3.7 Dual spaces

中文

Linear functionals

A linear functional $\phi$ on a vector space $V$ over a field $\mathbb{K}$ is a linear map $\phi :V\to \mathbb{K}$ , where the field ofscalars $\mathbb{K}$ is regarded as a one-dimensional vector space, spanned by the element 1,

$$\phi (au+bv)=a\phi (u)+b\phi (v).\text{\hspace{1em}(3.24)}$$

The use ofthe phrase linearfunctional in place of‘linear function’ is largely adopted with infinite dimensional spaces in mind. For example, if $V$ is the space of continuous functions on the interval $[0,1]$ and $K(x)$ is an integrable function on [0, 1], let ${\phi }_K:V\to \mathbb{K}$ be defined by

$${\phi }_K(f)={\int }_0^{1}K(y)f(y)dy.$$

As the linear map ${\phi }_K$ is a function whose argument is another function, the terminology ‘linear functional’ seems more appropriate. In this case it is common to write the action of ${\phi }_K$ on the function $f$ as ${\phi }_K[f]$ in place of ${\phi }_K(f)$

Theorem 3.8

If $\phi :V\to \mathbb{K}$ is any linearfunctional then its kernel $\ker \phi$ has codimension 1. Conversely, any subspace $W\subset V$ ofcodimension 1 defines a linearfunctional $\phi$ on $V$ uniquely up to a scalarfactor, such that $W=\ker \phi .$

Proof

The first part of the theorem follows from Example 3.22 and Eq. (3.3),

$$\mathrm{codim}(\ker \phi )=\dim (V/(\ker \phi ))=\dim (\mathrm{im}\phi )=\dim \mathbb{K}=1.$$

To prove the converse, let $u$ be any vector not belonging to $W$ – if no such vector exists then $W=V$ and the codimension is 0. The set ofcosets $au+W=a(u+W)$ where $a\in \mathbb{K}$ form a one-dimensional vector space that must be identical with all of $V/W$ . Every vector $v\in V$ therefore has a unique decomposition $v=au+w$ where $w\in W$ , since

$$v=a^{\prime }u+w^{\prime }⟹(a-a^{\prime })u=w-w^{\prime }⟹a=a^{\prime }\text{ and }w=w^{\prime }.$$

A linear functional $\phi$ having kernel $W$ has $\phi (W)=0\mathrm{and}\phi (u)=c\ne 0.$ , for $\mathrm{if}c=0$ then the kernel of$\phi$ is all of $V$. Furthermore, given any non-zero scala $c\in \mathbb{K}$ , these two requirements define a linear functional on $V$ as its value on any vector $v\in V$ is uniquely determined by

$$\phi (v)=\phi (au+w)=ac.$$

If ${\phi }^{\prime }$ is any other such linear functional, having $c^{\prime }={\phi }^{\prime }(u)\ne 0$ , then ${\phi }^{\prime }=(c^{\prime }/c)\phi$ since ${\phi }^{\prime }(v)=a{c}^{\prime }=(c^{\prime }/c)\phi (v)$ -

This proofeven applies, as it stands, to infinite dimensional spaces, although in that case it is usual to impose the added stipulation that linear functionals be continuous (see Section 10.9 and Chapter 13).

Exercise

If $\omega$ and $\rho$ are linear functionals on $V,$ , show that

$$\ker \omega =\ker \rho ⟺\omega =a\rho \text{ for some }a\in \mathbb{K}.$$

Example 3.26

Let $V={\mathbb{K}}^n$ , where $n$ is any integer $\ge 2$ or possibly $n=\infty$ . For conve nience, we will take $V$ to be the space ofrow vectors oflength $n$ here. Let $W$ be the subspace of vectors

$$W=\{(x_1,x_2,\dots )\in {\mathbb{K}}^n\mid x_1+x_2=0\}.$$

This is a subspace of codimension 1, for if we set $\mathbf{u}=(1,0,0,\dots )$ , then any vector $v$ can be written

$$v=(v_1,v_2,v_3,\dots )=au+w$$

where $a=v_1+v_2$ and $\mathbf{w}=(-v_2,v_2,v_3,...)$ . This decomposition is unique, for if $au+w=a^{\prime }u+w^{\prime }$ then $(a-a^{\prime })u=w^{\prime }-w\in W$ . Hence $a=a^{\prime }$ and $w=w^{\prime }$ , since $u\notin W$ . Every coset $v+W$ can therefore be uniquely expressed as $a(u+W)$ , and $W$ has codimension 1.

Let $\phi :V\to \mathbb{K}$ be the linear functional such that $\phi (W)=0\mathrm{and}\phi (u)=1$ . Then $\phi (v)=$ $\phi (au+w)=a\phi (u)+\phi (w)=a$ , so that

$$\phi ((x_1,x_2,x_3,\dots ))=x_1+x_2.$$

The kernel of $\phi$ is evidently $W,$ , and every other linear functional ${\phi }^{\prime }$ having kernel $W$ is of the form $\phi ((x_1,x_2,x_3,\dots ))=c(x_1+x_2)$

The dual space of a vector space

As for general linear maps, it is possible to add linear functionals and multiply them by scalars,

$$(\phi +\omega )(u)=\phi (u)+\omega (u),(a\omega )(u)=a\omega (u).$$

With respect to these operations the set oflinear functionals on $V$ forms a vector space over $\mathbb{K}$ called the dual space of $V_{;}$ , usually denoted $V^{\ast }$ . In keeping with earlier conventions, other possible notations for this space are $L(V,\mathbb{K})$ or $\mathrm{Hom}(V,K)$. Frequently, linear functionals on $V$ will be called covectors, and in later chapters we will have reason to refer to them as 1-forms.

Let $V$ be a finite dimensional vector space, $\dim V=n$ , and $\{e_1,e_2,\dots ,e_n\}$ any basis for $V$. A linear functional $\omega$ on $V$ is uniquely defined by assigning its values on the basis vectors,

$$w_1=\omega (e_1),w_2=\omega (e_2),\dots ,w_n=\omega (e_n),$$

since the value on any vector $\begin{array}{r}u=u^i{e}_i\equiv {\sum }_{i=1}^n{u}^i{e}_i\end{array}$ can be determined using linearity Eq. (3.24),

$$\omega (u)=\omega (u^i{e}_i)=u^i\omega (e_i)=w_i{u}^i\equiv \sum _{i=1}^n{w}_i{u}^i.\text{\hspace{1em}(3.25)}$$

Define $n$ linear functionals ${\epsilon }^1,{\epsilon }^2,\dots ,{\epsilon }^n$ by

$${\epsilon }^i(e_j)={\delta }_j^i\text{\hspace{1em}(3.26)}$$

where ${\delta }_j^i$ is the Kronecker delta defined in Eq. (3.14). Note that these equations uniquely define each linear functional ${\epsilon }^i$ , since their values are assigned on each basis vector $e_j$ in turn.

Theorem 3.9

The $n$-linearfunctionals $\{{\epsilon }^1,{\epsilon }^2,\dots ,{\epsilon }^n\}$ form a basis of $V^{\ast }$, called the dual basis to $\{e_1,\dots ,e_n\}$ . Hence $\dim V^{\ast }=n=\dim V$ .

Proof

Firstly, suppose there are scalars $a_i\in \mathbb{K}$ such that

$$a_i{\epsilon }^i=0.$$

Applying the linear functional on the left-hand side of this equation to an arbitrary basis vector $e_j$

$$0=a_i{\epsilon }^i(e_j)=a_i{\delta }_j^i=a_j,$$

shows that the linear functionals $\{{\epsilon }^1,\dots ,{\epsilon }^n\}$ are linearly independent. Furthermore these linear functions span $V^{\ast }$ since every linear functional $\omega$ on $V$ can be written

$$\omega =w_i{\epsilon }^i\text{ where }{w}_i=\omega (e_i).\text{\hspace{1em}(3.27)}$$

This follows from

$$\begin{array}{rl}{w}_i{\epsilon }^i(u)& =w_i{\epsilon }^i(u^j{e}_j)\\ & =w_i{u}^j{\epsilon }^i(e_j)\\ & =w_i{u}^j{\delta }_j^i\\ & =w_i{u}^i\\ & =\omega (u)\text{ by Eq. (3.25) }.\end{array}$$

Thus $\omega$ and ${\omega }^{\prime }=w_i{\epsilon }^i$ have the same effect on every vector $u=u^i{e}_i;$ ; they are therefore identical linear functionals. The proposition that $\dim V^{\ast }=n$ follows from Corollary 3.4.

Exercise

Show that the expansion Eq. (3.27) is unique; i.e., if $\omega =w_i^{\prime }{\epsilon }^i$ then $w_i^{\prime }=w_i$ for each $i=$ $1,\dots ,n.$

Given a basis $E=\{e_i\}$ , we will frequently refer to the $n$ numbers $w_i=\omega (e_i)$ as the components of the linear functional $\omega$ in this basis. Alternatively, we can think ofthem as the components of$\omega$ with respect to the dual basis in $V^{\ast }$ . The formula Eq. (3.25) has a somewha deceptive ‘dot product’ feel about it. In Chapter 5 a dot product will be correctly defined as a product between vectors from the same vector space, while Eq. (3.25) is a product between vectors from different spaces $V$ and $V^{\ast }$ . It is, in fact, better to think of the components $u^i$ of a vector $u$ from $V$ as forming a column vector, while the components $w_j$ of a linea functional $\omega$ form a row vector. The above product then makes sense as a matrix product between a $1\times n$ row matrix and an $n\times 1$ column matrix. While a vector is often thought of geometrically as a directed line segment, often represented by an arrow, this is not a good way to think of a covector. Perhaps the best way to visualize a linear functional is as a set of parallel planes of vectors determined by $\omega (v)=\mathrm{const}.$ . (see Fig. 3.3).

Dual of the dual

We may enquire whether this dualizing process can be continued to generate further vector spaces such as the dual of the dual space $V^{\ast \ast }$ , etc. For finite dimensional spaces the process essentially stops at the first dual, for there is a completely natural way in which $V$ can be identified with $V^{\ast \ast }$ . To understand how $V$ itself can be regarded as the dual space of $V^{\ast }$ , define a linear map $\bar{v}:V^{\ast }\to \mathbb{K}$ corresponding to any vector $v\in V$ by

Figure 3.3 Geometrical picture of a linear functional

$$\bar{v}(\omega )=\omega (v)\text{ for all }\omega \in V^{\ast }.\text{\hspace{1em}(3.28)}$$

The map $\bar{\mathbf{v}}$ is a linear functional on $V^{\ast }$ , since

$$\begin{array}{rl}& \bar{v}(a\omega +b\rho )=(a\omega +b\rho )(v)\\ & =a\omega (v)+b\rho (v)\\ & =a\bar{v}(\omega )+b\bar{v}(\rho ).\end{array}$$

The map $\beta :V\to V^{\ast \ast }$ defined by $\beta (v)=\bar{v}$ is linear, since

$$\begin{array}{rl}\beta (au+bv)(\phi )& =\overline{au+bv}(\phi )(\phi \in V^{\ast })\\ & =\phi (au+bv)\\ & =a\phi (u)+b\phi (v)\\ & =a\bar{u}(\phi )+b\bar{v}(\phi )\\ & =a\beta (u)(\phi )+b\beta (v)(\phi ).\end{array}$$

As this holds for arbitrary covectors $\phi$ we have $\beta (au+bv)=a\beta (u)+b\beta (v)$ . Furthermore if $e_1,e_2,\dots ,e_n$ is a basis of $V$ with dual basis ${\epsilon }^1,{\epsilon }^2,\dots ,{\epsilon }^n$ , then

$${\bar{e}}_i({\epsilon }^j)={\epsilon }^j(e_i)={\delta }_i^j,$$

and it follows from Theorem 3.9 that $\{\beta (e_i)={\bar{e}}_i\}$ is the basis of $V^{\ast \ast }$ dual to the basis $\{{\epsilon }^j\}$ of $V^{\ast }$ . The map $\beta :V\to V^{\ast \ast }$ is therefore onto since every $f\in V^{\ast \ast }$ can be written in the form

$$f=u^i\beta (e_i)=\beta (u)=\bar{u}\text{ where }u=u^i{e}_i.$$

Since $\{\beta (e_i)\}$ is a basis, it follows from Theorem 3.5 that the components $u^i$ , and therefore the vector $u$, are uniquely determined by $f.$ . The map $\beta$ is thus a vector space isomorphism, as it is both onto and one-to-one.

We have shown that $V\cong V^{\ast \ast }$ . In itself this is to be expected since these spaces have the same dimension, but the significant thing to note is that since the defining Eq. (3.28) makes no mention of any particular choice of basis, the correspondence between $V$ and $V^{\ast \ast }$ is totally natural. There is therefore no ambiguity in identifying $\bar{v}$ with $v,$ and rewriting Eq. (3.28) as

$$v(\omega )=\omega (v).$$

This reciprocity between the two spaces $V$ and $V^{\ast }$ lends itself to the following alternative notations, which will be used interchangeably throughout this book:

$$⟨\omega ,v⟩\equiv ⟨v,\omega ⟩\equiv v(\omega )\equiv \omega (v).\text{\hspace{1em}(3.29)}$$

However, it should be pointed out that the identification of $V$ and $V^{\ast \ast }$ will only work for finite dimensional vector spaces. In infinite dimensional spaces, every vector may be regarded as a linear functional on $V^{\ast }$ in a natural way, but the converse is not true – there exist linear functionals on $V^{\ast }$ that do not correspond to vectors from $V$

Transformation law of covector components

By Theorem 3.9 the spaces $V$ and $V^{\ast }$ are in one-to-one correspondence since they have the same dimension, but unlike that described above between $V$ and $V^{\ast \ast }$ this correspondence is not natural. For example, if $v=v^i{e}_i$ is a vector in $V$ let $\upsilon$ be the linear functional whose components in the dual basis are exactly the same as the components of the original vector, $\mathbf{\upsilon }={\mathbf{v}}^i{\epsilon }^i$ . While the map $v\mapsto \upsilon$ is a vector space isomorphism, the same rule applied with respect to a different basis $\{e_i^{\prime }\}$ will generally lead to a different correspondence between vectors and covectors. Thus, given an arbitrary vector $v\in V$ there is no basis-independent way ofpointing to a covector partner in $V^{\ast }$ . Essentially this arises from the fact that the law of transformation of components of a linear functional is different from the transformation law of components for a vector.

We have seen in Section 3.6 that the transformation of a basis can be written by Eqs. Eq. (3.16) and Eq. (3.20),

$$e_i=A_i^j{e}_j^{\prime },e_j^{\prime }=A_j^{\prime k}{e}_k\text{\hspace{1em}(3.30)}$$

where $[A_k^i]$ and $[A^{\prime i}{}_k]$ are related through the inverse matrix equations Eq. (3.19). Let $\{{\epsilon }^i\}$ and $\{{\epsilon }^{\prime i}\}$ be the dual bases corresponding to the bases $\{{\mathbf{e}}_j\}$ and $\{e_j^{\prime }\}$ respectively of $V$ ,

$${\epsilon }^i(e_j)={\delta }_j^i,{\epsilon }^{\prime i}(e_j^{\prime })={\delta }_j^i.\text{\hspace{1em}(3.31)}$$

Set

$${\epsilon }^i=B_j^i{\epsilon }^{\prime j},$$

and substituting this and Eq. (3.30) into the first identity of Eq. (3.31) gives, after replacing the index $j$ by $k$,

$$\begin{array}{rl}& {\delta }_k^i=B_j^i{\epsilon }^{\prime j}(e_k)\\ & =B_j^i{\epsilon }^{\prime j}(A_k^l{e}_l^{\prime })\\ & =B_j^i{A}_k^l{\delta }_l^j\\ & =B_j^i{A}_k^j.\end{array}$$

Hence $B_j^i={A^{\prime }}_j^i$ and the transformation of the dual basis is

$${\epsilon }^i=A_j^{\prime i}{\epsilon }^{\prime j}.\text{\hspace{1em}(3.32)}$$

Exercise

Show the transform inverse to Eq. (3.32)

$${\epsilon }^{\prime j}=A_k^j{\epsilon }^k.\text{\hspace{1em}(3.33)}$$

If $\omega =w_i{\epsilon }^i$ is a linear functional having components $w_i$ with respect to the first basis, then

$$\omega =w_i{\epsilon }^i=w_i{A}_j^{{}^{\prime }i}{\epsilon }^{{}^{\prime }j}=w_i^{\prime }{\epsilon }^{{}^{\prime }i}$$

where

$$w_i^{\prime }=A_i^{\prime j}{w}_j.\text{\hspace{1em}(3.34)}$$

This is known as the covariant vector transformation law of components. Its inverse is

$$w_j=A_j^k{w}_k^{\prime }.\text{\hspace{1em}(3.35)}$$

These equations are to be compared with the contravariant vector transformation law of components of a vector $v=v^i{e}_i$ , given by Eqs. Eq. (3.17) and Eq. (3.21),

$$v^{\prime j}=A_i^j{v}^i,v^i=A_j^{\prime i}{v}^j.\text{\hspace{1em}(3.36)}$$

Exercise

Verify directly from Eq. (3.34) and Eq. (3.36) that Eq. (3.25) is basis-independent,

$$\omega (u)=w_i{u}^i=w_j^{\prime }{u}^{\prime j}.$$

Exercise

Show that if the components of $\omega$ are displayed as a $1\times n$ row matrix ${\mathbf{w}}^T=$ $\left(w_1,w_2,\dots ,w_n\right)$ then the transformation law Eq. (3.34) can be written as a matrix equation

$${\mathbf{w}}^{\prime T}={\mathbf{w}}^T\mathsf{A}.$$

Problems

Problem 3.14

Find the dual basis to the basis of ${\mathbb{R}}^3$ having column vector representation

$${\mathbf{e}}_1=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right),{\mathbf{e}}_2=\left(\begin{array}{c}1\\ 0\\ -1\end{array}\right),{\mathbf{e}}_3=\left(\begin{array}{c}0\\ -1\\ 1\end{array}\right).$$

Problem 3.15

Let $\mathcal{P}(x)$ be the vector space of real polynomials $f(x)=a_0+a_{1}x+\cdot \cdot \cdot +a_n{x}^n$ If $(b^0,b^1,b^2,\dots )$ is any sequence of real numbers, show that the map $\beta :\mathcal{P}(x)\mathbb{R}$ given by

$$\beta (f(x))=\sum _{i=0}^n{b}^i{a}_i$$

is a linear functional on $\mathcal{P}(x)$

Show that every linear functional $\beta$ on $\mathcal{P}(x)$ can be obtained in this way from such a sequence and hence that $({\hat{\mathbb{R}}}^{\infty }{)}^{\ast }={\mathbb{R}}^{\infty }$

Problem 3.16

Define the annihilator $S^{\perp }$ of a subset $S\subseteq V$ as the set of all linear functionals tha vanish on $S$,

$$S^{\perp }=\{\omega \in V^{\ast }\mid \omega (u)=0\forall u\in S\}.$$

(a) Show that for any subset $S,S^{\perp }$ is a vector subspace of $V^{\ast }$ .

(b) If $T\subseteq S,$ , show that $S^{\perp }\subseteq T^{\perp }$

(c) If $U$ is a vector subspace of $V$, show that $(V/U{)}^{\ast }\cong U^{\perp }$ . [Hint: For each $\omega$ in $U^{\perp }$ define the element $\bar{\omega }\in (V/U{)}^{\ast }$ by $\bar{\omega }(v+U)=\omega (v).]$

(d) Show that $U^{\ast }\cong V^{\ast }/U^{\perp }$

(e) If $V$ is finite dimensional with $\dim V=n$ and $W$ is any subspace of $V$ with $\dim W=m$ , show that $\dim W^{\perp }=n-m$ . [Hint: Use a basis adapted to the subspace $W$ by Theorem 3.7 and conside its dual basis in $V^{\ast }.]$

(f) Adopting the natural identification of $V$ and $V^{\ast \ast }$ , show that $(W^{\perp }{)}^{\perp }=W$

Problem 3.17

Let $u$ be a vector in the vector space $V$ of dimension $n$.

(a) If $\omega$ is a linear functional on $V$ such that $a=\omega (u)\ne 0$ , show that a basis $e_1,\dots ,e_n$ can be chosen such that

$$u=e_1\mathrm{and}\omega =a{\epsilon }^1$$

where $\{{\epsilon }^1,\dots ,{\epsilon }^n\}$ is the dual basis. [Hint: Apply Theorem 3.7 to the vector $u$ and try a further basis transformation of the form $e_1^{\prime }=e_1,e_2^{\prime }=e_2+a_2{e}_1,\dots ,e_n^{\prime }=e_n+a_n{e}_1.$

(b) If $a=0,$ , show that the basis may be chosen such that

$$u=e_1\text{ and }\omega ={\epsilon }^2.$$

Problem 3.18

For the three-dimensional basis transformation of Problem 3.13 evaluate the ${\epsilon }^{\prime j}$ dual to $e_i^{\prime }$ in terms of the dual basis ${\epsilon }^j$ . What are the components of the linear functiona $\omega ={\epsilon }^1+{\epsilon }^2+{\epsilon }^3$ with respect to the new dual basis?

Problem 3.19

If $A:V\to V$ is a linear operator, define its transpose to be the linear map $A^{\prime }$ : $V^{\ast }\to V^{\ast }$ such that

$$A^{\prime }\omega (u)=\omega (Au),\forall u\in V,\omega \in V^{\ast }.$$

Show that this relation uniquely defines the linear operator $A^{\prime }$ and that

$$O^{\prime }=O,({\mathrm{id}}_V{)}^{\prime }={\mathrm{id}}_{V^{\ast }},(aB+bA{)}^{\prime }=a{B}^{\prime }+b{A}^{\prime },\forall a,b\in \mathbb{K}.$$

(a) Show that $(BA{)}^{\prime }=A^{\prime }{B}^{\prime }$

(b) If $A$ is an invertible operator then show that $(A^{\prime }{)}^{-1}=(A^{-1}{)}^{\prime }$

(c) If $V$ is finite dimensional show that $A^{\prime \prime }=A.$ , if we make the natural identification of $V^{\ast \ast }$ and $V$.

(d) Show that the matrix of components of the transpose map $A$ with respect to the dual basis is the transpose of the matrix of $A,{\mathsf{A}}^{\prime }={\mathsf{A}}^T$

(e) Using Problem 3.16 show that $\ker A^{\prime }=(\mathrm{im}A{)}^{\perp }$

(f) Use Eq. (3.10) to show that the rank of $A^{\prime }$ equals the rank of $A$.

Problem 3.20

The row rank ofa matrix is defined as the maximum number oflinearly independen rows, while its column rank is the maximum number of linearly independent columns.

(a) Show that the rank of a linear operator $A$ on a finite dimensional vector space $V$ is equal to the column rank of its matrix $A$ with respect to any basis of $V.$ .

(b) Use parts (d) and (f) of Problem 3.19 to show that the row rank of a square matrix is equal to its column rank.

Problem 3.21

Let $S$ be a linear operator on a vector space $V.$

(a) Show that the rank of $S$ is one, $\rho (S)=1$ , if and only if there exists a non-zero vector $u$ and a non-zero linear functional $\alpha$ such tha

$$S(v)=u\alpha (v).$$

(b) With respect to any basis $\{e_i\}$ of $V$ and its dual basis $\{{\epsilon }^j\}$ , show that

$$S_j^i=u^i{a}_j\text{ where }u=u^i{e}_i,\alpha =a_j{\epsilon }^j.$$

(c) Show that every linear operator A of rank $r$ can be written as a sum of $r$ linear operators of rank one.

(d) Show that the last statement is equivalent to the assertion that for every matrix $A$ of rank $r$ there exist column vectors ${\mathbf{u}}_i$ and ${\mathbf{a}}_i(i=1,\dots ,r)$ such that

$$\mathsf{S}=\sum _{i=1}^r{\mathbf{u}}_i{\mathbf{a}}_i^T.$$