13 Affine spaces and quotient vector spaces
13.1 Affine mappings and affine spaces
Previously we saw that we can take the sum of subspaces of a vector space. In this final chapter of the Linear Algebra I module we introduce the concept of a quotient of a vector space by a subspace.
Translations are among the simplest non-linear mappings.
Definition 13.1 • Translation
Let \(V\) be a \(\mathbb{K}\)-vector space and \(v_0 \in V.\) The mapping \[T_{v_0} : V \to V,\qquad v \mapsto v+v_0\] is called the translation by the vector \(v_0.\)
Remark 13.2
Notice that for \(v_0\neq 0_V,\) a translation is not linear, since \(T_{v_0}(0_V)=0_V+v_0=v_0\neq 0_V.\)
Taking \(s_1=1\) and \(s_2=-1\) in (6.1), we see that a linear map \(f : V \to W\) between \(\mathbb{K}\)-vector spaces \(V,W\) satisfies \(f(v_1-v_2)=f(v_1)-f(v_2)\) for all \(v_1,v_2 \in V.\) In particular, linear maps are affine maps in the following sense:
Definition 13.3 • Affine mapping
A mapping \(f : V \to W\) is called affine if there exists a linear map \(g : V \to W\) so that \(f(v_1)-f(v_2)=g(v_1-v_2)\) for all \(v_1,v_2 \in V.\) We call \(g\) the linear map associated to \(f\).
Affine mappings are compositions of linear mappings and translations:
Proposition 13.4
A mapping \(f : V \to W\) is affine if and only if there exists a linear map \(g : V \to W\) and a translation \(T_{w_0} : W \to W\) so that \(f=T_{w_0}\circ g.\)
Proof. \(\Leftarrow\) Let \(g : V \to W\) be linear and \(T_{w_0} : W \to W\) be a translation for some vector \(w_0 \in W\) so that \(T_{w_0}(w)=w+w_0\) for all \(w \in W.\) Let \(f=T_{w_0}\circ g\) so that \(f(v)=g(v)+w_0\) for all \(v \in V.\) Then \[f(v_1)-f(v_2)=g(v_1)+w_0-g(v_2)-w_0=g(v_1)-g(v_2)=g(v_1-v_2),\] hence \(f\) is affine.
\(\Rightarrow\) Let \(f : V \to W\) be affine and \(g : V \to W\) its associated linear map. Since \(f\) is affine we have for all \(v \in V\) \[f(v)-f(0_V)=g(v-0_V)=g(v)-g(0_V)=g(v)\] where we use the linearity of \(g\) and Lemma 6.6. Writing \(w_0=f(0_V)\) we thus have \[f(v)=g(v)+w_0\] so that \(f\) is the composition of the linear map \(g\) and the translation \(T_{w_0} : W \to W,\) \(w \mapsto w+w_0.\)
Example 13.5
Let \(\mathbf{A}\in M_{m,n}(\mathbb{K}), \vec{b} \in \mathbb{K}^m\) and \[f_{\mathbf{A},\vec{b}} : \mathbb{K}^n \to \mathbb{K}^m, \quad \vec{x}\mapsto \mathbf{A}\vec{x}+\vec{b}.\] Then \(f_{\mathbf{A},\vec{b}}\) is an affine map whose associated linear map is \(f_\mathbf{A}.\) Conversely, combining Lemma 7.4 and Proposition 13.4, we see that every affine map \(\mathbb{K}^n \to \mathbb{K}^m\) is of the form \(f_{\mathbf{A},\vec{b}}\) for some matrix \(\mathbf{A}\in M_{m,n}(\mathbb{K})\) and vector \(\vec{b} \in \mathbb{K}^m.\)
An affine subspace of a \(\mathbb{K}\)-vector space \(V\) is a translation of a subspace by some fixed vector \(v_0.\)
Definition 13.6 • Affine subspace
Let \(V\) be a \(\mathbb{K}\)-vector space. An affine subspace of \(V\) is a subset of the form \[U+v_0=\{u+v_0 : u \in U\},\] where \(U\subset V\) is a subspace and \(v_0 \in V.\) We call \(U\) the associated vector space to the affine subspace \(U+v_0\) and we say that \(U+v_0\) is parallel to \(U.\)
Example 13.7
Let \(V=\mathbb{R}^2\) and \(U=\operatorname{span}\{\vec{e}_1+\vec{e}_2 \}=\left\{s(\vec{e}_1+\vec{e}_2)| s \in \mathbb{R}\right\}\) where here, as usual, \(\{\vec{e}_1,\vec{e}_2\}\) denotes the standard basis of \(\mathbb{R}^2.\) So \(U\) is the line through the origin \(0_{\mathbb{R}^2}\) defined by the equation \(y=x.\) By definition, for all \(\vec{v} \in \mathbb{R}^2\) we have \[U+\vec{v}=\left\{\vec{v}+s\vec{w} : s \in \mathbb{R}\right\},\] where we write \(\vec{w}=\vec{e}_1+\vec{e}_2.\) So for each \(\vec{v} \in \mathbb{R}^2,\) the affine subspace \(U+\vec{v}\) is a line in \(\mathbb{R}^2,\) the translation by the vector \(\vec{v}\) of the line defined by \(y=x.\)
13.2 Quotient vector spaces
Let \(U\) be a subspace of a \(\mathbb{K}\)-vector space \(V.\) We want to make sense of the notion of dividing \(V\) by \(U.\) It turns out that there is a natural way to do this and moreover, the quotient \(V/U\) again carries the structure of a \(\mathbb{K}\)-vector space. The idea is to define \(V/U\) to be the set of all translations of the subspace \(U,\) that is, we consider the set of subsets \[V/U=\{U+v | v \in V\}.\] We have to define what it means to add affine subspaces \(U+v_1\) and \(U+v_2\) and what it means to scale \(U+v\) by a scalar \(s\in \mathbb{K}.\) Formally, it is tempting to define \(0_{V/U}=U+0_V\) and \[\tag{13.1} (U+v_1)+_{V/U}(U+v_2)=U+(v_1+v_2)\] for all \(v_1,v_2 \in V\) as well as \[\tag{13.2} s\cdot_{V/U}(U+v)=U+(s v)\] for all \(v \in V\) and \(s \in \mathbb{K}.\) However, we have to make sure that these operations are well defined. We do this with the help of the following lemma.
Lemma 13.8
Let \(U\subset V\) be a subspace. Then any vector \(v \in V\) belongs to a unique affine subspace parallel to \(U,\) namely \(U+v.\) In particular, two affine subspaces \(U+v_1\) and \(U+v_2\) are either equal or have empty intersection.
Proof. Since \(0_V \in U,\) we have \(v \in (U+v),\) hence we only need to show that if \(v \in (U+\hat{v})\) for some vector \(\hat{v},\) then \(U+v=U+\hat{v}.\) Assume \(v \in (U+\hat{v})\) so that \(v=u+\hat{v}\) for some vector \(u \in U.\) Suppose \(w \in (U+\hat{v}).\) We need to show that then also \(w \in (U+v).\) Since \(w \in (U+\hat{v})\) we have \(w=\hat{u}+\hat{v}\) for some vector \(\hat{u} \in U.\) Using that \(\hat{v}=v-u,\) we obtain \[w=\hat{u}+v-u=\hat{u}-u+v\] Since \(U\) is a subspace we have \(\hat{u}-u \in U\) and hence \(w \in (U+v).\)
Conversely, suppose \(w \in (U+v),\) it follows exactly as before that then \(w \in (U+\hat{v})\) as well.
We are now going to show that (13.1) and (13.2) are well defined. We start with (13.1). Let \(v_1,v_2 \in V\) and \(w_1,w_2 \in V\) such that \[U+v_1=U+w_1 \qquad \text{and} \qquad U+v_2=U+w_2.\] We need to show that \(U+(v_1+v_2)=U+(w_1+w_2).\) By Lemma 13.8 it suffices to show that \(w_1+w_2\) is an element of \(U+(v_1+v_2).\) Since \(U+w_1=U+v_1\) it follows that \(w_1 \in (U+v_1)\) so that \(w_1=u_1+v_1\) for some element \(u_1 \in U.\) Likewise it follows that \(w_2=u_2+v_2\) for some element \(u_2 \in U.\) Hence \[w_1+w_2=u_1+u_2+v_1+v_2.\] Since \(U\) is a subspace, we have \(u_1+u_2 \in U\) and thus it follows that \(w_1+w_2\) is an element of \(U+(v_1+v_2).\)
For (13.2) we need to show that if \(v \in V\) and \(w \in V\) are such that \(U+v=U+w,\) then \(U+(s v)=U+(s w)\) for all \(s \in \mathbb{K}.\) Again, applying Lemma 13.8 we only need to show that \(s w \in U+(s v).\) Since \(U+v=U+w\) it follows that there exists \(u \in U\) with \(w=u+v.\) Hence \(s w=s u+s v\) and \(U\) being a subspace, we have \(s u \in U\) and thus \(s w\) lies in \(U+(s v),\) as claimed.
Having equipped \(V/U\) with addition \(+_{V/U}\) defined by (13.1) and scalar multiplication \(\cdot_{V/U}\) defined by (13.2), we need to show that \(V/U\) with zero vector \(U+0_V\) is indeed a \(\mathbb{K}\)-vector space. All the properties of Definition 4.2 for \(V/U\) are however simply a consequence of the corresponding property for \(V.\) For instance commutativity of vector addition in \(V/U\) follows from the commutativity of vector in addition in \(V,\) that is, for all \(v_1,v_2 \in V\) we have \[(U+v_1)+_{V/U}(U+v_2)=U+(v_1+v_2)=U+(v_2+v_1)=(U+v_2)+_{V/U}(U+v_1).\] The remaining properties follow similarly.
Notice that we have a surjective mapping \[p : V \to V/U, \quad v \mapsto U+v.\] which satisfies \[p(v_1+v_2)=U+(v_1+v_2)=(U+v_1)+_{V/U}(U+v_2)=p(v_1)+_{V/U} p(v_2)\] for all \(v_1,v_2 \in V\) and \[p(s v)=U+(s v)=s\cdot_{V/U}(U+v)=s\cdot_{V/U}p(v).\] for all \(v \in V\) and \(s \in \mathbb{K}.\) Therefore, the mapping \(p\) is linear.
Definition 13.9 • Quotient vector space
The vector space \(V/U\) is called the quotient (vector) space of \(V\) by \(U\). The linear map \(p : V \to V/U\) is called the canonical surjection from \(V\) to \(V/U.\)
The mapping \(p : V \to V/U\) satisfies \[p(v)=0_{V/W}=U+0_V \quad \iff \quad v \in U\] and hence \(\operatorname{Ker}(p) = U.\) This gives:
Proposition 13.10
Suppose the \(\mathbb{K}\)-vector space \(V\) is finite dimensional. Then \(V/U\) is finite dimensional as well and \[\dim(V/U)=\dim(V)-\dim(U).\]
Proof. Since \(p\) is surjective it follows that \(V/U\) is finite dimensional as well. Hence we can apply Theorem 6.20 and obtain \[\dim V=\dim \operatorname{Ker}(p)+\dim \operatorname{Im}(p)=\dim U+\dim(V/U),\] where we use that \(\operatorname{Im}(p)=V/U\) and \(\operatorname{Ker}(p)=U.\)
Example 13.11 • Special cases
In the case where \(U=V\) we obtain \(V/U=\{0_{V/U}\}.\)
In the case where \(U=\{0_V\}\) we obtain that \(V/U\) is isomorphic to \(V.\)
Exercises
Exercise 13.1
Show that the image of an affine subspace under an affine map is again an affine subspace; and that the preimage of an affine subspace under an affine map is either an affine subspace, or is empty (cf. Proposition 6.10).
Solution
Let \(V\) be a \(\mathbb{K}\)-vector space and let \(U\) be a subspace of \(V.\) We first show that the image of an affine subspace \(U+v_0,\) \(v_0\in V\) under an affine map \(g:V\to V\) is again an affine subspace. According to Proposition 13.4, \(g=T_{w_0}\circ f\) for some linear map \(f:V\to V\) and \(T_{w_0}:w\mapsto w+w_0.\) Let \(v\in U+v_0\) i.e. \(v=u+v_0\) for some \(u\in U.\) Then \[g(v) = (T_{w_0}\circ f)(u+v_0) = T_{w_0}(f(u)+f(v_0)) = f(u) + f(v_0)+w_0\] and hence \[g(U+v_0) = f(U)+f(v_0)+w_0,\] which is an affine subspace since \(f(U)\) is a subspace of \(V.\) Regarding the preimage, observe that \(g^{-1}(U+v_0)\) is either empty or, if not, it contains an element \(v_1\) such that \(g(v_1) = v_0.\) In this case, we claim that \[g^{-1}(U+v_0)=f^{-1}(U)+v_1.\] Let \(v\in f^{-1}(U)+v_1,\) then \(v=\hat{u} + v_1\) for some \(\hat{u}\in f^{-1}(U).\) Then \[g(v) = g(\hat{u} + v_1) = f(\hat{u}+v_1)+w_0 = f(\hat{u})+f(v_1)+w_0 = f(\hat{u}) + g(v_1)\in U+v_0.\] Conversely, if \(v\in g^{-1}(U+v_0),\) then we might write \(v=(v-v_1)+v_1\) and we claim that \(v-v_1\in f^{-1}(U)\): \[f(v-v_1) = g(v)-g(v_1)=g(v)-v_0\in U+v_0-v_0= U\]
Exercise 13.2
Show that in \(P_2(\mathbb{R}),\) the set of polynomials \(f\) with \(f(1) = 1\) is an affine subspace. What is the associated vector subspace?
Exercise 13.3
Let \(A\) be an affine subspace of a \(\mathbb{K}\)-vector space \(V.\) Show that for all integers \(n \geqslant 1,\) all \(v_1, \dots, v_n \in A,\) and all scalars \(s_1, \dots, s_n \in \mathbb{K}\) with \(\sum_i s_i = 1,\) we have \(\sum_i s_i v_i \in A.\)
Show, conversely, that if \(A\) is a non-empty subset of \(V\) with this property, then \(A\) is an affine subspace.
(Hint: Choose an \(a_0 \in A\) and show that \(U = \{a - a_0 : a \in A\}\) is a vector subspace.)
Exercise 13.4
(hard!) Let \(f : \mathbb{Q}^2 \to \mathbb{Q}^2\) be a map which preserves co-linearity, i.e. if \(P, Q, R\) lie on a straight line then \(f(P), f(Q), f(R)\) lie on a straight line. Show that \(f\) is an affine map.
Exercise 13.5
Let \(P\) be the space of all polynomial functions \(\mathbb{R}\to \mathbb{R}.\) Define \(W = \{p \in P : p(0) = p(1)\}\) and \(V = \{ p \in P : p(0) = p(1) = 0\}.\) What is the dimension of the quotient spaces \(P / W\) and \(P / V\)?
Exercise 13.6
Let \(f : V \to V'\) be a linear map, and \(W \subset V,\) \(W' \subset V'\) vector subspaces such that \(f(W) \subset W'.\)
Show that there is a uniquely determined linear map \(\bar{f} : V/W \to V'/W'\) satisfying \[f(v + W) = f(v) + W'\] for all \(v \in V.\)
Show that if \(V = V'\) and \(W = W',\) then we have \[\det f = \det (\bar{f}) \det(f|_{W}),\] where \(f|_{W} : W \to W\) denotes the restriction of \(f\) to \(W.\)
MCQ 13.1
Let \(\mathbf{A}\in M_{m,n}(\mathbb{K})\) and let \(\vec b\in\mathbb{R}^m.\) The set of solutions of the linear system \(\mathbf{A}\vec x=\vec b\) forms an affine subspace of \(\mathbb{K}^n.\)
- True
- False
MCQ 13.2
Let \(\mathbf{A}\in M_{m,n}(\mathbb{K}).\) Then \(\operatorname{Im}(\mathbf{A})\) is isomorphic to \(\mathbb{K}^n/\operatorname{Ker}(\mathbf{A}).\)
- True
- False
MCQ 13.3
Suppose \(U\subset V\subset W.\) Then \(\dim(W/V)\leqslant\dim(W/U).\)
- True
- False
MCQ 13.4
Suppose \(U\subset V\subset W.\) Then \(\dim(V/U)\leqslant\dim(W/V).\)
- True
- False
MCQ 13.5
Let \(V\subset \mathbb{K}^\infty\) be defined as \(V = \{(0,x_2,x_3,\ldots) : x_i\in\mathbb{K}\}.\) Then \(\dim(\mathbb{K}^\infty/V)=1.\)
- True
- False
MCQ 13.6
Let \(V\subset \mathbb{R}^n\) be an affine subspace. If \(\vec x,\vec y\in V,\) then \((1-s)\vec x+s\vec y\in V\) for all \(s\in\mathbb{R}.\)
- True
- False