Mathematical Methods of Physics/Vector Spaces

As is no doubt seen in elementary Physics, the notion of vectors, quantities that have a "magnitude" and a "direction" (whatever these may be) is very convenient in several parts of Physics. Here, we wish to put this idea on the rigorous foundation of Linear Algebra, to facilitate its further use in Physics. The interested reader is encouraged to look up the Wikibook Linear Algebra for details regarding the intricacies of the topic.

Vector Spaces

Let $F$ be field and let $V$ be a set. $V$ is said to be a Vector Space over $F$ along with the binary operations of addition and scalar product iff

$\mathrm{\forall }𝐀,𝐁,𝐂\in V$

(i) $𝐀+𝐁=𝐁+𝐀$ ...(Commutativity)

(ii) $𝐀+(𝐁+𝐂)=(𝐀+𝐁)+𝐂$ ...(Associativity)

(iii) $\mathrm{\exists }\mathrm{𝟎}\in V$ such that $𝐀+\mathrm{𝟎}=𝐀$ ...(Identity)

(iv)$\mathrm{\exists }(\mathbf{-}𝐀)\in V$ such that $𝐀+\mathbf{-}𝐀=\mathrm{𝟎}$ ...(Inverse)

$\mathrm{\forall }$ $a,b\in F$

(v) $(a+b)𝐀=a𝐀+b𝐀$

(vi)$a(𝐀+𝐁)=a𝐀+a𝐁$

(vii)$a(b𝐀)=(ab)𝐀$

The elements of $V$ are called vectors while the elements of $F$ are called scalars. In most problems of Physics, the field $F$ of scalars is either the set of real numbers $ℝ$ or the set of complex numbers $ℂ$.

Examples of vector spaces:

(i) The set ${ℝ}^n$ over $ℝ$ can be visualised as the space of ordinary vectors "arrows" of elementary Physics.

(ii) The set of all real polynomials $P(x)=a_0+a_{1}x+\dots +a_n{x}^n$ is a vector space over $ℝ$

(iii) Indeed, the set of all functions $f:[a,b]\to ℝ$ is also a vector spaces over $ℝ$, with addition and scalar multiplication defined as is usual.

Although the idea of vectors as "arrows" works well in most examples of vector spaces and is useful in solving problems, the latter two examples were deliberately provided as cases where this intuition fails to work.

Basis

A set $E\subset V$ is said to be linearly independent if and only if

$a_1{𝐄}_1+a_2{𝐄}_2+\dots +a_n{𝐄}_n=\mathrm{𝟎}$ implies that $a_1=a_2=\dots =0$, whenever ${𝐄}_1,{𝐄}_2,\dots ,{𝐄}_n\in E$

A set $E\subset V$ is said to cover $V$ if for every $𝐀\in V$ there exist $e_1,e_2,\dots$ such that $𝐀=e_1{𝐄}_1+e_2{𝐄}_2+\dots$. (we leave the question of finiteness of the number of terms open at this point)

A set $B\subset V$ is said to be a basis for $V$ if $B$ is linearly independent and if $B$ covers $V$.

If a vector space has a finite basis with $n$ elements, the vector space is said to be n-dimensional

As an example, we can consider the vector space ${ℝ}^3$ over reals. The vectors $(1,0,0);(0,1,0);(0,0,1)$ form one of the several possible basis for ${ℝ}^3$. These vectors are often denoted as $\hat{i},\hat{j},\hat{k}$ or as $\hat{x},\hat{y},\hat{z}$

Theorem

Let $V$ be a vector space and let $B=\{{𝐛}_1,{𝐛}_2,\dots ,{𝐛}_n\}$ be a basis for $V$. Then any subset of $V$ with $n+1$ elements is linearly dependent.

Proof

Let $E\subset V$ with $E=\{{𝐮}_1,{𝐮}_2,\dots ,{𝐮}_{n+1}\}$

By definition of basis, there exist scalars $a_{1i},a_{2i},\dots ,a_{ni}$ such that $u_i=\sum _m{a}_{mi}{𝐛}_m$

Hence we can write $\sum _{i=1}^{n+1}{c}_i{𝐮}_i=\mathrm{𝟎}$ as $\sum _{m=1}^n\sum _{i=1}^{n+1}{c}_i{a}_{mi}{𝐛}_m=\mathrm{𝟎}$ that is

$\sum _{m=1}^n{c}_1{a}_{1m}=0$
$\sum _{m=1}^n{c}_2{a}_{2m}=0$
$\dots$
$\sum _{m=1}^n{c}_{n+1}{a}_{3(n+1)}=0$

Which has a nontrivial solution for $c_i$. Hence $E$ is linearly dependent.

If a vector space has a finite basis of $n$ elements, we say that the vector space is n-dimensional

Inner Product

An in-depth treatment of inner-product spaces will be provided in the chapter on Hilbert Spaces. Here we wish to provide an introduction to the inner product using a basis.

Let $V$ be a vector space over $ℝ$ and let $B=\{{𝐛}_1,{𝐛}_2,\dots ,{𝐛}_n\}$ be a basis for $V$. Thus for every member $𝐮$ of $V$, we can write $𝐮=\sum _{i=1}^n{b}_i{𝐛}_i$. $b_i$ are called the components of $𝐮$ with respect to the basis $B$.

We define the inner product as a binary operation $(\cdot ):V\times V\to ℝ$ as $𝐱\cdot 𝐲=\sum _i{x}_i{y}_i$, where $x_i,y_i$ are the components of $𝐱,𝐲$ with respect to $B$

Note here that the inner product so defined is intrinsically dependent on the basis. Unless otherwise mentioned, we will assume the basis $\hat{x},\hat{y},\hat{z}$ while dealing with inner product of ordinary "vectors".

Linear Transformations

Let $U$, $V$ be vector spaces over $F$. A function $T:U\to V$ is said to be a Linear transformation if for all ${𝐮}_1,{𝐮}_2\in U$ and $c\in F$ if

(i)$T({𝐮}_1+{𝐮}_2)=T({𝐮}_1)+T({𝐮}_2)$

(ii)$T(c{𝐮}_1)=cT({𝐮}_1)$

Now let $E=\{{𝐞}_1,{𝐞}_2,\dots ,{𝐞}_m\}$ and $F=\{{𝐟}_1,{𝐟}_2,\dots ,{𝐟}_n\}$ be bases for $U,V$ respectively.

Let ${{𝐞}^{\prime }}_i=T({𝐞}_i)$. As $F$ is a basis, we can write ${{𝐞}^{\prime }}_i=\sum _j{a}_{ij}{𝐟}_j$.

Thus, by linearity we can say that if $T(𝐮)=𝐯$, we can write the components $v_j$ of $𝐯$ in terms of those of $𝐮$ as $v_j=\sum _i{u}_i{a}_{ij}$

The collection of coefficients $a_{ij}$ is called a matrix, written as

$𝐀=\left(\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& \dots & a_{1m}\\ a_{21}& a_{22}& a_{23}& \dots & a_{2m}\\ a_{31}& a_{32}& a_{33}& \dots & a_{3m}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ a_{n1}& a_{n2}& a_{n3}& \dots & a_{nm}\end{array}\right)$ and we can say that $T$ can be represented as a matrix $𝐀$ with respect to the bases $E,F$

Eigenvalue Problems

Let $V$ be a vector space over reals and let $T:V\to V$ be a linear transformations.

Equations of the type $T𝐮=\lambda 𝐮$, to be solved for $𝐮\in V$ and $\lambda \in ℝ$ are called eigenvalue problems. The solutions $\lambda$ are called eigenvalues of $T$ while the corresponding $𝐮$ are called eigenvectors or eigenfunctions. (Here we take $T𝐮=T(𝐮)$)