3.4 Vector subspaces and quotient spaces

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A (vector) subspace $W$ of a vector space $V$ is a subset that is a vector space in its own right, with respect to the operations of vector addition and scalar multiplication defined on $V$. There is a simple criterion for determining whether a subset is a subspace:

A subset $W$ is a subspace of $V$ if and only if $u+av\in W$ for all $a\in \mathbb{K}$ and all $u,v\in W$

For, setting $a=1$ shows that $W$ is closed under vector addition, while $u=0$ implies that it is closed with respect to scalar multiplication. Closure with respect to these two operations is sufficient to demonstrate that $W$ is a vector subspace: the zero vector $0\in W$ since $0=0v\in W$ for any $v\in W$ ; the inverse vector $-u=(-1)u\in W$ for every $u\in W$ and the remaining vector space axioms (VS1)–(VS5) are all satisfied by $W$ since they are inherited from $V$.

Example 3.11

Let $U=\{(u_1,u_2,\dots ,u_m,0,\dots ,0)\}\subseteq {\mathbb{R}}^n$ be the subset of $n$-vectors whose last $n-m$ components all vanish. $U$ is a vector subspace of ${\mathbb{R}}^n$ , since

$$\begin{array}{c}(u_1,\dots ,u_m,0,\dots ,0)+a(v_1,\dots ,v_m,0,\dots ,0)\\ =(u_1+a{v}_1,\dots ,u_m+a{v}_m,0,\dots ,0)\in U.\end{array}$$

This subspace is isomorphic to ${\mathbb{R}}^m$ , through the isomorphism

$$T:(u_1,\dots ,u_m,0,\dots ,0)\mapsto (u_1,\dots ,u_m).$$

Exercise

Show that ${\mathbb{R}}^n$ is isomorphic to a subspace of ${\mathbb{R}}^{\infty }$ for every $n>0$

Example 3.12

Let $V$ be a vector space over a field $\mathbb{K}$, and $u\in V$ any vector. The set $U=\{cu\mid c\in \mathbb{K}\}$ is a subspace of $V$, since for any pair of scalars $c,c^{\prime }\in \mathbb{K}$

$$cu+a(c^{\prime }u)=cu+(a{c}^{\prime })u=(c+a{c}^{\prime })u\in U.$$

Exercise

Show that the set $\{(x,y,z)|x+2y+3z=0\}$ forms a subspace of ${\mathbb{R}}^3$ , while the subse $\{(x,y,z)|x+2y+3z=1\}$ does not.

Example 3.13

The set of all continuous real-valued functions on ${\mathbb{R}}^n$ , denoted $\mathcal{C}({\mathbb{R}}^n)$ , is a subspace of $\mathcal{F}({\mathbb{R}}^n)$ defined in Example 3.8, for if $f$ and $g$ are any pair of continuous functions on ${\mathbb{R}}^n$ then so is any linear combination $f+ag$ where $a\in \mathbb{R}$

Exercise

Show that the vector space ofall real polynomials $\mathcal{P}(x)$ , defined in Example 3.10, is a vector subspace of $\mathcal{C}(\mathbb{R})$.

Given two subspaces $U$ and $W$ of a vector space $V$, their set-theoretical intersection $U\cap W$ forms a vector subspace of $V$, for if $u,w\in U\cap W$ then any linear combination $u+aw$ belongs to each subspace $U$ and $W$ separately. This argument can easily be extended to show that the intersection ${\bigcap }_{i\in I}{U}_i$ of any family of subspaces is a subspace of $V$

Complementary subspaces and quotient spaces

While the intersection of any pair of subspaces $U$ and $W$ is a vector subspace of $V$, this is not true of their set-theoretical union $U\cup V$ – consider, for example, the union of the two subspaces $\{(c,0)|c\in \mathbb{R}\}$ and $\{(0,c)|c\in \mathbb{R}\}$ of ${\mathbb{R}}^2$ . Instead, we can define the sum $U+W$ of any pair of subspaces to be the ‘smallest’ vector space that contains $U\cup W$ ,

$$U+W=\{u+w\mid u\in U,w\in W\}.$$

This is a vector subspace, for if $u=u_1+w_1$ and $v=u_2+w_2$ belong to $U+W$ , then

$$u+av=\left(u_1+w_1\right)+a\left(u_2+w_2\right)=\left(u_1+a{u}_2\right)+\left(w_1+a{w}_2\right)\in U+W.$$

Two subspaces $U$ and $W$ of $V$ are said to be complementary ifevery vector $v\in V$ has a unique decomposition $v=u+w$ where $u\in U$ and $w\in W$ . $V$ is then said to be the direct sum of the subspaces $U$ and $W$, written $V=U\oplus W$

Theorem 3.1

$U$ and $W$ are complementary subspaces of $V$ ifand only if(i) $V=U+W$ and $(ii)U\cap W=\{0\}$

Proof

If $U$ and $W$ are complementary subspaces then (i) is obvious, and if there exists a non-zero vector $u\in U\cap V$ then the zero vector would have alternative decompositions $0=$ $0+0$ and $0=u+(-u)$. Conversely, if (i) and (ii) hold then the decomposition $v=u+w$ is unique, for if $v=u^{\prime }+w^{\prime }\mathtt{then}u-u^{\prime }=w-w^{\prime }\in U\cap W$ . Hence $\begin{array}{r}u-u^{\prime }=w-w^{\prime }=0,\end{array}$ $\mathrm{so}u=u^{\prime }$ and $w=w^{\prime }$

Example 3.14

Let ${\mathbb{R}}_1$ be the subspace of ${\mathbb{R}}^n$ consisting of vectors of the form $\{(x_1,0,0,\dots ,0)|x_1\in \mathbb{R}\}$ , and $S_1$ the subspace

$$S_1=\{(0,x_2,x_3,\dots ,x_n)|(x_i\in \mathbb{R})\}.$$

Then ${\mathbb{R}}^n={\mathbb{R}}_1\oplus S_1$ . Continuing in a similar way $S_1$ may be written as a direct sum of ${\mathbb{R}}_2=$ $\{(0,x_2,0,\dots ,0)\}$ and a subspace $S_2$ . We eventually arrive at the direct sum decomposition

$${\mathbb{R}}^n={\mathbb{R}}_1\oplus {\mathbb{R}}_2\oplus \cdots \oplus {\mathbb{R}}_n\cong \mathbb{R}\oplus \mathbb{R}\oplus \cdots \oplus \mathbb{R}.$$

If $U$ and $W$ are arbitrary vector spaces it is possible to define their direct sum in a constructive way, sometimes called their external direct sum, by setting

$$U\oplus W=U\times W=\{(u,w)|u\in U,w\in W\}$$

with vector addition and scalar multiplication defined by

$$(u,w)+(u^{\prime },w^{\prime })=(u+u^{\prime },w+w^{\prime }),a(u,w)=(au,aw).$$

The map $\phi :U\to \hat{U}=\{(u,0)|u\in U\}\subset U\oplus W$ defined by $\phi (u)=(u,0)$ is clearly an isomorphism. Hence we may identify $U$ with the subspace $\hat{U}$ , and similarly $W$ is identifiable with $\hat{W}=\{(0,w)|w\in W\}$ . With these identifications the constructive notion ofdirect sum is equivalent to the ‘internally defined’ version, since $U\oplus W=\hat{U}\oplus \hat{W}$

The real number system $R$ can be regarded as a real vector space in which scalar multi plication is simply multiplication ofreal numbers – vectors and scalars are indistinguishable in this instance. Since the subspaces ${\mathbb{R}}_i$ defined in Example 3.14 are clearly isomorphic to $R$ for each $i=1,\dots ,n$ , the decomposition given in that example can be written

$${\mathbb{R}}^n\cong \underset{n}{\underbrace{\mathbb{R}\oplus \mathbb{R}\oplus \cdots \oplus \mathbb{R}}}.$$

For any given vector subspace there always exist complementary subspaces. We give the proof here as an illustration of the use of Zorn’s lemma, Theorem 1.6, but it is somewha technical and the reader will lose little continuity by moving on if they feel so inclined.

Theorem 3.2

Given a subspace $W\subseteq V$ there always exists a complementary subspace $U\subseteq V$ such that $V=U\oplus W.$

Proof

Given a vector subspace $W$ of $V$, let $\mathcal{U}$ be the collection of all vector subspaces $U\subseteq V$ such that $U\cap W=\{0\}$ . The set $\mathcal{U}$ can be partially ordered by set inclusion as in Example 1.5. Furthermore, if $\{U_i\mid i\in I\}$ is any totally ordered subset of $\mathcal{U}$ such that for every pair $i,j\in I$ we have either $U_i\subseteq U_j$ or $U_j\subseteq U_i$ , then their union is bounded above by

$$\tilde{U}=\bigcup _{i\in I}{U}_i.$$

The set $\tilde{U}$ is a vector subspace of $V$, for if $u\in \tilde{U}$ and $v\in \tilde{U}$ then there exists a member $U_i$ of the totally ordered family such that both vectors must belong to the same member $U_i-$ if $u\in U_j$ and $v\in U_k$ then set $i=k$ if $U_j\subseteq U_k$ , else set $i=j$ . Hence $u+av\in U_i\subseteq \tilde{U}$ for all $a\in \mathbb{K}$ . By Zorn’s lemma we conclude that there exists a maximal subspace $U\in \mathcal{U}.$

It remains to show that $U$ is complementary to $W$. Suppose not; then there exists a vector ${\upsilon }^{\prime }\in V$ that cannot be expressed in the form $v^{\prime }=u+w$ where $u\in U,w\in W$ . Let $U^{\prime }$ be the vector subspace defined by

$$U^{\prime }=\{a{v}^{\prime }+u|u\in U\}=\{a{v}^{\prime }\}\oplus U.$$

It belongs to the family $\mathcal{U}$, for if $\cdot U^{\prime }\cap W\ne \{0\}$ then there would exist a non-zero vector $w^{\prime }=$ $a{v}^{\prime }+u$ belonging to $W.$ . This implies $v^{\prime }=a^{-1}(w^{\prime }-u)$ , in contradiction to the requirement that $v^{\prime }$ cannot be expressed as a sum ofvectors from $U$ and $W$. Hence we have strict inclusion $U\subset U^{\prime }$ , contradicting the maximality of $U$. Thus $U$ is a subspace complementary to $W_{;}$ as required. -

This proofhas a distinctly non-constructive feel to it, which is typical ofproofs invoking Zorn’s lemma. A more direct way to arrive at a vector space complementary to a given subspace $W$ is to define an equivalence $\mathrm{relation}{\equiv }_W$ on $V$ by

$$u{\equiv }_{W}v\text{ iff }u-v\in W.$$

Checking the equivalence properties is easy:

Reflexive: $u-u=0\in W$ for all $u\in V$

Symmetric: $u-v\in W⟹v-u=-(u-v)\in W,$

Transitive: $u-v\in W\mathrm{and}v-w\in W⟹u-w=(u-v)+(v-w)\in W.$

The equivalence class to which $u$ belongs is written $u+W$ , where

$$u+W=\{u+w\mid w\in W\},$$

and is called a coset of $W$. This definition is essentially identical to that given in Section 2.5 for the case of an abelian group. It is possible to form the sum of cosets and multiply them by scalars, by setting

$$(u+W)+(v+W)=(u+v)+W,a(u+W)=(au)+W.$$

For consistency, it is necessary to show that these definitions are independent of the choice of coset representative. For example, if $u{\equiv }_W{u}^{\prime }$ and $\mathbf{v}{\equiv }_{\mathbf{W}}{\mathbf{v}}^{\prime }$ then $(u^{\prime }+v^{\prime }){\equiv }_W(u+v)$ , for

$$(u^{\prime }+v^{\prime })-(u+v)=(u^{\prime }-u)+(v^{\prime }-v)\in W.$$

Hence

$$(u^{\prime }+W)+(v^{\prime }+W)=(u^{\prime }+v^{\prime })+W=(u+v)+W=(u+W)+(v+W).$$

Similarly $\mathbf{a}{\mathbf{u}}^{\prime }{\equiv }_W$ $au$ since $a{u}^{\prime }-au=a(u^{\prime }-u)\in W$ and

$$a(u^{\prime }+W)=(a{u}^{\prime })+W=(au)+W=a(u+W).$$

The task of showing that the set of cosets is a vector space with respect to these operations is tedious but undemanding. For example, the distributive law (VS2) follows from

$$\begin{array}{rl}a((u+W)+(v+W))& =a((u+v)+W)\\ & =a(u+v)+W\\ & =(au+av)+W\\ & =((au)+W)+((av)+W)\\ & =a(u+W)+a(v+W).\end{array}$$

The rest of the axioms follow in like manner, and are left as exercises. The vector space of cosets of $W$ is called the quotient space of $V$ by $W$, denoted $V/W$

To picture a quotient space let $U$ be any subspace of $V$ that is complementary to $W.$ Every element of $V/W$ can be written uniquely as a coset $u+W$ where $u\in U$ . For, if $v+W$ is any coset, let $v=u+w$ be the unique decomposition of $v$ into vectors from $U$ and $W$ respectively, and it follows that $\mathbf{v}+\mathbf{W}=\mathbf{u}+\mathbf{W}$ since $v{\equiv }_{W}u$ . The map $T:U\to V/W$ defined by $T(u)=u+W$ describes an isomorphism between $U$ and $V/W$ . For, if $u+W=u^{\prime }+W$ where $u$, $u^{\prime }\in U$ then $u-u^{\prime }\in W\cap U$ , whence $u=u^{\prime }$ since $U$ and $W$ are complementary subspaces.

Exercise

Complete the details to show that the map $T:U\to V/W$ is linear, one-to-one and onto, so that $U\cong V/W$

This argument also shows that all complementary spaces to a given subspace $W$ are iso morphic to each other. The quotient space $V/W$ is a method for constructing the ‘canonica complement’ to $W.$

Example 3.15

While $V/W$ is in a sense complementary to $W$ it is not a subspace of $V$ and, indeed, there is no natural way ofidentifying it with any subspace complementary to $W$. For example, let $W=\{(x,y,0)|x,y\in \mathbb{R}\}$ be the subspace $z=0\mathrm{of}{\mathbb{R}}^3$ . Its cosets are planes $z=a$ , parallel to the $x$–$y$ plane, and it is these planes that constitute the ‘vectors’ of $V/W$ The subspace $U=\{(0,0,z)|z\in \mathbb{R}\}$ is clearly complementary to $W$ and is isomorphic to $V/W$ using the map

$$(0,0,a)\mapsto (0,0,a)+W=\{(x,y,a)|x,y\in \mathbb{R}\}.$$

However, there is no natural way of identifying $V/W$ with a complementary subspace such as $U$. For example, the space $U^{\prime }=\{(0,2z,z)|z\in \mathbb{R}\}$ is also complementary to $W$ since $U^{\prime }\cap W=\{0\}$ and every vector $(a,b,c)\in {\mathbb{R}}^3$ has the decomposition

$$(a,b,c)=(a,b-2c,0)+(0,2c,c),(a,b-2c,0)\in W,(0,2c,c)\in U^{\prime }.$$

Again, $U^{\prime }\cong V/W$ , under the map

$$(0,2c,c)\mapsto (0,2c,c)+W=(0,0,c)+W.$$

Note how the ${}^{\circ }\mathsf{W}-\mathsf{componen}{\mathsf{t}}^{\circ }$ of $(a,b,c)$ depends on the choice of complementary subspace; $(a,b,0)$ with respect to $U,$ , and $(a,b-2c,0)$ with respect to $U^{\prime }$

Images and kernels of linear maps

The image of a linear map $T:V\to W$ is defined to be the set

$$\mathrm{im}T=T(V)=\{w\mid w=Tv\}\subseteq W.$$

The set $\mathrm{im}T$ is a subspace of $W$, for if $w$, $w^{\prime }\in$ $\mathrm{im}T$ then

$$w+a{w}^{\prime }=Tv+aT{v}^{\prime }=T(v+a{v}^{\prime })\in \mathrm{im}T.$$

The kernel of the map $T$ is defined as the set

$$\ker T=T^{-1}(0)=\{v\in V\mid Tv=0\}\subseteq V.$$

This is also a subspace of $V$, for if $v,v^{\prime }\in \ker T$ then $T(v+a{v}^{\prime })=Tv+aT{v}^{\prime }=$ $0+0=0$ . The two spaces are related by the identity

$$\mathrm{im}T\cong V/\ker T.\text{\hspace{1em}(3.3)}$$

Proof

Define the map $\tilde{T}:V/\ker T\to \mathrm{im}T$ by

$$\tilde{T}(v+\ker T)=Tv.$$

This map is well-defined since it is independent of the choice of coset representative $v$,

$$\begin{array}{rl}v+\ker T=v^{\prime }+\ker T& ⟹v-v^{\prime }\in \ker T\\ & ⟹T(v-v^{\prime })=0\\ & ⟹Tv=T{v}^{\prime }\end{array}$$

and is clearly linear. It is onto and one-to-one, for every element of $\mathrm{im}T$ is of the form $Tv=\tilde{T}(v+\ker T)$ and

$$\begin{array}{c}\tilde{T}(v+\ker T)=\tilde{T}(v^{\prime }+\ker T)⟹Tv=T{v}^{\prime }\\ ⟹v-v^{\prime }\in \ker T\\ ⟹v+\ker T=v^{\prime }+\ker T.\end{array}$$

Hence $\tilde{T}$ is a vector space isomorphism, which proves Eq. (3.3).

Example 3.16

Let $V={\mathbb{R}}^3$ and $W={\mathbb{R}}^2$ , and define the map $T:V\to W$ by

$$T(x,y,z)=(x+y+z,2x+2y+2z).$$

The subspace $\mathrm{im}T\mathrm{of}{\mathbb{R}}^2$ consists of the set of all vectors of the form $(a,2a)$ , where $a\in \mathbb{R}$ while $\ker T$ is the subset of all vectors $(x,y,z)\in V$ such that $x+y+z=0.$ – check that these do form a subspace of $V$. If $\mathbf{v}=(x,y,z)$ and $a=x+y+z$ , then $v-ae\in \ker T$ where $e=(1,0,0)\in V$ , since

$$T(v-ae)=T(x,y,z)-T(x+y+z,0,0)=(0,0).$$

Furthermore $a$ is the unique value having this property, for if ${\mathbf{a}}^{\prime }\ne \mathbf{a}$ then $\mathbf{v}-{\mathbf{a}}^{\prime }\mathbf{e}\notin \ker T$. Hence every coset of $\ker T$ has a unique representative of the form $ae$ and may be written uniquely in the form $ae+\ker T$. The isomorphism $\tilde{T}$ defined in the above proof is given by

$$\tilde{T}(ae+\ker T)=(a,2a)=a(1,2).$$

Problems

Problem 3.5

If $L$, $M$ and $N$ are vector subspaces of $V$ show that

$$L\cap (M+(L\cap N))=L\cap M+L\cap N$$

but it is not true in general that

$$L\cap (M+N)=L\cap M+L\cap N.$$

Problem 3.6

Let $V=U\oplus W$ , and let $v=u+w$ be the unique decomposition of a vector $v$ into a sum ofvectors from $u\in U$ and $w\in W$ . Define theprojection operators $P_U:V\to U$ and $P_W:V\to$ $W$ by

$$P_U(v)=u,P_W(v)=w.$$

Show that

(a) $P_U^2=P_U$ and $P_W^2=P_W$

(b) Show that if $P:V\to V$ is an operator satisfying $P^2=P$ , said to be an idempotent operator, then there exists a subspace $U$ such that $P=P_U.[Hint;$ Set $U=\{u|Pu=u\}$ and $\mathcal{W}=\{w|Pw=0\}$ and show that these are complementary subspaces such that $P=P_U$ and $P_{\mathcal{W}}={\mathrm{id}}_V-P.]$