7 Quotient vector spaces
7.1 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 7.1 • 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 7.2
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 (3.6), 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 7.3 • 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 7.4
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.
Lemma 3.15 ➔
. 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 \(T_{w_0} : W \to W,\) \(w \mapsto w+w_0.\)Let \(f : V \to W\) be a linear map, then \(f(0_V)=0_W.\)
Example 7.5 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 3.18
Lemma 3.18 ➔
and Proposition 7.4A mapping \(g : \mathbb{K}^m \to \mathbb{K}^n\) is linear if and only if there exists a matrix \(\mathbf{B}\in M_{n,m}(\mathbb{K})\) so that \(g=f_\mathbf{B}.\)
Proposition 7.4 ➔
, 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 \(\vec{b} \in \mathbb{K}^m.\)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 \(f=T_{w_0}\circ g.\)
An affine subspace of a \(\mathbb{K}\)-vector space \(V\) is a translation of a subspace by some fixed vector \(v_0.\)
Definition 7.6 • 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 7.7
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 \(y=x.\)
7.2 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 \[\tag{7.1} (U+v_1)+_{V/U}(U+v_2)=U+(v_1+v_2)\] for all \(v_1,v_2 \in V\) as well as \[\tag{7.2} s\cdot_{V/U}(U+v)=U+(sv)\] 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 7.8
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.
Lemma 7.8 ➔
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 \(U+(v_1+v_2).\)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.
Lemma 7.8 ➔
we only need to show that \(sw \in U+(sv).\) Since \(U+v=U+w\) it follows that there exists \(u \in U\) with \(w=u+v.\) Hence \(sw=su+sv\) and \(U\) being a subspace, we have \(su \in U\) and thus \(sw\) lies in \(U+(sv),\) as claimed.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.
Definition 3.1 • Vector space ➔
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.A \(\mathbb{K}\)-vector space, or vector space over \(\mathbb{K}\) is a set \(V\) with a distinguished element \(0_V\) (called the zero vector) and two operations \[\begin{aligned} +_V : V \times V \to V& &(v_1,v_2) \mapsto v_1+_Vv_2& &(\text{vector addition}) \end{aligned}\] and \[\begin{aligned} \cdot_V : \mathbb{K}\times V \to V& &(s,v) \mapsto s\cdot_V v& &(\text{scalar multiplication}), \end{aligned}\] so that the following properties hold:
Commutativity of vector addition \[v_1+_Vv_2=v_2+_Vv_1\quad (\text{for all}\; v_1,v_2 \in V);\]
Associativity of vector addition \[v_1+_V(v_2+_Vv_3)=(v_1+_Vv_2)+_Vv_3 \quad (\text{for all}\; v_1,v_2,v_3 \in V);\]
Identity element of vector addition \[\tag{3.4} 0_V+_Vv=v+_V0_V=v\quad (\text{for all}\; v \in V);\]
Identity element of scalar multiplication \[1\cdot_V v=v\quad (\text{for all}\; v \in V);\]
Scalar multiplication by zero \[\tag{3.5} 0\cdot_{V}v=0_V \quad (\text{for all}\; v \in V);\]
Compatibility of scalar multiplication with field multiplication \[(s_1s_2)\cdot_V v=s_1\cdot_V(s_2\cdot_V v) \quad (\text{for all}\; s_1,s_2 \in \mathbb{K}, v \in V);\]
Distributivity of scalar multiplication with respect to vector addition \[s\cdot_V(v_1+_Vv_2)=s\cdot_Vv_1+_Vs\cdot_V v_2\quad (\text{for all}\; s\in \mathbb{K}, v_1,v_2 \in V);\]
Distributivity of scalar multiplication with respect to field addition \[(s_1+s_2)\cdot_Vv=s_1\cdot_Vv+_Vs_2\cdot_Vv \quad (\text{for all}\; s_1,s_2 \in \mathbb{K}, v \in V).\] The elements of \(V\) are called vectors.
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 \[p(sv)=U+(sv)=s\cdot_{V/U}(U+v)=s\cdot_{V/U}p(v).\] for all \(v \in V\) and \(s\in \mathbb{K}.\) Therefore, the mapping \(p\) is linear.
Definition 7.9 • 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 7.10
Suppose the \(\mathbb{K}\)-vector space \(V\) is finite dimensional. Then \(V/U\) is finite dimensional as well and \[\dim(V/U)=\dim(V)-\dim(U).\]
Theorem 3.76 • Rank–nullity theorem ➔
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 \(\operatorname{Ker}(p)=U.\)Let \(V,W\) be finite dimensional \(\mathbb{K}\)-vector spaces and \(f : V \to W\) a linear map. Then we have \[\dim(V)=\dim \operatorname{Ker}(f)+\dim \operatorname{Im}(f)=\operatorname{nullity}(f)+\operatorname{rank}(f).\]
Example 7.11 • 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 7.12 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 3.26
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 7.4
Proposition 3.26 ➔
).Let \(V,W\) be \(\mathbb{K}\)-vector spaces, \(U\subset V\) and \(Z\subset W\) be vector subspaces and \(f : V \to W\) a linear map. Then the image \(f(U)\) is a vector subspace of \(W\) and the preimage \(f^{-1}(Z)\) is a vector subspace of \(V.\)
Solution
Proposition 7.4 ➔
, \(g=T_{w_0}\circ f\) 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 \(v-v_1\in f^{-1}(U)\): \[f(v-v_1) = g(v)-g(v_1)=g(v)-v_0\in U+v_0-v_0= U\]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 \(f=T_{w_0}\circ g.\)