(V\).
The quotient map \(q:V\to V/U\) is defined by
\(\seteqnumber{0}{2.}{0}\)
\begin{equation*}
q(v)=v+U.
\end{equation*}
4 Recall from Algebra 1A that the power set of a set \(A\) is the set of all subsets of \(A\).
The quotient map \(q\) will be important to us. It has two key properties:
(1) \(q\) is surjective.
(2) \(q(v)=q(v')\) if and only if \(v\equiv v'\mod U\), that is, \(v-v'\in U\).
We can add and scalar multiply cosets to make \(V/U\) into a vector space and \(q\) into a linear map:
Theorem2.11. Let \(U\leq V\). Then, for \(v,w\in V\), \(\lambda \in \F \),
\(\seteqnumber{0}{2.}{0}\)
\begin{align*}
(v+U)+(w+U)&:=(v+w)+U\\ \lambda (v+U)&:=(\lambda v)+U
\end{align*}
give well-defined operations of addition and scalar multiplication on \(V/U\) with respect to which \(V/U\) is a vector space and \(q:V\to V/U\) is a linear map.
Moreover, \(\ker q=U\) and \(\im q=V/U\).
Proof. We phrase everything in terms of \(q\) to keep the notation under control. Since \(q\) surjects, we lose nothing by doing this: any element of \(V/U\) is of the form \(q(v)\) for some \(v\in V\).
With this understood, the proposed addition and scalar multiplication in \(V/U\) read
\(\seteqnumber{0}{2.}{0}\)
\begin{align*}
q(v)+q(w)&:=q(v+w)\\\lambda q(v)&:=q(\lambda v)
\end{align*}
so that \(q\) is certainly linear so long as these operations make sense. Here the issue is that if \(q(v)=q(v')\) and \(q(w)=q(w')\), we must show that
However, in this case, we have \(v-v'\in U\) and \(w-w'\in U\) so that
\(\seteqnumber{0}{2.}{1}\)
\begin{gather*}
(v+w)-(v'+w')=(v-v')+(w-w')\in U\\ \lambda v-\lambda v'=\lambda (v-v')\in U,
\end{gather*}
since \(U\) is a subspace, and this establishes (2.1).
As for the vector space axioms, these follow from those of \(V\). For example:
Here the first and third equalities are the definition of addition in \(V/U\) and the middle one comes from commutativity of addition in \(V\). The zero element is \(q(0)=0+U=U\) while the additive inverse of \(q(v)\) is \(q(-v)\).
The linearity of \(q\) comes straight from how we defined our addition and scalar multiplication while \(v\in \ker q\) if and only if \(q(v)=q(0)\) if and only if \(v=v-0\in U\) so that \(\ker q=U\). □
Figure 2.5: The quotient map \(q\).
Corollary2.12. Let \(U\leq V\). If \(V\) is finite-dimensional then so is \(V/U\) and
\(\seteqnumber{0}{2.}{1}\)
\begin{equation*}
\dim V/U=\dim V-\dim U.
\end{equation*}
Proof. Apply rank-nullity to \(q\) using \(\ker q=U\) and \(\im q=V/U\). □
Commentary. Many people find the quotient space \(V/U\) difficult to think about: its elements are (special) subsets of \(V\) and this can be confusing.
An alternative, perhaps better way, to proceed is to concentrate instead on the properties of \(V/U\) in much that same way that, in Analysis, we deal with real numbers via the axioms of a complete ordered field without worrying too much what a real number
actually is!
From this point of view, the quotient \(V/U\) of \(V\) by \(U\) is a vector space along with a linear map \(q:V\to V/U\) such that
• \(q\) surjects;
• \(\ker q=U\)
and this is really all you need to know!
The content of Theorem 2.11, from this perspective, is simply that quotients exist!
Theorem2.13 (First Isomorphism Theorem). Let \(\phi :V\to W\) be a linear map of
vector spaces.
Then \(V/\ker \phi \cong \im \phi \).
In fact, define \(\bar {\phi }:V/\ker \phi \to \im \phi \) by
where \(q:V\to V/\ker \phi \) is the quotient map.
Then \(\bar {\phi }\) is a well-defined linear isomorphism.
Proof. First we show that \(\bar {\phi }\) is well-defined: \(q(v)=q(v')\) if and only if \(v-v'\in \ker \phi \) if and only if \(\phi (v-v')=0\), or, equivalently, \(\phi (v)=\phi (v')\). We also get a
bit more: \(\bar {\phi }\) injects since if \(\bar {\phi }(q(v))=\bar {\phi }(q(v'))\) then \(\phi (v)=\phi (v')\) which implies that \(q(v)=q(v')\).
To see that \(\bar {\phi }\) is linear, we compute using the linearity of \(q\) and \(\phi \):
It remains to show that \(\bar {\phi }\) is surjective: but if \(w\in \im \phi \), then \(w=\phi (v)=\bar {\phi }(q(v))\), for some \(v\in V\), and we are done. □
Remarks.
(1) Let \(q:V\to V/\ker \phi \) be the quotient map and \(i:\im \phi \to W\) the inclusion. Then the First Isomorphism Theorem shows that we may write \(\phi \) as the composition \(i\circ \bar {\phi }\circ q\) of a quotient map, an isomorphism and an
inclusion.
(2) This whole story of cosets, quotients and the First Isomorphism Theorem has versions in many other contexts such as group theory and ring theory (see MA22017).
