2.2 Vector spaces

2.2.1 Vectors and vector spaces

A vector space (or a linear space) $V$ over a field $K$ is a set in which two operations, addition and multiplication by an element of $K$ (called a scalar), are defined. (In this book we are mainly interested in $K=\mathbb{R}$ and $\mathbb{C}$ .) The elements (called vectors) of $V$ satisfy the following axioms:

(i) $u+v=v+u$ .
(ii) $(u+v)+w=u+(v+w).$
(iii) There exists a zero vector $0$ such that $v+0=v$
(iv) For any $u$ , there exists $-u$ , such that $u+(-u)=0$ .
(v) $c(u+v)=cu+cv$
(vi) $(c+d)u=cu+du$
(vii) $(cd)u=c(du)$
(viii) $1u=u$

Here $u,v,w\in V$ and $c,d\in K$ and 1 is the unit element of $K$

Let $\{v_i\}$ be a set of $k(>0)$ vectors. If the equation

$$x_1{\mathbf{v}}_1+x_2{\mathbf{v}}_2+\cdots +x_k{\mathbf{v}}_k=0\text{\hspace{1em}(2.6)}$$

has a non-trivial solution, $x_i\ne 0$ for some $i$ , the set of vectors $\{v_j\}$ is called linearly dependent, while if (2.6) has only a trivial solution, $x_i=0$ for any $i$ , $\{v_i\}$ is said to be linearly independent. If at least one of the vectors is a zero vector $0$ , the set is always linearly dependent.

A set of linearly independent vectors $\{e_i\}$ is called a basis of $V$ , if any element $v\in V$ is written uniquely as a linear combination of $\{e_i\}$ :

$$\mathbf{v}=v^1{\mathbf{e}}_1+v^2{\mathbf{e}}_2+\cdots +v^n{\mathbf{e}}_n.\text{\hspace{1em}(2.7)}$$

The numbers $v^i\in K$ are called the components of $\mathbf{v}$ with respect to the basis $\{e_j\}$ . If there are $n$ elements in the basis, the dimension of $V$ is $n$ , denoted by $\dim V=n$ . We usually write the $n$ -dimensional vector space over $K$ as $V(n,K)$ (or simply $V$ if $n$ and $K$ are understood from the context). We assume $n$ is finite.

2.2.2 Linear maps, images and kernels

Given two vector spaces $V$ and $W$ , a map $f:V\to W$ is called a linear map if it satisfies $f(a_1{v}_1+a_2{v}_2)=a_{1}f(v_1)+a_{2}f(v_2)$ for any $a_1,a_2\in K$ and $v_1,v_2\in V$ . A linear map is an example of a homomorphism that preserves the vector addition and the scalar multiplication. The image of $f$ is $f(V)\subset W$ and the kernel of $f$ is $\{v\in V|f(v)=0\}$ and denoted by $\mathrm{im}f$ and $\ker f$ respectively. $\ker f$ cannot be empty since $f(0)$ is always $0$ . If $W$ is the field $K$ itself, $f$ is called a linear function. If $f$ is an isomorphism, $V$ is said to be isomorphic to $W$ and vice versa, denoted by $V\cong W$ . It then follows that $\dim V=\dim W$ . In fact, all the $n$ -dimensional vector spaces are isomorphic to $K^n$ , and they are regarded as identical vector spaces. The isomorphism between the vector spaces is an element of $\mathrm{GL}(n,K)$ .

Theorem 2.1

If $f:V\to W$ is a linear map, then

$$\dim V=\dim (\ker f)+\dim (\mathrm{im}f).\text{\hspace{1em}(2.8)}$$

Proof

Since $f$ is a linear map, it follows that $\ker f$ and $\mathrm{im}f$ are vector spaces, see exercise 2.8. Let the basis of $\ker f$ be $\{g_1,\dots ,g_r\}$ and that of $\mathrm{im}f$ be $\{h_1^{\prime },\dots ,h_s^{\prime }\}$ . For each $i$ ( $1\le i\le s$ ), take $h_i\in V$ such that $f(h_i)=h_i^{\prime }$ and consider the set of vectors $\{g_1,\dots ,g_r,h_1,\dots ,h_s\}$ .

Now we show that these vectors form a linearly independent basis of $V$ . Take an arbitrary vector $v\in V$ . Since $f(v)\in \mathrm{im}f$ , it can be expanded as $f(v)=c^i{h}_i^{\prime }=c^{i}f(h_i)$ . From the linearity of $f$ , it then follows that $f(v-c^i{h}_i)=0$ , that is $v-c^i{h}_i\in \ker f$ . This shows that an arbitrary vector $v$ is a linear combination of $\{g_1,\dots ,g_r,h_1,\dots ,h_s\}$ . Thus, $V$ is spanned by $r+s$ vectors. Next let us

assume $a^i{g}_i+b^i{h}_i=0$ . Then $0=f(0)=f(a^i{g}_i+b^i{h}_i)=b^{i}f(h_i)=b^i{h}_i^{\prime }$ , which implies that $b^i=0$ . Then it follows from $a^i{g}_i=0$ that $a^i=0$ , and the set $\{g_1,\dots ,g_r,h_1,\dots ,h_s\}$ is linearly independent in $V$ . Finally we find $\dim V=r+s=\dim (\ker f)+\dim (\mathrm{im}f)$ .

[Remark: The vector space spanned by $\{h_1,\dots ,h_s\}$ is called the orthogonal complement of $\ker f$ and is denoted by $(\ker f{)}^{\perp }$ .]

Exercise 2.8

(1) Let $f:V\to W$ be a linear map. Show that both $\ker f$ and $\mathrm{im}f$ are vector spaces.

(2) Show that a linear map $f:V\to V$ is an isomorphism if and only if $\ker f=\{0\}$ .

2.2.3 Dual vector space

The dual vector space has already been introduced in section 1.2 in the context of quantum mechanics. The exposition here is more mathematical and complements the materials presented there.

Let $f:V\to K$ be a linear function on a vector space $V(n,K)$ over a field $K$ . Let $\{e_i\}$ be a basis and take an arbitrary vector $v=v^1{e}_1+\cdots +v^n{e}_n$ . From the linearity of $f$ , we have $f(v)=v^{1}f(e_1)+\cdots +v^{n}f(e_n)$ . Thus, if we know $f(e_i)$ for all $i$ , we know the result of the operation of $f$ on any vector. It is remarkable that the set of linear functions is made into a vector space, namely a linear combination of two linear functions is also a linear function.

$$\left(a_1{f}_1+a_2{f}_2\right)(\mathbf{v})=a_1{f}_1(\mathbf{v})+a_2{f}_2(\mathbf{v})\text{\hspace{1em}(2.9)}$$

This linear space is called the dual vector space to $V(n,K)$ and is denoted by $V^{\ast }(n,K)$ or simply by $V^{\ast }$ . If $\dim V$ is finite, $\dim V^{\ast }$ is equal to $\dim V$ . Let us introduce a basis $\{e^{\ast i}\}$ of $V^{\ast }$ . Since $e^{\ast i}$ is a linear function it is completely specified by giving $e^{\ast i}(e_j)$ for all $j$ . Let us choose the dual basis,

$$e^{\ast i}\left({\mathbf{e}}_j\right)={\delta }_j^i.\text{\hspace{1em}(2.10)}$$

Any linear function $f$ , called a dual vector in this context, is expanded in terms of $\{e^{\ast i}\}$ ,

$$f=f_i{e}^{\ast i}.\text{\hspace{1em}(2.11)}$$

The action of $f$ on $v$ is interpreted as an inner product between a column vector and a row vector,

$$f(\mathbf{v})=f_i{e}^{\ast i}\left(v^j{\mathbf{e}}_j\right)=f_i{v}^j{e}^{\ast i}\left({\mathbf{e}}_j\right)=f_i{v}^i.\text{\hspace{1em}(2.12)}$$

We sometimes use the notation $⟨,⟩:V^{\ast }\times V\to K$ to denote the inner product.

Let $V$ and $W$ be vector spaces with a linear map $f:V\to W$ and let $g:W\to K$ be a linear function on $W$ ( $g\in W^{\ast }$ ). It is easy to see that the

composite map $g\circ f$ is a linear function on $V$ . Thus, $f$ and $g$ give rise to an element $h\in V^{\ast }$ defined by

$$h(\mathbf{v})\equiv g(f(\mathbf{v}))\mathbf{v}\in V.\text{\hspace{1em}(2.13)}$$

Given $g\in W^{\ast }$ , a map $f:V\to W$ has induced a map $h\in V^{\ast }$ . Accordingly, we have an induced map $f^{\ast }:W^{\ast }\to V^{\ast }$ defined by $f^{\ast }:g\mapsto h=f^{\ast }(g)$ , see figure 2.9. The map $h$ is called the pullback of $g$ by $f^{\ast }$ .

Figure 2.9. The pullback of a function $g$ is a function $f^{\ast }(g)=g\circ f$ .

Since $\dim V^{\ast }=\dim V$ , there exists an isomorphism between $V$ and $V^{\ast }$ . However, this isomorphism is not canonical; we have to specify an inner product in $V$ to define an isomorphism between $V$ and $V^{\ast }$ and vice versa, see the next section. The equivalence of a vector space and its dual vector space will appear recurrently in due course.

Exercise 2.9

Suppose $\{f_j\}$ is another basis of $V$ and $\{f^{\ast i}\}$ the dual basis. In terms of the old basis, $f_i$ is written as $f_i=A_i^j{e}_j$ where $A\in \mathrm{GL}(n,K)$ . Show that the dual bases are related by $e^{\ast i}=f^{\ast j}{A}_j^i$ .

2.2.4 Inner product and adjoint

Let $V=V(m,K)$ be a vector space with a basis $\{e_i\}$ and let $g$ be a vector space isomorphism $g:V\to V^{\ast }$ , where $g$ is an arbitrary element of $\mathrm{GL}(m,K)$ . The component representation of $g$ is

$$g:v^j\to g_{ij}{v}^j.\text{\hspace{1em}(2.14)}$$

Once this isomorphism is given, we may define the inner product of two vectors $v_1,v_2\in V$ by

$$g\left({\mathbf{v}}_1,{\mathbf{v}}_2\right)\equiv ⟨g{\mathbf{v}}_1,{\mathbf{v}}_2⟩.\text{\hspace{1em}(2.15)}$$

Let us assume that the field $K$ is a real number $\mathbb{R}$ . For definiteness. Then equation (2.15) has a component expression,

$$g\left({\mathbf{v}}_1,{\mathbf{v}}_2\right)=v_1^i{g}_{ji}{v}_2^j.\text{\hspace{1em}(2.16)}$$

We require that the matrix $(g_{ij})$ be positive definite so that the inner product $g(v,v)$ has the meaning of the squared norm of $v$ . We also require that the metric be symmetric: $g_{ij}=g_{ji}$ so that $g(v_1,v_2)=g(v_2,v_1)$ .

Next, let $W=W(n,\mathbb{R})$ be a vector space with a basis $\{f_{\alpha }\}$ and a vector space isomorphism $G:W\to W^{\ast }$ . Given a map $f:V\to W$ , we may define the adjoint of $f$ , denoted by $\tilde{f}$ , by

$$G(\mathbf{w},f\mathbf{v})=g(\mathbf{v},\tilde{f}\mathbf{w})\text{\hspace{1em}(2.17)}$$

where $v\in V$ and $w\in W$ . It is easy to see that $(\tilde{f})=f$ . The component expression of equation (2.17) is

$$w^{\alpha }{G}_{\alpha \beta }{f}_i^{\beta }{v}^i=v^i{g}_{ij}{\tilde{f}}_{\alpha }^j{w}^{\alpha }\text{\hspace{1em}(2.18)}$$

where $f_i^{\beta }$ and ${\tilde{f}}_{\alpha }^j$ are the matrix representations of $f$ and $\tilde{f}$ respectively. If $g_{ij}={\delta }_{ij}$ and $G_{\alpha \beta }={\delta }_{\alpha \beta }$ , the adjoint $\tilde{f}$ reduces to the transpose $f^{\mathrm{t}}$ of the matrix $f$ .

Let us show that $\dim \mathrm{im}f=\dim \mathrm{im}\tilde{f}$ . Since (2.18) holds for any $v\in V$ and $w\in W$ , we have $G_{\alpha \beta }{f}_i^{\beta }=g_{ij}{\tilde{f}}_{\alpha }^j$ , that is

$$\tilde{f}=g^{-1}{f}^{\mathrm{t}}{G}^{\mathrm{t}}.\text{\hspace{1em}(2.19)}$$

Making use of the result of the following exercise, we obtain $\mathrm{rank}f=\mathrm{rank}\tilde{f}$ , where the rank of a map is defined by that of the corresponding matrix (note that $g\in \mathrm{GL}(m,\mathbb{R})$ and $G\in \mathrm{GL}(n,\mathbb{R})$ ). It is obvious that $\dim \mathrm{im}f$ is the rank of a matrix representing the map $f$ and we conclude $\dim \mathrm{im}f=\dim \mathrm{im}\tilde{f}$ .

Exercise 2.10

Let $V=V(m,\mathbb{R})$ and $W=W(n,\mathbb{R})$ and let $f$ be a matrix corresponding to a linear map from $V$ to $W$ . Verify that $\mathrm{rank}f=\mathrm{rank}{f}^{\mathrm{t}}=\mathrm{rank}(M{f}^{\mathrm{t}}N)$ , where $M\in \mathrm{GL}(m,\mathbb{R})$ and $N\in \mathrm{GL}(n,\mathbb{R})$ .

Exercise 2.11

Let $V$ be a vector space over $\mathbb{C}$ . The inner product of two vectors $v_1$ and $v_2$ is defined by

$$g\left({\mathbf{v}}_1,{\mathbf{v}}_2\right)={\bar{v}}_1^i{g}_{ij}{v}_2^j\text{\hspace{1em}(2.20)}$$

where $\bar{\cdot }$ denotes the complex conjugate. From the positivity and symmetry of the inner product, $g(v_1,v_2)=\overline{g(v_2,v_1)}$ , the vector space isomorphism $g:V\to V^{\ast }$ is required to be a positive-definite Hermitian matrix. Let $f:V\to W$ be a (complex) linear map and $G:W\to W^{\ast }$ be a vector space isomorphism. The adjoint of $f$ is defined by $g(v,\tilde{f}w)=\overline{G(w,fv)}$ . Repeat the analysis to show that

(a) $\tilde{f}=g^{-1}{f}^{†}{G}^{†}$ , where $†$ denotes the Hermitian conjugate, and
(b) $\dim \mathrm{im}f=\dim \mathrm{im}\tilde{f}$

Theorem 2.2 · Toy index theorem

Let $V$ and $W$ be finite-dimensional vector spaces over a field $K$ and let $f:V\to W$ be a linear map. Then

$$\dim \ker f-\dim \ker \tilde{f}=\dim V-\dim W.\text{\hspace{1em}(2.21)}$$

Proof

Theorem 2.1 tells us that

$$\dim V=\dim \ker f+\dim \mathrm{im}f$$

and, if applied to $\tilde{f}:W\to V$

$$\dim W=\dim \ker \tilde{f}+\dim \mathrm{im}\tilde{f}.$$

We saw earlier that $\dim \mathrm{im}f=\dim \mathrm{im}\tilde{f}$ , from which we obtain

$$\dim V-\dim \ker f=\dim W-\dim \ker \tilde{f}.$$

Note that in (2.21), each term on the LHS depends on the details of the map $f$ . The RHS states, however, that the difference in the two terms is independent of $f$ ! This may be regarded as a finite-dimensional analogue of the index theorems, see chapter 12.

2.2.5 Tensors

A dual vector is a linear object that maps a vector to a scalar. This may be generalized to multilinear objects called tensors, which map several vectors and dual vectors to a scalar. A tensor $T$ of type $(p,q)$ is a multilinear map that maps $p$ dual vectors and $q$ vectors to $\mathbb{R}$ ,

$$T:\stackrel{p}{⨂}{V}^{\ast }\stackrel{q}{⨂}V\to \mathbb{R}.\text{\hspace{1em}(2.22)}$$

For example, a tensor of type $(0,1)$ maps a vector to a real number and is identified with a dual vector. Similarly, a tensor of type $(1,0)$ is a vector. If $\omega$ maps a dual vector and two vectors to a scalar, $\omega :V^{\ast }\times V\times V\to \mathbb{R}$ , $\omega$ is of type $(1,2)$ .

The set of all tensors of type $(p,q)$ is called the tensor space of type $(p,q)$ and denoted by ${\mathfrak{T}}_q^p$ . The tensor product $\tau =\mu \otimes \nu \in {\mathfrak{T}}_q^p\otimes {\mathfrak{T}}_{q^{\prime }}^{p^{\prime }}$ is an element of ${\mathfrak{T}}_{q+q^{\prime }}^{p+p^{\prime }}$ defined by

$$\begin{array}{rl}& \tau \left({\omega }_1,\dots ,{\omega }_p,{\xi }_1,\dots ,{\xi }_{p^{\prime }};u_1,\dots ,u_q,v_1,\dots ,v_{q^{\prime }}\right)\\ & =\mu \left({\omega }_1,\dots ,{\omega }_p;u_1,\dots ,u_q\right)v\left({\xi }_1,\dots ,{\xi }_{p^{\prime }};v_1,\dots ,v_{q^{\prime }}\right).\end{array}\text{\hspace{1em}(2.23)}$$

Another operation in a tensor space is the contraction, which is a map from a tensor space of type $(p,q)$ to type $(p-1,q-1)$ defined by

$$\tau (\dots ,e^{\ast i},\dots ;\dots ,e_i,\dots )\text{\hspace{1em}(2.24)}$$

where $\{e_i\}$ and $\{e^{\ast i}\}$ are the dual bases.

Exercise 2.12

Let $V$ and $W$ be vector spaces and let $f:V\to W$ be a linear map. Show that $f$ is a tensor of type (1, 1).

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