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 6.7. 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 6.21 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 13.12
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 again an affine subspace or empty (cf. Proposition 6.11).
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 \(v-v_1\in f^{-1}(U)\): \[f(v-v_1) = g(v)-g(v_1)=g(v)-v_0\in U+v_0-v_0= U\]
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 \(\dim(W/V)\le\dim(W/U).\)
- True
- False
MCQ 13.4
Suppose \(U\subset V\subset W.\) Then \(\dim(V/U)\le\dim(W/V).\)
- 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