5.1 Real inner product spaces

中文

Let $V$ be a real finite dimensional vector space with $\dim$ $V=n$ . A real inner product, often referred to simply as an inner product when there is no danger of confusion, on the vector space $V$ is a map $V\times V\to \mathbb{R}$ that assigns a real number $u\cdot v\in \mathbb{R}$ to every pair of vectors $u$, $v\in V$ satisfying the following three conditions:

(RIP1) The map is symmetric in both arguments, $\mathbf{u}\cdot \mathbf{v}=\mathbf{v}\cdot \mathbf{u}.$

(RIP2) The distributive law holds, $u\cdot (av+bw)=au\cdot v+bu\cdot w.$

(RIP3) If $u\cdot v=0$ for all $v\in V$ then $u=0.$

A real vector space $V$ together with an inner product defined on it is called a real inner product space. The inner product is also distributive on the first argument for, by conditions (RIP1) and (RIP2),

$$(au+bv)\cdot w=w\cdot (au+bv)=aw\cdot u+bw\cdot v=au\cdot w+bv\cdot w.$$

We often refer to this linearity in both arguments by saying that the inner product is bilinear.

As a consequence of property (RIP3) the inner product is said to be non-singular and is often referred to as pseudo-Euclidean. Sometimes (RIP3) is replaced by the stronger condition

(RIP3’) $u\cdot u>0$ for all vectors $u\ne 0.$

In this case the inner product is said to be positive definite or Euclidean, and a vector space $V$ with such an inner product defined on it is called a Euclidean vector space. Condition (RIP3’) implies condition (RIP3), for if there exists a non-zero vector $u$ such that $u\cdot v=0$ for all $v\in V$ then $u\cdot u=0$ (on setting $v=u)$ , which violates (RIP3’). Positive definiteness is therefore a stronger requirement than non-singularity.

Example 5.1

The space of ordinary 3-vectors $a$, $b$, etc. is a Euclidean vector space, often denoted ${\mathbb{E}}^3$ , with respect to the usual scalar product

$$\mathbf{a}\cdot \mathbf{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3=|\mathbf{a}||\mathbf{b}|\cos \theta$$

where $|\mathbf{a}|$ is the length or magnitude of the vector $a$ and $\theta$ is the angle between $a$ and $b$. Conditions (RIP1) and (RIP2) are simple to verify, while (RIP3’) follows from

$$\mathbf{a}\cdot \mathbf{a}={\mathbf{a}}^2=(a_1{)}^2+(a_2{)}^2+(a_3{)}^2>0\text{ if }\mathbf{a}\ne 0.$$

This generalizes to a positive definite inner product on ${\mathbb{R}}^n$ ,

$$\mathbf{a}\cdot \mathbf{b}=a_1{b}_1+a_2{b}_2+\cdots +a_n{b}_n=\sum _{i=1}^n{a}_i{b}_i,$$

the resulting Euclidean vector space denoted by ${\mathbb{E}}^n$ .

The magnitude of a vector $w$ is defined as $w\cdot w$ . Note that in a pseudo-Euclidean space the magnitude of a non-vanishing vector may be negative or zero, but in a Euclidean space it is always a positive quantity. The length of a vector in a Euclidean space is defined to be the square root of the magnitude.

Two vectors $u$ and $v$ are said to be orthogonal if $u\cdot v=0$ . By requirement (RIP3) there is no non-zero vector $u$ that is orthogonal to every vector in $V$. A pseudo-Euclidean inner product may allow for the existence of self-orthogonal or null vectors $u\ne 0$ having zero magnitude $u\cdot u=0$ , but this possibility is clearly ruled out in a Euclidean vector space.

In Chapter 9 we shall see that Einstein’s special theory of relativity postulates a pseudo-Euclidean structure for space-time known as Minkowski space, in which null vectors play a significant role.

Components of a real inner product

Given a basis $\{e_1,...,e_n\}$ of an inner product space $V$ , set

$$g_{ij}=e_i\cdot e_j=g_{ji},\text{\hspace{1em}(5.1)}$$

called the components of the inner product with respect to the basis $\{e_i\}$ . The inner product is completely specified by the components of the symmetric matrix, for if $u=$ $u^i{e}_i,v=v^j{e}_j$ are any pair of vectors then, on using (RIP1) and (RIP2), we have

$$\mathbf{u}\cdot \mathbf{v}=g_{ij}{u}^i{v}^j.\text{\hspace{1em}(5.2)}$$

If we write the components of the inner product as a symmetric matrix

$$\mathsf{G}=[g_{ij}]=[e_i\cdot e_j]=[e_j\cdot e_i]=[g_{ji}]={\mathsf{G}}^T,$$

and display the components of the vectors $u$ and $v$ in column form as $\mathbf{u}=[u^i],\mathbf{v}=[v^j]$ , then the inner product can be written in matrix notation,

$$\mathbf{u}\cdot \mathbf{v}={\mathbf{u}}^T\mathbf{Gv}.$$

Theorem 5.1 · Nondegeneracy and the Gram matrix

The matrix $G$ is non-singular if and only if condition (RIP3) holds

Proof

To prove the ifpart, assume that $G$ is singular, $\det [g_{ij}]=0$ . Then there exists a non-trivial solution $u^j$ to the linear system of equations

$$g_{ij}{u}^j\equiv \sum _{j=1}^n{g}_{ij}{u}^j=0.$$

The vector $\mathbf{u}={\mathbf{u}}^j{\mathbf{e}}_j$ is non-zero and orthogonal to all $v=v^i{e}_i$

$$u\cdot v=g(u,v)=g_{ij}{u}^i{v}^j=0,$$

in contradiction to (RIP3).

Conversely, assume the matrix $G$ is non-singular and that there exists a vector $u$ violating (RIP3); $u\ne 0$ and $u\cdot v=0$ for all $v\in V$ . Then, by Eq. (5.2), we have

$$u_j{v}^j=0\text{ where }{u}_j=g_{ij}{u}^i=g_{ji}{u}^i$$

for arbitrary values of $v^j$ . Hence $u_j=0(j=1,\dots ,n)$ . However, this implies a non-trivial solution to the set of linear equations $g_{ji}{u}^i=0$ , which is contrary to the non-singularity assumption, $\det$ $[g_{ij}]\ne 0$

Orthonormal bases

Under a change of basis

$$e_i=A_i^j{e}_j^{\prime },e_j^{\prime }=A_j^{\prime k}{e}_k,\text{\hspace{1em}(5.3)}$$

the components $g_{ij}$ transform by

$$\begin{array}{r}{g}_{ij}=e_i\cdot e_j=(A_i^k{e}_k^{\prime })\cdot (A_j^l{e}_l^{\prime })\\ =A_i^k{g}_{kl}^{\prime }{A}_j^l,\end{array}\text{\hspace{1em}(5.4)}$$

where $g_{kl}^{\prime }=e_k^{\prime }\cdot e_l^{\prime }$ . In matrix notation this equation reads

$$\mathsf{G}={\mathsf{A}}^T{\mathsf{G}}^{\prime }\mathsf{A}.\text{\hspace{1em}(5.5)}$$

Using ${\mathsf{A}}^{\prime }=[A^{\prime }{}_k^j]={\mathsf{A}}^{-1}$ , the transformed matrix $G^{\prime }$ can be written

$${\mathsf{G}}^{\prime }=({\mathsf{A}}^{\prime }{)}^T{\mathsf{GA}}^{\prime }.\text{\hspace{1em}(5.6)}$$

An orthonormal basis $\{e_1,e_2,\dots ,e_n\}$ , for brevity written ‘o.n. basis’, consists of vectors all of magnitude $\pm 1$ and orthogonal to each other in pairs,

$$g_{ij}=e_i\cdot e_j={\eta }_i{\delta }_{ij}\text{ where }{\eta }_i=\pm 1,\text{\hspace{1em}(5.7)}$$

where the summation convention is temporarily suspended. We occasionally do this when a relation is referred to a specific class of bases.

Theorem 5.2 · Existence of an orthonormal basis

In any finite dimensional real inner product space $(V,\cdot )$ , with $\dim$ $V=n$ there exists an orthonormal basis $\{e_1,e_2,\dots ,e_n\}$ satisfying Eq. (5.7).

Proof

The method is by a procedure called Gram–Schmidt orthonormalization, an algorithmic process for constructing an o.n. basis starting from any arbitrary basis $\left\{u_1,u_2,\dots ,u_n\right\}$ . For Euclidean inner products the procedure is relatively straightforward, but the possibility of vectors having zero magnitudes in general pseudo-Euclidean spaces makes for added complications.

Begin by choosing a vector $u$ such that $\mathbf{u}\cdot \mathbf{u}\ne 0$ . This is always possible because if $u\cdot u=0$ for all $u\in V$ , then for any pair of vectors $u$, $v$

$$0=(u+v)\cdot (u+v)=u\cdot u+2u\cdot v+v\cdot v=2u\cdot v,$$

which contradicts the non-singularity condition (RIP3). For the first step of the Gram–Schmidt procedure we normalize this vector,

$$e_1=\frac{u}{\sqrt{|u\cdot u)|}}\text{ and }{\eta }_1=e_1\cdot e_1=\pm 1.$$

In the Euclidean case any non-zero vector $u$ will do for this first step, and $e_1\cdot e_1=1$

Let $V_1$ be the subspace of $V$ consisting of vectors orthogonal to $e_1$ ,

$$V_1=\{w\in V\mid w\cdot e_1=0\}.$$

This is a vector subspace, for if $w$ and $w^{\prime }$ are orthogonal to $e_1$ then so is any linear combination of the form $w+a{w}^{\prime }$ ,

$$(w+a{w}^{\prime })\cdot e_1=w\cdot e_1+a{w}^{\prime }\cdot e_1=0.$$

For any $v\in V$ , the vector $v^{\prime }=v-a{e}_1\in V_1$ where $a={\eta }_1(v\cdot e_1)$ , since $v^{\prime }\cdot e_1=v\cdot e_1-$ $({\eta }_1{)}^2\mathbf{v}\cdot {\mathbf{e}}_1=0$ . Furthermore, the decomposition $\mathbf{v}=a{\mathbf{e}}_1+{\mathbf{v}}^{\prime }$ into a component parallel to $e_1$ and a vector orthogonal to $e_1$ is unique, for if $\mathbf{v}=a^{\prime }{\mathbf{e}}_1+{\mathbf{v}}^{\prime \prime }$ where $v^{\prime \prime }\in V_1$ then

$$(a^{\prime }-a)e_1=v^{\prime \prime }-v^{\prime }.$$

Taking the inner product ofboth sides with $e_1$ gives firstly ${\mathbf{a}}^{\prime }=\mathbf{a}$ , and consequently ${\mathbf{v}}^{\prime \prime }={\mathbf{v}}^{\prime }$

The inner product restricted to $V_1$ , as a map $V_1\times V_1\to \mathbb{R}.$ , is an inner product on the vector subspace $V_1$ . Conditions (RIP1) and (RIP2) are trivially satisfied if the vectors $u,$ $v$ and $w$ are restricted to vectors belonging to $V_1$ . To show (RIP3), that this inner product is non-singular, let $v^{\prime }\in V_1$ be a vector such that ${\mathbf{v}}^{\prime }\cdot {\mathbf{w}}^{\prime }=0$ for all $w^{\prime }\in V_1$ . Then $v^{\prime }$ is orthogonal to every vector in $w\in V$ for, by the decomposition

$$w={\eta }_1(w\cdot e_1)e_1+w^{\prime },$$

we have ${\mathbf{v}}^{\prime }\cdot \mathbf{w}=0$ . By condition (RIP3) for the inner product on $V$ this implies $v^{\prime }=0$ , as required.

Repeating the above argument, there exists a vector $u^{\prime }\in V_1$ such that ${\mathbf{u}}^{\prime }\cdot {\mathbf{u}}^{\prime }\ne 0$ . Set

$$e_2=\frac{u^{\prime }}{\sqrt{|u^{\prime }\cdot u^{\prime }|}}$$

and ${\eta }_2=e_2\cdot e_2=\pm 1$ . Clearly $e_2\cdot e_1=0$ since $e_2\in V_1$ . Defining the subspace $V_2$ of vectors orthogonal to $e_1$ and $e_2$ , the above argument can be used again to show that the restriction of the inner product to $V_2$ satisfies (RIP1)–(RIP3). Continue this procedure until $n$ orthonormal vectors $\{e_1,e_2,...,e_n\}$ have been produced. These vectors must be linearly independent, for if there were a vanishing linear combination ${\mathbf{a}}^i{\mathbf{e}}_i=0$ , then performing the inner product of this equation with any $e_j$ gives $a^j=0$ . By Theorem 3.3 these vectors form a basis of $V$. At this stage of the orthonormalization process $V_n=\{0\}$ , as there can be no vector that is orthogonal to every $e_1,\dots ,e_n$ , and the procedure comes to an end.

The following theorem shows that for a fixed inner product space, apart from the order in which they appear, the coefficients ${\eta }_i$ are the same in all orthonormal frames.

Theorem 5.3 · Sylvester’s law of inertia

(Sylvester) The number of $+$ and $-$ signs among the ${\eta }_i$ is independent of the choice of orthonormal basis.

Proof

Let $\{e_i\}$ and $\{f_j\}$ be two orthonormal bases such that

$$e_1\cdot e_1=\cdots =e_r\cdot e_r=+1,e_{r+1}\cdot e_{r+1}=\cdots =e_n\cdot e_n=-1,$$ $$f_1\cdot f_1=\cdots =f_s\cdot f_s=+1,f_{s+1}\cdot f_{s+1}=\cdots =f_n\cdot f_n=-1.$$

If $s>r$ then the vectors $f_1,\dots ,f_s$ and $e_{r+1},\dots ,e_n$ are a set of $s+n-r>n=\dim V$ vectors and there must be a non-trivial linear relation between them,

$$a^1{f}_1+\cdots +a^s{f}_s+b^1{e}_{r+1}+\cdots +b^{n-r}{e}_n=0.$$

The $a^i$ cannot all vanish since the $e_i$ form an l.i. set. Similarly, not all the $b^j$ will vanish. Setting

$$u=a^1{f}_1+\cdots +a^s{f}_s=-b^1{e}_{r+1}-\cdots -b^{n-r}{e}_n\ne 0$$

we have the contradiction

$$u\cdot u=\sum _{i=1}^s(a^i{)}^2>0\text{ and }u\cdot u=-\sum _{j=1}^{n-r}(b^j{)}^2<0.$$

Hence $r=s$ and the two bases must have exactly the same number of $+$ and $-$ signs.

If $r$ is the number of $+$ signs and $s$ the number of $-$ signs then their difference $r-s$ is called the index of the inner product. Sylvester’s theorem shows that it is an invariant of the inner product space, independent of the choice of o.n. basis. For a Euclidean inner product, $r-s=n$ , although the word ‘Euclidean’ is also applied to the negative definite case, $r-s=-n$. If $r-s=\pm (n-2)$ , the inner product is called Minkowskian.

Example 5.2

In a Euclidean space the Gram–Schmidt procedure is carried out as follows:

$$\begin{array}{lll}{f}_1=u_1& e_1=\frac{f_1}{\sqrt{f_1\cdot f_1}}& {\eta }_1=e_1\cdot e_1=1,\\ f_2=u_2-(e_1\cdot u_2)e_1& e_2=\frac{f_2}{\sqrt{f_2\cdot f_2}}& {\eta }_2=e_2\cdot e_2=1,\\ f_3=u_3-(e_1\cdot u_3)e_1-(e_2\cdot u_3)e_2& e_3=\frac{f_3}{\sqrt{f_3\cdot f_3}}& {\eta }_3=e_3\cdot e_3=1,\text{ etc. }\end{array}$$

Since each vector has positive magnitude, all denominators $\sqrt{f_i\cdot f_i}>0,$ , and each step is well-defined. Each vector $e_i$ is a unit vector and is orthogonal to each previous $e_j\left(j<i\right)$

Example 5.3

Consider an inner product on a three-dimensional space having components in a basis $u_1,u_2,u_3$

$$\mathsf{G}=[g_{ij}]=[u_i\cdot u_j]=\left(\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right).$$

The procedure given in Example 5.2 obviously fails as each basis vector is a null vector, $u_1\cdot u_1=u_2\cdot u_2=u_3\cdot u_3=0$ , and cannot be normalized to a unit vector.

Firstly, we find a vector $u$ such that $\mathbf{u}\cdot \mathbf{u}\ne 0$ . Any vector of the form $\mathbf{u}={\mathbf{u}}_1+a{\mathbf{u}}_2$ with $a\ne 0$ will do, since

$$u\cdot u=u_1\cdot u_1+2a{u}_1\cdot u_2+u_2\cdot u_2=2a.$$

Setting $a=1$ gives $u=u_1+u_2$ and $u\cdot u=2$ . The first step in the orthonormalization process is then

$$e_1=\frac{1}{\sqrt{2}}(u_1+u_2),{\eta }_1=e_1\cdot e_1=1.$$

There is of course a significant element of arbitrariness in this as the choice of $u$ is by no means unique; for example, choosing $\begin{array}{r}a=\frac{1}{2}\end{array}$ leads to $\begin{array}{r}{e}_1=u_1+\frac{1}{2}{u}_2\end{array}$

The subspace $V_1$ of vectors orthogonal to $e_1$ consists of vectors of the form $v=a{u}_1+$ $b{u}_2+c{u}_3$ such that

$$v\cdot e_1\propto v\cdot u=(a{u}_1+b{u}_2+c{u}_3)\cdot (u_1+u_2)=a+b+2c=0.$$

Setting, for example, $c=0$ and $a=-b=1$ results in $v=u_1-u_2$ . The magnitude of $v$ is $v\cdot v=-2$ and normalizing gives

$$e_2=\frac{1}{\sqrt{2}}(u_1-u_2),{\eta }_2=e_2\cdot e_2=-1,e_2\cdot e_1=0.$$

Finally, we need a vector $w=a{u}_1+b{u}_2+c{u}_3$ that is orthogonal to both $e_1$ and $e_2$ . These two requirements imply that $a=b=-c$ , and setting $c=1$ results in $w=u_1+u_2-u_3$ Normalizing $w$ results in

$$\begin{array}{c}w\cdot w=(u_1+u_2-u_3)\cdot (u_1+u_2-u_3)=-2\\ ⟹e_3=\frac{1}{\sqrt{2}}(u_1+u_2-u_3),{\eta }_3=e_3\cdot e_3=-1.\end{array}$$

The components of the inner product in this o.n. basis are therefore

$${\mathsf{G}}^{\prime }=[g_{ij}^{\prime }]=[e_i\cdot e_j]=\left(\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right).$$

The index of the inner product is $1-2=-1$

Any pair of orthonormal bases $\left\{{\mathbf{e}}_i\right\}$ and $\{e_i^{\prime }\}$ are connected by a basis transformation

$$e_i=L_i^j{e}_j^{\prime },$$

such that

$$g_{ij}=e_i\cdot e_j=e_i^{\prime }\cdot e_j^{\prime }=g_{ij}^{\prime }=\mathrm{diag}({\eta }_1,\dots ,{\eta }_n).$$

From $Eq.$ . Eq. (5.4) we have

$$g_{ij}=g_{kl}{L}_i^k{L}_j^l,\text{\hspace{1em}(5.8)}$$

or its matrix equivalent

$$\mathsf{G}={\mathsf{L}}^T\mathsf{GL}.\text{\hspace{1em}(5.9)}$$

For a Euclidean metric $\mathsf{G}=\mathsf{I}.$ , and $L$ is an orthogonal transformation, while for a Minkoswkian metric with $n=4$ the transformations are Lorentz transformations discussed in Section 2.7. As was shown in Chapter 2, these transformations form the groups $O(n)$ and $O$(3, 1) respectively. The general pseudo-orthogonal inner product results in a group $O(p,q)$ of pseudo-orthogonal transformations of type $(p,q)$

Problems

Problem 5.1

Let $(V,\cdot )$ be a real Euclidean inner product space and denote the length of a vecto $x\in V$, $|x|=\sqrt{x\cdot x}$ . Show that two vectors $u$ and $v$ are orthogonal if $|u+v{|}^2=|u{|}^2+|v{|}^2$

Problem 5.2

Let

$$\mathsf{G}=[g_{ij}]=[u_i\cdot u_j]=\left(\begin{array}{ccc}0& 1& 0\\ 1& 0& -1\\ 0& -1& 1\end{array}\right)$$

be the components of a real inner product with respect to a basis $u_1,u_2,u_3$ . Use Gram–Schmid orthogonalization to find an orthonormal basis $e_1,e_2,e_3$ , expressed in terms of the vectors $u_i$ , and find the index of this inner product.

Problem 5.3

Let $G$ be the symmetric matrix of components of a real inner product with respect to a basis $u_1,u_2,u_3$ ,

$$\mathsf{G}=[g_{ij}]=[u_i\cdot u_j]=\left(\begin{array}{ccc}1& 0& 1\\ 0& -2& 1\\ 1& 1& 0\end{array}\right).$$

Using Gram–Schmidt orthogonalization, find an orthonormal basis $e_1,e_2,e_3$ expressed in terms of the vectors $u_i$

Problem 5.4

Define the concept of a ‘symmetric operator $S:V\to V$ as one that satisfies

$$(Su)\cdot v=u\cdot (Sv)\text{ for all }u,v\in V.$$

Show that this results in the component equation

$$S_i^k{g}_{kj}=g_{ik}{S}_j^k,$$

equivalent to the matrix equation

$${\mathsf{S}}^T\mathsf{G}=\mathsf{GS}.$$

Show that for an orthonormal basis in a Euclidean space this results in the usual notion of symmetry, but fails for pseudo-Euclidean spaces.

Problem 5.5

Let $V$ be a Minkowskian vector space of dimension $n$ with index $n-2$ and let $k\ne 0$ be a null vector $(k\cdot k=0)$ in $V$.

(a) Show that there is an orthonormal basis $e_1,\dots ,e_n$ such that

$$k=e_1-e_n.$$

(b) Show that if $u$ is a ‘timelike’ vector, defined as a vector with negative magnitude $u\cdot u<0$ , then $u$ is not orthogonal to $k$.

(c) Show that if $v$ is a null vector such that $v\cdot k=0$ , then $v\propto k.$

(d) If $\cdot n\ge 4$ which of these statements generalize to a space of index $n-42$