VECTOR SPACES

The structure of euclidean space (as outlined in Theorem 3.1.1) is important for other ``spaces'' used in applied mathematics. For example, consider the mass-spring arrangement in Fig. 3.2.1a. If the body of mass $m$ is displaced from rest position and released or pushed, oscillatory motion results. If this is done in the absence of external forces and friction is ignored, the displacement $y(t)$ of the body from rest is given by

$$y(t) = C_1 \cos \omega t + C_2\sin \omega t$$

where $\omega = \sqrt{k/m}$ with $k$ being the spring constant. Constants $C_1$ and $C_2$ are determined by how far the body is started from rest and how fast and in what direction it is pushed initially. The set of functions

$$V = \{f\vert f(t) = C_1 \cos \omega t + C_2 \sin \omega t, C_1 \in \mathbb{R}, C_2\in \mathbb{R}\}$$

along with the usual definition (from calculus) of addition of functions and multiplication of a function by a constant turns out to have the properties of $E^n$ outlined in Sec. 3.1.

Because there are many important structures with the properties of euclidean spaces, we put them all under the umbrella concept of vector space.

Definition 3.2.1   A real vector space consists of the following.

(a)

A set $V$ of objects. These objects are called vectors even though they may be functions or matrices in a specific case.

(b)

An operation denoted by $+$ which associates with each pair of vectors v,w in $V$ a vector ${\bf v+w}$ in $V$, called the sum of v and w.

(c)

An operation called scalar multiplication which associates with each real number $r$ and vector v in $V$ a vector $r{\bf v}$ in $V$ that is called the product of $r$ and v.

The operations must be defined in such a way that

$~~~\left.\begin{matrix} {\text{1. Addition is commutative:}} \ {\bf v+w=w+v}.\... ...f v+(w+u) = (y+w) + u}. \hfill \end{matrix}\right\} {\bf v,w,u} {\text{ in }} V$

3.

There exists a zero vector $\theta$ in $V$ such that ${\bf u} + \theta = {\bf u}$ for all vectors u in $V$. Vector $\theta$ is called an additive identity.

4.

For each vector v in $V$ there exists an additive inverse $-{\bf v}$ in $V$ such that ${\bf v} + (-{\bf v}) = \theta$.

$~~~\left.\begin{matrix} {\text{5. }} (rs){\bf v} = r(s{\bf v})\hfill \\ {\te... ... r({\bf v+w}) = r{\bf v} + r{\bf w}\hfill \end{matrix}\right\} r,s\in\mathbb{R}$

8.

$1{\bf v}={\bf v}$ for every v in $V$.

It is important that a real vector space consist of the set of vectors and the two operations with certain properties. The same set of vectors with different operations may not satisfy the required properties.

Note that the properties which we derived for $E^n$ have become the defining properties, or axioms, for a vector space. The reason that this works well is that $E^n$, although a specific example, possesses the important qualities for generalization. This happens often in applied mathematics: A specific problem leads to a specific solution--yet the solution actually solves many more problems when it is seen in a larger context.

To show that an object is a vector space, we must show that closure for both operations holds [parts (b) and (c) of the definition] and that properties 1 through 8 hold. Altogether these 10 properties are called the axioms for a real vector space. To show that an object is not a vector space, we need only show that 1 of the 10 axioms fails to hold.

Example 1   Is the set of ordered pairs $V = \{(x_1,x_2)\vert x_1\in \mathbb{R}, x_2\in \mathbb{R}\}$ a vector space?

Solution Until operations of vector addition and scalar multiplication are specified, we cannot test for vector space structure. Therefore, we do not have a vector space.

Example 2   Now $E^n$ is a vector space, because the definitions of the operations imply closure and Theorem 3.1.1 shows that properties 1 to 8 hold.

At this point, we recall that in $E^n$ only the vector $(0,0,\ldots, 0)$ had the property of being an additive identity. Also, given $(x_1,\ldots, x_n)$ in $E^n$, only $(-x_1,\ldots, -x_n)$ is an additive inverse. In general, this uniqueness holds in any vector space.

Theorem 3.2.1   Let $V$ be a vector space.

(a)

There exists only one additive identity in $V$.

(b)

Given x in $V$, there is only additive inverse of x in $V$.

Proof. (a)  Let $\theta_1$ and $\theta_2$ be additive identities for $V$. We will show that they are equal. By the additive-identity axiom

$$\theta_1+\theta_2 = \theta_1$$

However,

$$\theta_1 + \theta_2 = \theta_2+\theta_1 = \theta_2$$

$${\text{\scriptsize ~~~~~~~~~~~~~~~~~Commutativity}}^{\displaystyl... ...~~~~~~~~{}^{\displaystyle \nwarrow}{\text{\scriptsize Additive-identity axiom}}$$

Therefore $\theta_1=\theta_1 +\theta_2 = \theta_2$.

(b)  Let x be in $V$. Let w and u be additive inverses. We will show that ${\bf w=u}$. We have

$${\bf x+w} = \theta ={\bf x+u}$$

By commutativity, ${\bf w+x=u+x}$, and upon adding w to both sides of this last equation, we have

$${\bf (w+x) + w = (u+x) + w}$$

Finally, by associativity

$${\bf w} + ({\bf x+w}) = {\bf u+(x+w)}$$

$${\bf w} + \theta = {\bf u} + \theta$$

$${\bf w} = {\bf u}$$

$\qedsymbol$

Example 3   Let $V = \mathcal{M}_{mn} = \{m\times n$ matrices with real entries$\}$, let vector addition be the addition of matrices, and let scalar multiplication be the multiplication of matrices by scalars. Is $V$ with these operations a vector space?

Solution From our work with matrices we know that closure, commutativity, associativity, and distributivity hold. The additive identity is the $m\times n$ matrix with all entries zero. The additive inverse of $A_{m\times n}$ is $-A_{m\times n}$. Clearly $1A_{m\times n} = A_{m\times n}$. Therefore this is a vector space.

Example 4   Let $\mathcal{P}_2 = \{$polynomials $f(x) = a_2x^2 + a_1x+a_0$, where $a_0,a_1,a_2\in \mathbb{R}\}$, and define addition and scalar multiplication as follows:

(a)

Addition. Let $f(x) = a_2x^2 + a_1x+a_0$ and $g(x) = b_2x^2 + b_1x + b_0$. Then define

$$(f+g)(x) = (a_2+b_2)x^2 + (a_1+b_1)x + (a_0+b_0)$$

(b)

Scalar multiplication. Let $r\in\mathbb{R}, f(x) = a_2x^2 + a_1x+a_0$. Define

$$(rf)(x) = (ra_2)x^2 + (ra_1)x + (ra_0)$$

Show that $\mathcal{P}_2$ is a vector space.

Solution Closure follows easily from the definitions since the right-hand sides of the equations in (a) and (b) are polynomials of degree $\le 2$. For commutativity

$$(f+g)(x) = (a_2+b_2)x^2 + (a_1+b_1)x + (a_0+b_0)$$

$$= (b_2+a_2)x^2 + (b_1+a_1)x + (b_0+a_0)$$

$$= (g+f)(x)$$

Associativity follows similarly. The $\theta$ is $0x^2 + 0x+0$, which is the function $f(x)\equiv 0$. The additive inverse $-f$ is simply $-a_2x^2 - a_1x-a_0$. The other axioms are easily verified.

Example 5   Let $\mathcal{P}_n =\{$polynomials $f(x) = a_nx^n + a_{n-1}x^{n-1} +\cdots+ a_2x^2 + a_1x+a_0\}$ with operations defined by [let $g(x) = b_nx^n +\cdots+ b_1x + b_0$ and $r\in \mathbb{R}$]

$$(f+g)(x) = (a_n+b_n)x^n +\cdots+ (a_1+b_1)x + (a_0+b_0)$$

$$(rf)(x) = (ra_n)x^n +\cdots+ (ra_1)x + (ra_0)$$

So $\mathcal{P}_n$ is a real vector space.

As we see more examples of vector spaces, we will be led to theorems about their structure. Theorems are formed by considering examples. For instance, in calculus after showing

$$\frac{d}{dx} x^2 = 2x\qquad \frac{d}{dx} x^3 = 3x^2$$

by using the definition of derivative, we can guess that

$$\frac{d}{dx} x^4 = 4x^3$$

and so on until we formulate the theorem:

$$\frac{d}{dx} x^n = nx^{n-1}\qquad n\ge 1, n {\text{ an integer}}$$

In proving theorems we can use axioms, previous theorems, and facts from earlier mathematics courses.

Example 6   In $E^n, \mathcal{M}_{mn}$, and $\mathcal{P}_n$, what is the result of multiplying a vector by the scalar $r=0$? State a possible theorem.

Solution In $E^n$,

$$0(x_1,\ldots, x_n) = (0x_1,\ldots, 0x_n) = (0,0,\ldots, 0) = \theta$$

In $\mathcal{M}_{mn}$,

$$0 \left(\begin{matrix} a_{11}&\cdots&a_{1n}\\ \noalign{\smallsk... ...p } \multispan3\dotfill\\ \noalign{\smallskip } 0&\cdots&0\end{matrix}\right)$$

And in $\mathcal{P}_n$,

$$0(a_nx^n +\cdots+ a_1x+a_0) = 0x^n +\cdots+ 0x+0=\theta$$

A possible theorem is: If $V$ is a vector space and ${\bf x}\in V$, then $0{\bf x} =\theta$.

The guess in the solution to Example 6 is actually correct.

Theorem 3.2.2   If $V$ is a real vector space and ${\bf x}\in V$, then $0{\bf x} =\theta$.

Proof.

$$0{\bf x} = (0+0){\bf x} = 0{\bf x} +0{\bf x}$$

$$\theta = 0{\bf x} + [-(0{\bf x})] = (0{\bf x}+0{\bf x}) + [-(0{\bf x})]$$

$$= 0{\bf x} + \{0{\bf x} + [-(0{\bf x})]\}$$

$$= 0{\bf x}+\theta$$

$$= 0{\bf x}$$

Therefore, $\theta=0{\bf x}$. $\qedsymbol$

Example 7   This is an important example because it shows that vector addition need not be related to ordinary addition and the zero vector $\theta$ need not involve the real number 0. Let $V = \{x\vert x\in\mathbb{R}, x>0\}$. Define addition and scalar multiplication as follows:

Addition. For $x\in V, y\in V$, define

$$\underbrace{x\oplus y} = \underbrace{xy}$$

$$\stackrel{\hfill\nearrow}{\stackrel{{\text{\scriptsize Vector sum... ...t{\scriptsize Ordinary}}\hfill} {{\text{\scriptsize multiplication}}\hfill}}~~~$$

Multiplication. For $r\in\mathbb{R}, x\in V$, define

$$\underbrace{r\odot x} = \underbrace{x^r}$$

$$\stackrel{{\text{\scriptsize Scalar}}^{\displaystyle\nearrow}\hfi... ...ext{\scriptsize Ordinary}}\hfill} {{\text{\scriptsize exponentiation}}\hfill}}~$$

Show that $V$ with these operations is a vector space. We have used $\oplus$ and $\odot$ to denote the vector space operations, to distinguish them from the ordinary operations in the solution.

Solution Before showing that $V$ with these operations is a vector space, we look at some specific vector sums and scalar multiples. First, note that the vectors are just positive real numbers. So, for addition we have

$$2\oplus 3 = 2 \cdot 3 = 6$$

$$\stackrel{\phantom{Vec}} {\stackrel{{\text{\scriptsize Vector}}\hfill}{{\text{\scriptsize sum}}\hfill}}\quad ~\stackrel{\displaystyle \nwarrow\hfill}{\stackrel{{\text{\scriptsize From}}\hfill}{ {\text{\scriptsize definition}}\hfill}}$$

$$\begin{matrix} \hfill 4\oplus 6 =24&\quad&10\oplus 10 = 100\hfill... ...t{\scriptsize Vector}}&\quad&{\text{\scriptsize definition}}\hfill \end{matrix}$$

Now to show that we have a vector space, we must show that all the properties of the definition are satisfied.

Closure of addition. Let $x\in V, y\in V$. Then $x\oplus y = xy$. Since $x$ and $y$ are positive real numbers and the product of positive real numbers is a positive real number, $xy\in V$. Therefore, $x\oplus y\in V$, and we have closure for addition.

Closure for scalar multiplication. Let $r\in \mathbb{R}$ and $x\in V$. Then $r\odot x=x^r$. Since $x$ is a positive real number, any real power of it is also a positive real number. Therefore $r\odot x\in V$, and we have closure for addition.

1. Commutativity for addition. Let $x\in V, y\in V$. Then

   $$x\oplus y = xy = yx = y\oplus x$$

   $$\begin{matrix}{\text{\scriptsize By definition}}\nearrow\!\!\!\!&... ...oalign{\vspace{-5pt}} &{\text{\scriptsize for real numbers}}\hfill \end{matrix}$$

   
   
   Therefore $x\oplus y = y\oplus x$.
2. Associativity for addition. Let $x\in V, y\in V, z\in V$. Then

   $$(x\oplus y) \oplus z = (xy) \oplus z = (xy)z = x(yz) = x\oplus (yz) = x\oplus (y\oplus z)$$

   $$\begin{matrix}\Big\uparrow~~~~&&\Big\uparrow~~~~~~~~~~~~~~~~\\ {\... ...noalign{\vspace{-5pt}} &{\text{\scriptsize of real numbers}}\hfill \end{matrix}$$

   
   

3. Additive identity. To determine $\theta$, we ask for the equation
   $$x\oplus \theta = x$$
   to hold. The left-hand side is $x\theta$ by definition of $\oplus$. Therefore, $\theta=1$. That is, the positive real number 1 is the only additive identity or ``zero'' for this space. That is,
   $$x\oplus 1 = x$$

4. Additive inverse. For any $x\in V,-x$ must satisfy $x\oplus (-x) = \theta$. Now we must be very careful: $-x$ is not ``minus $x$''; it is the symbol for additive inverse, and $\theta$ is not the real number zero, it is 1. So the equation is
   $$\begin{matrix} &x\oplus (-x)=\theta\\ {\text{\scriptsize By definition}}&&{\text{\scriptsize From 3 above}}\\ &x\cdot (-x)=1 \end{matrix}$$
   Therefore the unique solution for $-x$ is $(-x)=1/x$. Since $x>0$, we know that $1/x>0$, so $-x\in V$.
5. Associative law for scalar multiplication. Let $r\in \mathbb{R}, s\in \mathbb{R}, x\in V$. Then
   $$(rs)\odot x = x^{rs}\quad {\text{and}}\quad r\odot (s\odot x) = r\odot (x^s) = (x^s)^r$$
   By laws of exponents, $(x^s)^r = x^{sr} = x^{rs}$. Therefore, $(rs) \odot x = r \odot (s\odot x)$.

6,7.

Distributive laws. Let $r\in \mathbb{R}, s\in \mathbb{R}, x\in V, y\in V$. Then

$$(r+s) \odot x = x^{r+s} = x^rx^s = x^r \oplus x^s = (r\odot x) \oplus (s\odot x)$$

$$\begin{matrix}\nearrow&\nwarrow&\nwarrow\hfill&\nwarrow\hfill\\ {... ...ultiplication}}\hfill&&&{\text{\scriptsize multiplication}} \hfill \end{matrix}$$

$$r\odot (x\oplus y) = r\odot (xy) = (xy)^r = x^ry^r = x^r\oplus y^r = (r\odot x) \oplus (r\odot y)$$

8.

Let $x\in V$. Then $1\odot x = x^1=x$.

Therefore, $V$ with the operations as defined is a vector space.

Example 8   Let $V = \{x\vert x\in\mathbb{R}, x>0\}$, and define addition and scalar multiplication as follows:

1. Addition. Let $x\in V, y\in V$. Define

   $$x+y = x+y$$

   $$\begin{matrix}~~~~\nearrow&\nwarrow\\ ~~~~{\text{\scriptsize Vect... ...ptsize addition}}\hfill&{\text{\scriptsize of real numbers}}\hfill \end{matrix}$$

   
   

2. Scalar multiplication. Let $r\in\mathbb{R}, x\in V$. Define

   $$rx =r \cdot x$$

   $$~~~~~~\nwarrow$$

   $$~~{\text{\scriptsize Ordinary multiplication}}$$

   
   

Show that $V$ is not a vector space.

Solution To show that the structure is not a vector space, all we have to show is that at least one of the axioms fails to hold true. In this case closure for scalar multiplication fails to hold since if $r<0$, $rx=r\cdot x<0$ and $rx\notin V$. Again we see that the truth of the axioms depends on the set $V$ and the operations.

Example 9   Regarding the mass spring apparatus in Fig. 3.2.1 and the related discussion,

$$V = \{f\vert f(t) = c_1 \cos wt + c_2 \sin wt, t\in \mathbb{R}\}$$

with the usual definition of addition of functions and multiplication of functions is a vector space.

Solution Closure. Let $f(t) = c_1 \cos wt + c_2 \sin wt$ and $g(t) = c_3 \cos wt + c_4 \sin wt$. We have

$$(f+g)(t) = (c_1+c_3)(\cos wt) + (c_2+c_4)(\sin wt)$$

$$(rf)(t) = rc_1 \cos wt + rc_2 \sin wt$$

which are in $V$. Commutative and associative properties hold. The zero vector is 0 $\cos wt + 0 \sin wt$, which is the identically zero function. The additive inverse of $f$ is

$$-c_1 \cos wt - c_2 \sin wt$$

The other properties are left to the reader.

Complex Vector Spaces If in the definition of real vector space we replace real numbers by complex numbers, we have the notion of a complex vector space. The complex vector spaces we use most often in this text are $\mathbb{C}^n$ and $\mathcal{C}_{mn}$, which are defined below.

Definition 3.2.2   Vector space $\mathbb{C}^n$ is the complex vector space consisting of $n$-tuples $(z_1,z_2,\ldots, z_n)$ of complex numbers with the operations

$$(z_1,\ldots, z_n) + (w_1,\ldots, w_n) = (z_1+w_1,\ldots, z_n+w_n)$$

$$c(z_1,\ldots, z_n) = (cz_1,\ldots, cz_n)$$

where $(w_1,\ldots, w_n)$ is a complex $n$-tuple and $c$ is any complex number.

Definition 3.2.3   $\mathcal{C}_{mn}$ is the complex vector space of $m\times n$ matrices with complex number entries along with the standard matrix operations of addition and scalar multiplication.

The zero vectors for $\mathbb{C}^n$ and $\mathcal{C}_{mn}$ are, respectively, the same as the zero vectors for $E^n$ and $\mathcal{M}_{mn}$, as can be verified directly.

In the remainder of the text, the term vector space will mean the real vector space unless we are working specifically with a complex vector space. When you are working with complex vector spaces, it is important to remember that the vectors can be constructed by using complex numbers and that the scalars for scalar multiplication can be any complex number.

Example 10   Let $V = \{$hermitian $n\times n$ matrices$\}$, and give $V$ the usual matrix operations. Is $V$ a vector space?

Solution Since there is no restriction on the entries of the matrices, we are checking to see whether $V$ with these operations is a complex vector space. Recall that the main-diagonal entries of a hermitian matrix are real numbers. Thus, in general, if $A$ is hermitian, the scalar multiple $iA$ is not hermitian. Therefore $V$ is not closed under scalar multiplication and is not a vector space.