# 1.1 Vector spaces

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Description of vector spaces, subspaces, bases and spans.

## Vector spaces

Definition 1 A linear vector space $\left(X,R,+,·\right)$ is given by a signal space $X$ (called vectors), a set of scalars $R$ , an addition operation $+:X×X\to X$ , and a multiplication operation $·:R×X\to X$ , such that:

1. $X$ forms a group under addition:
1. $\forall \phantom{\rule{3.33333pt}{0ex}}x,y\in X\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\exists !\phantom{\rule{3.33333pt}{0ex}}x+y\in X,\phantom{\rule{1.em}{0ex}}$ (closed under addition)
2. $\exists \phantom{\rule{3.33333pt}{0ex}}0\in X$ such that $0+X=X+0=X.\phantom{\rule{1.em}{0ex}}$ (additive identity)
3. $\forall \phantom{\rule{3.33333pt}{0ex}}x\in X\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\exists \phantom{\rule{3.33333pt}{0ex}}y\in X$ such that $x+y=0,\phantom{\rule{1.em}{0ex}}$ (additive inverse)
4. $\forall \phantom{\rule{3.33333pt}{0ex}}x,y,z\in X\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}x+\left(y+z\right)=\left(x+y\right)+z.\phantom{\rule{1.em}{0ex}}$ (associative law)
2. Multiplication has the following properties: for any $x,y\in X$ and $a,b\in R$ :
1. $a·x\in X,\phantom{\rule{1.em}{0ex}}$ (closure in $X$ under multiplication)
2. $a·\left(b·x\right)=\left(a·b\right)·x,\phantom{\rule{1.em}{0ex}}$ (compatibility)
3. $\left(a+b\right)·x=a·x+b·x,\phantom{\rule{1.em}{0ex}}$ (distributive law over $R$ )
4. $a·\left(x+y\right)=a·x+a·y.\phantom{\rule{1.em}{0ex}}$ (distributive law over $X$ )
3. The set $R$ has the following properties:
1. There exists $1\in R$ s.t. $1·x=x\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\forall \phantom{\rule{3.33333pt}{0ex}}x\in X,\phantom{\rule{1.em}{0ex}}$ (multiplication identity)
2. There exists $0\in R$ s.t. $0·x=0\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\forall \phantom{\rule{3.33333pt}{0ex}}x\in X.\phantom{\rule{1.em}{0ex}}$ (multiplicative null element)

Example 1 Here are some examples of vector spaces:

• $X={\mathbb{R}}^{n}$ (space of all vectors of length $n$ ) over $R=\mathbb{R}$ is a vector space.
• $X={\mathbb{C}}^{n}$ ( $\mathbb{C}$ is complex numbers) over $R=\mathbb{C}$ is a vector space.
• $X={\mathbb{R}}^{n}$ over $R=\mathbb{C}$ is not a vector space, because closure in $X$ under multiplication is not met.
• $X=C\left[T\right]$ (continuous functions in $T$ ) over $R=\mathbb{R}$ is a vector space.

## Subspaces

Definition 2 A subset $M\subseteq X$ is a linear subspace of $X$ if $M$ itself is a linear vector space. Note that, in particular, this implies that any subspace $M$ must obey $0\in M$ .

Example 2 Here are some examples of subspaces:

• In $X={\mathbb{R}}^{2}$ over $R=\mathbb{R}$ , any line that passes through the origin is a subspace of $X$ :
$M=\left\{\left(x,y\right),\in ,{\mathbb{R}}^{2},\phantom{\rule{3.33333pt}{0ex}},\phantom{\rule{3.33333pt}{0ex}},\text{such},\phantom{\rule{4.pt}{0ex}},\text{that},\phantom{\rule{3.33333pt}{0ex}},\phantom{\rule{3.33333pt}{0ex}},\frac{y}{x},=,c\right\}.$
• In $X=C\left[T\right]$ over $R=\mathbb{R}$ , the followings are subspaces of $X$ :
$\begin{array}{cc}\hfill {M}_{1}& =\left\{f\left(x\right)=a{x}^{2}+bx+c:a,b,c\in \mathbb{R}\right\}\hfill \\ \hfill {M}_{2}& =\left\{f\left(x\right):f\left({x}_{0}\right)=0\right\}.\hfill \end{array}$
In contrast, the set ${M}_{3}=\left\{f\left(x\right):f\left({x}_{0}\right)=a\ne 0\right\}$ is not a subspace.

Proposition 1 If $M$ and $N$ are subspaces of $X$ , then $M\cap N$ is also a subspace.

Proof: We assume that $M$ and $N$ hold properties of linear vector space, and show that so does $M\cap N$ :

1. $x,y\in M\cap N\phantom{\rule{1.em}{0ex}}⇒\phantom{\rule{1.em}{0ex}}\left\{\begin{array}{c}x,y\in M\phantom{\rule{1.em}{0ex}}⇒\phantom{\rule{1.em}{0ex}}x+y\in M\\ x,y\in N\phantom{\rule{1.em}{0ex}}⇒\phantom{\rule{1.em}{0ex}}x+y\in N\end{array}\right\}\phantom{\rule{1.em}{0ex}}⇒\phantom{\rule{1.em}{0ex}}x+y\in M\cap N$
2. $\left\{\begin{array}{c}M\phantom{\rule{1.em}{0ex}}\text{linear}\phantom{\rule{4.pt}{0ex}}\text{vector}\phantom{\rule{4.pt}{0ex}}\text{space}⇒0\in M\\ N\phantom{\rule{1.em}{0ex}}\text{linear}\phantom{\rule{4.pt}{0ex}}\text{vector}\phantom{\rule{4.pt}{0ex}}\text{space}⇒0\in N\end{array}\right\}⇒0\in M\cap N$
3. $x\in M\cap N⇒\left\{\begin{array}{c}x\in M\phantom{\rule{1.em}{0ex}}\exists y\in M\phantom{\rule{1.em}{0ex}}\text{s.t.}\phantom{\rule{1.em}{0ex}}x+y=0\\ x\in N\phantom{\rule{1.em}{0ex}}\exists y\in N\phantom{\rule{1.em}{0ex}}\text{s.t.}\phantom{\rule{1.em}{0ex}}x+y=0\end{array}\right\}⇒y\in M\cap N$

The other properties are shown in a similar fashion.

Definition 3 A vector $x\in X$ , where $\left(X,R,+,·\right)$ is a vector space, is a linear combination of a set $\left\{{x}_{1},{x}_{2},...,{x}_{n}\right\}\subseteq X$ if it can be written as $x={\sum }_{i=1}^{n}{a}_{i}·{x}_{i}$ , ${a}_{i}\in R$ . The set of all linear combinations of a set of points $\left\{{x}_{1},{x}_{2},...,{x}_{n}\right\}$ builds a linear subspace of $X$ .

Example 3 $Q={\sum }_{i=0}^{2}{a}_{i}{x}^{i}$ is a linear subspace of $\left(C\left[T\right],\mathbb{R},+,·\right)$ containing the set of all quadratic functions, as it corresponds to all linear combinations of the set of functions $\left\{{x}^{2},x,1\right\}$ .

## Bases and spans

Definition 4 For the set $S=\left\{{x}_{1},{x}_{2},...,{x}_{n}\right\}\subseteq X$ , the span of $S$ is written as

$\left[S\right]=\text{span}\left(S\right)=\left\{x,:,x,=,\sum _{i=1}^{n},{a}_{i},{x}_{i},,,{a}_{i},\in ,R\right\}.$

Example 4 The space of quadratic functions $Q$ is written as $Q=\left[{S}_{1}\right]$ , with ${S}_{1}=\left\{{x}^{2},x,1\right\}$ . The space can also be written as $\left[{S}_{2}\right]$ with ${S}_{2}=\left\{1,x,{x}^{2}-2\right\}\right)$ , i.e. $\left[{S}_{1}\right]=\left[{S}_{2}\right]$ . To prove this, we need to show $\left[{S}_{2}\right]\subseteq \left[{S}_{1}\right]$ and $\left[{S}_{1}\right]\subseteq \left[{S}_{2}\right]$ . For the former case we have

$\begin{array}{cc}\hfill x& ={a}_{1}+{a}_{2}x+{a}_{3}\left({x}^{2}-2\right)\hfill \\ & =\left({a}_{1}-2{a}_{3}\right)+{a}_{2}x+{a}_{3}{x}^{2},\hfill \end{array}$

which means that every element that can be spanned by ${S}_{2}$ , can also be spanned by ${S}_{1}$ , and hence $\left[{S}_{2}\right]\subseteq \left[{S}_{1}\right]$ . The latter case can be shown in a similar manner.

Definition 5 A set $S$ is a linearly independent set if

$\sum _{i=1}^{n}{a}_{i}{x}_{i}=0⇔{a}_{i}=0,\forall \phantom{\rule{3.33333pt}{0ex}}i\in \left\{1,2,...,n\right\}.$

Otherwise, the set $S$ is linearly dependent .

Definition 6 A finite set $S$ of linearly independent vectors is a basis for the space $X$ if $\left[S\right]=X$ , i.e. if $X$ is spanned by $S$ .

Definition 7 The dimension of $X$ is the number of elements of its basis $|S|$ . A vector space for which a finite basis does not exist is called an infinite-dimensional space .

Theorem 1 Any two bases of a subspace have the same number of elements.

Proof: We prove by contradiction: assume that ${S}_{1}=\left\{{x}_{1},...,{x}_{n}\right\}$ and ${S}_{2}=\left\{{y}_{1},...,{y}_{m}\right\}$ , $m>n$ , are two bases for a subspace $X$ with different numbers of elements. We have that since ${y}_{1}\in X$ it can be written as a linear combination of ${S}_{1}$ :

${y}_{1}=\sum _{i=1}^{n}{a}_{i}{x}_{i}.$

Order the elements of ${S}_{1}$ above so that ${a}_{1}$ is nonzero; since ${y}_{1}$ must be nonzero then at least one such ${a}_{i}$ must exist. Solving the above equation for ${x}_{1}$ yields

${x}_{1}=\frac{1}{{a}_{1}}\left({y}_{1},-,\sum _{i=2}^{n},{a}_{i},{x}_{i}\right).$

Thus $\left\{{y}_{1},{x}_{2},{x}_{3},...,{x}_{n}\right\}$ is a basis, in terms of which we can write any vector of the space $X$ , including ${y}_{2}$ :

${y}_{2}={b}_{1}{y}_{1}+\sum _{i=2}^{n}{b}_{i}{x}_{i}.$

Since ${y}_{1},{y}_{2}$ are linearly independent, at least one of the values of ${b}_{i},\phantom{\rule{3.33333pt}{0ex}}i>2,$ must be nonzero. Sort the remaining ${x}_{i}$ so that ${b}_{2}$ is nonzero. Solving for ${x}_{2}$ results in

${x}_{2}=\frac{1}{{b}_{2}}{y}_{2}-\frac{{b}_{1}}{{b}_{2}}{y}_{1}+\sum _{i=3}^{n}\frac{{b}_{i}}{{b}_{2}}{x}_{i}.$

Therefore, $\left\{{y}_{1},{y}_{2},{x}_{3},...,{x}_{n}\right\}$ is a basis for $X$ . Continuing in this way, we can eliminate each ${x}_{i}$ , showing that $\left\{{y}_{1},{y}_{2},...,{y}_{n}\right\}$ is a basis for $X$ . Thus, we have ${y}_{n+1}={\sum }_{i=1}^{n}{c}_{i}{y}_{i}$ , or equivalently:

${c}_{n+1}{y}_{n+1}+\sum _{i=1}^{n}{c}_{i}{y}_{i}=0\phantom{\rule{1.em}{0ex}}\text{with}\phantom{\rule{1.em}{0ex}}{c}_{n+1}=-1.$

As a result, ${S}_{2}$ is linearly dependent and is not a basis. Therefore, all bases of $X$ must have the same number of elements.

## Basis representations

Having a basis in hand for a given subspace allows us to express the points in the subspace in more than one way. For each point $x\in \left[S\right]$ in the span of a basis $S=\left\{{S}_{1},{S}_{2}...{S}_{n}\right\}$ , that is,

$x=\sum _{i=1}^{n}{a}_{i}{S}_{i},$

there is a one-to-one map (i.e., an equivalence) between $x\in \left[S\right]$ and $a=\left\{{a}_{1},...,{a}_{n}\right\}\in {R}^{n}$ , that is, both $x$ and $a$ uniquely identify the point in $S$ . This is stated more formally as a theorem.

Theorem 2 If $S$ is a linearly independent set, then

$\sum _{i=1}^{n}{a}_{i}{S}_{i}=\sum _{i=1}^{n}{b}_{i}{S}_{i}$

if and only if ${a}_{i}={b}_{i}$ for $i=1,2...n$ .

Proof: Theorem 1 states that the scalars $\left\{{a}_{1},...,{a}_{n}\right\}$ are unique for $x$ . We begin by assuming that indeed

$\sum _{i=1}^{n}{a}_{i}{S}_{i}=\sum _{i=1}^{n}{b}_{i}{S}_{i}.$

This implies

$\begin{array}{cc}\hfill \sum _{i=1}^{n}{a}_{i}{S}_{i}-\sum _{i=1}^{n}{b}_{i}{S}_{i}& =0,\hfill \\ \hfill \sum _{i=1}^{n}\left({a}_{i}-{b}_{i}\right){S}_{i}& =0.\hfill \end{array}$

Since the elements of $S$ are linearly independent, each one of the scalars of the sum must be zero, that is, ${a}_{i}-{b}_{i}=0$ and so ${a}_{i}={b}_{i}$ for each $i=1,...,n$ .

Example 5 (Digital Communications) A transmitter sends two waveforms:

${S}_{1}\left(t\right)=\sqrt{2/T}cos\left(2\pi {f}_{c}t\right)\phantom{\rule{14.22636pt}{0ex}}t\in \left[0,T\right]\phantom{\rule{14.22636pt}{0ex}}\text{if}\phantom{\rule{4.pt}{0ex}}\text{bit}\phantom{\rule{4.pt}{0ex}}\text{1}\phantom{\rule{4.pt}{0ex}}\text{is}\phantom{\rule{4.pt}{0ex}}\text{transmitted,}$
${S}_{0}\left(t\right)=\sqrt{2/T}sin\left(2\pi {f}_{c}t\right)\phantom{\rule{14.22636pt}{0ex}}t\in \left[0,T\right]\phantom{\rule{14.22636pt}{0ex}}\text{if}\phantom{\rule{4.pt}{0ex}}\text{bit}\phantom{\rule{4.pt}{0ex}}\text{0}\phantom{\rule{4.pt}{0ex}}\text{is}\phantom{\rule{4.pt}{0ex}}\text{transmitted.}$

The signal $r\left(t\right)$ recorded by the receiver is continuous, that is, $r\left(t\right)\in C\left[T\right]$ . Assuming that the propagation delay is known and corrected at the receiver, we will have they the received signal must be in the span of the possible transmitted signals, i.e., $r\left(t\right)\in span\left({S}_{1}\left(t\right),{S}_{0}\left(t\right)\right)$ . One can check that ${S}_{1}\left(t\right)$ and ${S}_{2}\left(t\right)$ are linearly independent. Thus, one can use a unique choice of coefficients ${a}_{0}$ and ${a}_{1}$ that denote whether bit 0 or bit 1 is transmitted and contain the amount of attenuation caused by the transmission:

$r\left(t\right)={a}_{1}{S}_{1}\left(t\right)+{a}_{0}{S}_{0}\left(t\right).$

The uniqueness of this representation can only be obtained if the transmitted signals ${S}_{0}\left(t\right)$ and ${S}_{1}\left(t\right)$ are linearly independent. The waveforms above are used in in phase shift keying (PSK); other similar examples include frequency shift keying (FSK) and quadrature amplitude modulation (QAM).

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