Groups, Fields and Vector Spaces

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Class Notes For Advanced EngineeringMathematics I

(ENM510)

Michael A. Carchidi

September 20, 2010

Chapter 2 - Groups, Fields and Linear Vector Spaces

The following notes are written by Dr. Michael A. Carchidi at the Universityof Pennsylvania for the purpose of teaching the contents covered in the ENM510- Advanced Engineering Mathematics I Course.

2.1 Binary Operations and Groups

LetG be a set of elements and let G G be the set of all ordered pairs fromG, i.e.,

G G= {(x, y) |x G and y G}.Abinary operation defined onG is a rule that takes an element,(x, y),ofGGas input and returns an element, z, ofG as output. We write this operation asz= x y and this is shown schematically in the following diagram.

(x, y) (z= x y).

In other words, a binary operation on a set G is just a mapping from GG toG, i.e.,

T :G G G such that T(x, y) =x y.Note that by definition a set is always closed under a binary operation definedon the set, which means for any x andy, in G, x y is also in G. For example,


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division would not be a binary operation defined on the positive integers (Z+)

since a b

is not necessarily a positive integer, i.e., 2 3 / Z+. On the other hand multipli-cation would be a binary operation defined on the positive integers (Z+) since

a b

is a positive integer.

Note that a binary operation defined on a set G can also be viewed as a rulethat takes two elements ofG (in some order) as input and provides some element

ofG as output. Note that order is important since, in general,a b6=b a, i.e.,236= 3 2.

LetG be a set of elements with a binary operation defined onG and denotedby . The set G together with this binary operation is called a group, and isdenoted byhG, i, if the following 3 properties hold.

a.) For alla,b and c in G, it is true that (a b) c=a (b c), i.e., the binaryoperation isassociative.

b.) There exist an element inGcalled theidentity(and denoted bye) such that

a e= e a= a for all a inG.c.) For each a in G, there exist its inverse in G (denoted by a0) such that

a a0 =a0 a= e.

Note that the property of commutation (i.e., a b= b a) need not hold forhG, i to be a group and a group for which a b = b a for all a andb in G iscalled a commutative (or Abelian) group.

Example

The set of integers Zunder addition forms a group. In fact, it forms an Abeliangroup.

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Abelian group. This is called a Lie Group because the elements in G are all

functions of a continuous parameter .

Example: A Finite Group

Consider the set G = {1, +1} with multiplication as the binary operation.You can easily check that this forms an Abelian group. It contains only twoelements and if hence called a finite group. Consequently, this is not an exampleof a Lie group.

Example: A Non-Abelian Group

LetGn be the set of all invertiblennmatrices with matrix multiplication asthe binary operation. This set forms a group and in this case we have anon-abeliangroup since the multiplication of matrices is not commutative.

2.1 Fields

Let Fbe a set of elements with two binary operations defined on F, calledadditionand denoted by andmultiplicationand denoted by . The setFtogether with these operations is called a field, and is denoted by hF,,i, if thefollowing 10 properties hold.

a1.) For allaandbinF, it is true thatab= ba, i.e., addition iscommutative.

a2.) For alla,b and c in F, it is true that (a b) c= a (b c), i.e., additionisassociative.

a3.) There exist an element called the additive identity(and denoted by ) suchthata = a= a for all a inF.

a4.) For each a in F, there exist an additive inverse(denoted by a0) such thata a0 =a0 a= 0.

b1.) For all a and b in F, it is true that a

b = b

a, i.e., multiplication iscommutative.

b2.) For all a, b and c in F, it is true that (a b) c = a (b c), i.e.,multiplication isassociative.

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b3.) There exist an element called themultiplicative identity(and denoted by )

such that a = a= a for all a in F, and 6=.

b4.) For eachainFexcept for the additive identity0, there exist amultiplicativeinverse(denoted bya00) such that a a00 =a00 a= 1.

c1.) For all a, b andc inF, it is true that a (b c) = (a b) (a c), i.e.,right distributivity.

c2.) For alla,b andc in F, it is true that (a b) c= (a c) (b c), i.e.,leftdistributivity.

Note that we will drop the abstract symbols and and simply use +

for addition (i.e., write a+ b instead of a b), and we will use no symbol formultiplication, (i.e., write ab instead of a b). In addition, we shall use thesymbols0 in place of,1 in place of,a in place ofa0, and a1 in place ofa00.With these conventions, the above field properties become more familiar and arewritten as follows.

a1.) For allaandbinF, it is true thata+b= b +a, i.e., addition iscommutative.

a2.) For alla,b andc in F, it is true that (a + b) + c= a + (b + c), i.e., additionisassociative.

a3.) There exist an element called the additive identity(and denoted by0) suchthata + 0 = 0 + a=a for all a in f .

a4.) For each a inf, there exist an additive inverse (denoted bya) such thata + (a) = (a) + a= 0.

b1.) For all a and b in F, it is true that a b = b a, i.e., multiplication iscommutative.

b2.) For alla,b andc in F, it is true that(a b) c=a (b c), i.e., multiplicationisassociative.

b3.) There exist an element called themultiplicative identity(and denoted by1)such that a 1 = 1 a= a for alla in F, and1 6= 0.

b4.) For eachainFexcept for the additive identity0, there exist amultiplicativeinverse(denoted bya1) such that a a1 =a1 a= 1.

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c1.) For all a,b and c inF, it is true that a (b + c) = (a b) + (a c), i.e.,right

distributivity.

c2.) For all a, b andc in F, it is true that (a+b)c = (ac) + (bc), i.e., leftdistributivity.

We shall call the elements of a fieldscalars.

Some Examples of Fields

We shall assume that addition of two elements a and b in a field F is represented

bya + band multiplication of two elementsa and b in a fieldFis represented byab. Some examples of common fields F include:

F =R, the field of all realnumbers under ordinary real addition and realmultiplication,

F = C, the field of all complexnumbers under ordinary complex additionand complex multiplication, i.e.

C=

a + ib | a R,b R and i =

1with

(a1+ ib1) + (a2+ ib2) = (a1+ a2) + i(b1+ b2)

and(a1+ ib1)(a2+ ib2) = (a1a2 b1b2) + i(b1a2+ a1b2).

F = Q, the field of all rationalnumbers under ordinary fraction additionand fraction multiplication, i.e.,

Q=

p

q

p Z,q Zandq6= 0

with p1q1

+p2

q2= p

1q2+ q1p2q1q2

& p1

q1p

2

q2= p

1p2q1q2

.

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F =Fp, an extension field of rational numbers, which is defined by the set

Fp= {a + bp | a, b Q and pa prime integer}

with addition defined by

(a1+ b1

p) + (a2+ b2

p) = (a1+ a2) + (b1+ b2)

p

and multiplication defined by

(a1+ b1

p)(a2+ b2

p) = (a1a2+pb1b2) + (a1b2+ a2b1)

p.

F = Zp

, the field of all integers from 0 to p

1, inclusive (for p prime)under addition and multiplication mod p. This means for a and b in Zp,a + bequals the remainder you get when the sumofa and b is divided bypandab equals the remainder you get when the productofa and b is dividedbyp. That is,

a + b= (a + b)mod(p) and a b= (ab)mod(p).

For example, in the fieldhZ11, +, i,additionandmultiplicationare definedby the tables

+ 0 1 2 3 4 5 6 7 8 9 10

0 0 1 2 3 4 5 6 7 8 9 101 1 2 3 4 5 6 7 8 9 10 02 2 3 4 5 6 7 8 9 10 0 13 3 4 5 6 7 8 9 10 0 1 24 4 5 6 7 8 9 10 0 1 2 35 5 6 7 8 9 10 0 1 2 3 46 6 7 8 9 10 0 1 2 3 4 57 7 8 9 10 0 1 2 3 4 5 68 8 9 10 0 1 2 3 4 5 6 79 9 10 0 1 2 3 4 5 6 7 8

10 10 0 1 2 3 4 5 6 7 8 9

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and

0 1 2 3 4 5 6 7 8 9 100 0 0 0 0 0 0 0 0 0 0 01 0 1 2 3 4 5 6 7 8 9 102 0 2 4 6 8 10 1 3 5 7 93 0 3 6 9 1 4 7 10 2 5 84 0 4 8 1 5 9 2 6 10 3 75 0 5 10 4 9 3 8 2 7 1 66 0 6 1 7 2 8 3 9 4 10 57 0 7 3 10 6 2 9 5 1 8 48 0 8 5 2 10 7 4 1 9 6 39 0 9 7 5 3 1 10 8 6 4 2

10 0 10 9 8 7 6 5 4 3 2 1

respectively.

Note thatR,C,Q, andFp are fields consisting of an infinite number of elementswhileZpis a field consisting of a finite number (namelyp). In this course, we willbe concerned only with thefields R and C.

2.2 An Example of an Exotic Field

Consider the set of real numbersR with the usual definition of multiplication,

but not the usual defi

nition of addition. Let+and no symbol be ordinary additionand multiplication of real numbers and let use define the binary operation ofaddition as

x y= (x3 + y3)1/3

and the binary operation of multiplication as

x y = xy.

For example2 3 = (23 + 33)1/3 = (351/3

Let us show that hR,,i does forms a field. Toward this end we need to show

that propertiesa1 - a4,b1 - b4 and c1 and c2 all hold.

a1.) Commutativity of Addition: Starting with

x y= (x3 + y3)1/3

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we have

x y= (x3

+ y3

)1/3

= (y3

+ x3

)1/3

=y xfor allx and y inR.

a2.) Associativity of Addition: Starting with

x y= (x3 + y3)1/3

we have

(x y) z = (x3 + y3)1/3 z

= ((x3 + y3)1/3)3 + z3)1/3

= (x3 + y3 + z3)1/3

= (x3 + ((y3 + z3)1/3)3)1/3

= x (y3 + z3)1/3

= x (y z)

and so(x y) z= x (y z)

for allx,y andz inR.

a3.) The Existence of an Additive Identity: Note that = 0is an additive identity

sincex = (x3 +3)1/3 = (x3 + 03)1/3 = (x3)1/3 =x

and x= (3 + x3)1/3 = (03 + x3)1/3 = (x3)1/3 =x

for allx in R.

a4.) The Existence of an Additive Inverse for each element of V: For each ele-mentx, we note that

x (

x) = (x3 + (

x)3)1/3 = (x3

x3)1/3 = 0

and(x) x= ((x)3 + x3)1/3 = (x3 + x3)1/3 = 0

so that for each element x, there exist and additive inversex.

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b.) Since multiplication is defined in the usual way and the additive identity is

0, the four properties b1 through b4 all hold.

c1.) Right Distributivity: Using the definitions of addition and multiplication wenote that

x (y z) = x ((y3 + z3))1/3

= x((y3 + z3))1/3

= (x3(y3 + z3))1/3

= (x3y3 + x3z3)1/3

= ((xy)3 + (xz)3)1/3

= ((x

y)3

+ (x

z)3

)1/3

= (x y) (x z)

and sox (y z) = (x y) (x z)


c2.) Left Distributivity: Using the definitions of addition and multiplication wenote that

(x y) z = ((x3 + y3)1/3) z

= ((x3 + y3)1/3)z= ((x3 + y3)z3)1/3

= (x3z3 + y3z3)1/3

= ((xz)3 + (yz)3)1/3

= ((x z)3 + (y z)3)1/3

= (x z) (y z)

and so(x y) z= (x z) (y z)


The above calculations show thatR is a field of with the operations, and as defined above.

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2.3 Vector Spaces

Let hF, +, i be a field. A vector space over a field F is a non-empty set Vof elements (called vectors) together with two algebraic operations called vectoraddition () and scalar multiplication () such that the following 10 propertiesare true.

.) For all x andy in V, x y V, that is, the set V is closed under vectoraddition.

1.) For all x andy inV, it is true that x y= y x, i.e., vector addition iscommutative.

2.) For allx, y and z in V, it is true that (x y) z= x (y z), i.e., vectoraddition isassociative.

3.) There exist an element called the additive identity(and denoted by ) suchthatx 0= 0 x= x for all xin V.

4.) For each x inV, there exist an additive inverse (denoted by x0) such thatx x0 =x0 x= 0.

.) For all Fand all x inV, x V, that is, the set V isclosed underscalar multiplication over F.

1.) For all Fand all x and y inV, it is true that (x y) = ( x) ( y), i.e.,distributivityover vector addition.

2.) For all and inFand all x in V, it is true that ( + ) x= ( x) ( x), i.e.,distributivityover scalar addition.

3.) For all andinFand all x in V, it is true that ( )x= (x),i.e.,multiplicityof scalars.

4.) If1 is the multiplicative identity inF, then1 x= x for all xin V.

Note that we will drop the special symbols ,and, and write (for example)2 and 3 as

(+)x= x +x and ()x= (x),

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respectively, and it should not cause any confusion as long as we know what are

thescalars (italics font) and what are the vectors (bold font). In addition weshall use the symbol 0 for the additive identity in the field and the symbol 0 forthe additive identity in the vector space. Also we shall use the symbol xin placeofx0 for the additive inverse ofx.

We could denote the vector space by hV, ,; F, +, i, but we just use thesymbolVwith the field and operations understood. The fieldFunder the vectorspace V is called the coefficientfield for V. IfF = R, then V is called a realvector space, and ifF =C,V is called a complex vector space. In this course, weshall be concerned mostly with real and complex vector spaces.

Note that it can easily be shown in any vector space that

0 x= 0 or 0x= 0

for all x V, and(1) x= x

for all x V, and 0= 0 or 0= 0

for all F.

2.4 Numerous Examples

We now present examples of vector spaces, many of which we will encounterand need for this course.

Example: Mnm(F)

This vector space is given by

Mnm(F) =

[aij]nm | aij F

which is the set of alln mmatrices whose entries come from the fieldhF, +, i,with vector addition and scalar multiplication given in the usual way as

[aij]nm [bij]nm= [aij+ bij]nm

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and

[aij]nm= [ aij]nm

for F. Thedimension ofMnm(F) is given by nm. We will shortly definethe dimension of a vector space.

Example: Fn andFm

These vector spaces are special cases ofMnm(F)and are given by

Fn =Mn1(F) and Fm= M1m(F)

which are the sets of all n1columnmatrices and 1m rowmatrices whose entries

come from the fieldF, with vector addition and scalar multiplication defined asinMnm(F).The dimension ofF

n is given byn and the dimension ofFmis givenbym.

Example: Pn[x; F]

This vector space is defined by

Pn[x; F] =

( n

Xk=0akx

k

ak F)

,

which is the set of all polynomials in x havingdegree less than or equal to nandwhose coefficients come from the fieldF, with vector or polynomial addition andscalar multiplication defined by

nXk=0

akxk

nXk=0

bkxk =

nXk=0

(ak+ bk)xk

and

n

Xk=0akx

k =n

Xk=0(ak)x

k

for F. The dimension ofPn[x; F] is given byn + 1.

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Example: P[x; F]

This vector space is defined by

P[x; F] =

( Xk=0

akxk

ak F

)

which is the set of allpolynomials inx withno limitation on its degreeand whosecoefficients come from the fieldF, with vector or polynomial addition and scalarmultiplication defined by

Xk=0 akxk

Xk=0 bkxk =

Xk=0(ak+ bk)xkand

Xk=0

akxk =

Xk=0

(ak)xk

for F. The dimension ofP[x; F] is infinite.

Example: C[x, R; a, b]

This vector space consist of all continuous real-valued functions in x over the

intervala x b, with vector addition and scalar multiplication given by(f g)(x) =f(x) + g(x) and ( f)(x) =f(x)

for R. The dimension ofC[x, R; a, b] is infinite.

Example: B[x, R; a, b]

This vector space consist of all bounded real-valued functions in x over theintervala x b, with vector addition and scalar multiplication given by

(f g)(x) =f(x) + g(x) and ( f)(x) =f(x)for R. The dimension ofB [x, R; a, b]is infinite.

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Example: D[x, R; a, b]

This vector space consist of all differentiable real-valued functions in x overthe intervala < x < b, with vector addition and scalar multiplication given by

(f g)(x) =f(x) + g(x) and ( f)(x) =f(x)

for R. The dimension ofD[x, R; a, b]is infinite.

Example: D2[x, R; a, b]

This vector space consist of all twice differentiable real-valued functions in x

over the interval a < x < b, with vector addition and scalar multiplication givenby

(f g)(x) =f(x) + g(x) and ( f)(x) =f(x)for R. The dimension ofD2[x, R; a, b] is infinite.

Example: Dn[x, R; a, b]

This vector space consist of alln-times differentiable real-valued functions inxover the intervala < x < b, with vector addition and scalar multiplication givenby

(f g)(x) =f(x) + g(x) and ( f)(x) =f(x)for R. The dimension ofDn[x, R; a, b] is infinite.

Example: D[x, R; a, b]

This vector space consist of all infinitely differentiable real-valued functions inxover the intervala < x < b, with vector addition and scalar multiplication givenby

(f g)(x) =f(x) + g(x) and ( f)(x) =f(x)for

R. The dimension ofD[x, R; a, b] is infinite.

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Example: C[x,y,R; S]

This vector space consist of all continuous real-valued functions inxandyoversome two dimensional region S, with vector addition and scalar multiplicationgiven by

(f g)(x, y) =f(x, y) + g(x, y) and ( f)(x, y) =f(x, y)

for R. The dimension ofC[x,y,R; S] is infinite.

Example: B[x,y,R; S]

This vector space consist of all bounded real-valued functions in x and y oversome two dimensional region S, with vector addition and scalar multiplicationgiven by


for R. The dimension ofB [x,y,R; S]is infinite.

Example: D[x,y,R; S]

This vector space consist of all differentiable real-valued functions in x andy

over some two dimensional region S, with vector addition and scalar multiplicationgiven by


for R. This means that both f /x and f /y exist for a function f inD[x,y,R; S].The dimension ofD[x,y,R; S]is infinite.

Example: D2[x,y,R; S]

This vector space consist of all twice differentiable real-valued functions in

x and y over some two dimensional region S, with vector addition and scalarmultiplication given by


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for R. This means that: 2f /x2, 2f /y2 and2f /xy = 2f /yx allexist for a function f inD

2

[x,y,R; S].The dimension ofD2

[x,y,R; S] is infinite.

Note that we can define Dn[x,y,R; S] for any positive integer n and we candefineD[x,y,R; S], just like we did above for functions of a single variable.

Example: P[x, R; T]

This vector space consist of all piecewise continuous real-valued periodic func-tions in x having period T, so that f(x+T) = f(x) for all x R, with vectoraddition and scalar multiplication given by

(f g)(x) =f(x) + g(x) and ( f)(x) =f(x)for R. The dimension ofP[x, R; T]is infinite.

2.5 An Example of An Exotic Vector Space

LetVbe the set of the3 1 column matrices,

V =

xy

z

x,y,z R, withx >0

with vector addition and scalar multiplication defined as xy

z

x

0

y0

z0

=

xx

0

(y3 + y03)1/3

z+ z0 2

and

xy

z

=

x

1/3yz+ 2(1 )

,

respectively, let us show that Vis a vector space over the field of real numbers R.

.) Closure Under Vector Addition: First we must show thatV isclosedunder

vector addition,. For xy

z

and

x

0

y0

z0

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elements ofV wefind that xy

z

x

0

y0

z0

=

xx

0

(y3 + y03)1/3

z+ z0 2

is also an element ofV since x andx0 real and positive imply that xx0 isreal and positive. Alsoy, y0, z andz0 all real imply that (y3 +y03)1/3 andz+ z0 2are real. ThereforeV is closed under vector addition, . Next weconcentrate of the four properties of vector addition.

1.) Commutativity:

xy

z

x0y0

z0

=

xx0(y3 + y03)1/3

z+ z0 2

=

x0x(y03 + y3)1/3

z0 + z 2

=

x0y0

z0

xy

z

and so xy

z

x

0

y0

z0

=

x

0

y0

z0

xy

z

.

2.) Associativity:

xyz

x

0

y0

z0

x00

y00

z00

= xy

z

x0

x00

(y03 + y003)1/3

z0 + z00 2

=

x(x

0x00)(y3 + ((y03 + y003)1/3)3)1/3

z+ (z0 + z00 2) 2

=

xx

0x00

(y3 + y03 + y003)1/3

z+ z0 + z00 2 2

=

(xx0)x00

((y3 + y03)1/3)3 + y003)1/3

(z+ z0

2) + z00

2

=

xx0

(y3 + y03)1/3

z+ z0

2

x00

y00

z00

=

xy

z

x

0

y0

z0

x

00

y00

z00

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and so xy

z

x0y0

z0

x00y00

z00

=

xy

z

x0y0

z0

x00y00

z00

.

3.) The Existence of an Additive Identity (Zero Vector): To determine the can-didate for thezero vector (if it exists), we write

0= 0

xy

z

=

x

0

01/3y0z+ 2(1 0)

=

10

2

.

To check that this is the zero vector forV, we form xy

z

10

2

=

10

2

xy

z

=

1x(03 + y3)1/3

2 + z 2

=

xy

z

.

4.) The Existence of an Additive Inverse for each element of V: For each ele-ment,

xyz

inV, the additive inverse, (if it exists) should satisfy xy

z

0

= (1) xy

z

=

x

1

(1)1/3y(1)z+ 2(1 (1))

=

1/xy

4 z

.

To check that this is the additive inverse for xy

z

we calculate, xy

z

1/xy

4 z

=

1/xy

4 z

xy

z

=

(1/x)x((y)3 + y3)1/3

(4 z) + z 2

=

10

2

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and so xy

z

xy

z

0

=

xy

z

1/xy

4 z

=

10

2

=0.

which is the additive identity (or zero vector) for ..) Closure Under Scalar Multiplication: Next we check to see if V is closed

under scalar multiplication,. For

xy

z

inV and in R, we have

xy

z

=

x

1/3yz+ 2(1 )

which is also inV, sincex real and positive and real imply thatx is realand positive (i.e., one may always choose the positive value for x ), e.g.,

x1/2

= x.In addition, fory, zand all real, the quantities 1/3y andz+ 2(1 )are both real. ThereforeV is closed under scalar multiplication, . Nextwe concentrate of the four properties of scalar multiplication.

1.) Distributivity over Vector Addition:

xy

z

x

0

y0

z0

=

xx

0

(y3 + y03)1/3

z+ z0 2

=

(xx

0)

1/3(y3 + y03)1/3

(z+ z0 2) + 2(1 )

=

x

x0

(y3 +y03)1/3

z+z0 2+ 2 2

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=

xx0

((1/3

y)3

+ (1/3

y0

)3

)1/3

(z+ 2(1 )) + (z0 + 2(1 )) 2

=

x

1/3yz+ 2(1 )

x

0

1/3y0

z0 + 2(1 )

=

xy

z

x

0

y0

z0

and so

xy

z

x0

y0

z0

=

xyz

x0

y0

z0

.

2.) Distributivity over Scalar Addition:

(+)

xy

z

=

x

+

(+)1/3y(+)z+ 2(1 (+))

=

xx

(y3 +y3)1/3

z+z+ 2(1

)

=

x

x

((1/3y)3 + (1/3y)3)1/3

(z+ 2(1 )) + (z+ 2(1 )) 2

=

x

1/3yz+ 2(1 )

x

1/3yz+ 2(1 )

=

xy

z

xy

z

and so

(+)

xy

z

=

xy

z

xy

z

.

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3.) Multiplicity among Scalars:

xy

z

=

x

1/3yz+ 2(1 )

=

(x

)

1/3(1/3y)(z+ 2(1 )) + 2(1 )

=

x

()1/3yz+ 2 2+ 2 2

=

x

()1/3yz+ 2(1 )

= () xy

z

and so

xy

z

= ()

xy

z

4.) Scalar Multiplication by the Field Multiplicative Identity:

1 xy

z

= x1

11/3y1z+ 2(1 1)

= xyz

and so

1

xy

z

=

xy

z

The two closure properties along with the eight algebraic properties are allvalid and this show that V is a vector space over the field of real numbers Rusing the operations,

(vector addition) and (scalar multiplication) as defined

above.

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2.6 Subspaces

A subsetUof a vector space Vis a called a subspace ofV ifU(together withthe same operations of vector addition and scalar multiplication as defined onV),is itself a vector space. Since the validity of the algebraic properties1 through4and1through4(in the definition of a vector space) are true for all elementsinVand hence for all elements in U, we see that a non-emptysubset of a vectorspaceV is a subspace ofV if it is:

.) closed under vector addition, and

.) closed under scalar multiplication.

Note that one way to show that U is not empty is to show that the additiveinverse 0 ofV is in U. We denote the fact thatUis a subspace ofV by writingU V, and its is always true for any vector space V that {0} V andV V.

Example: Null Spaces of a Matrix

LetA= [aij]nm be an n m matrix whose entries come from some fieldF.

The right-null space of A, denoted by Rnull{A}, is the subspace of Fm

consisting of allm 1matrix solutions to the equation AnmXm1= 0n1.

The left-null space ofA, denoted by Lnull{A}, is the subspace ofFn con-sisting of all1 nmatrix solutions to the equation X1nAnm= 01m.

Note that since AX= 0 if and only ifXTAT =0, we see that

Rnull{A}= Lnull{AT} and Lnull{A}= Rnull{AT}.

In addition, most books define only the right-null space of a matrix Aandthey call this the null space of the matrix A and they denote it by Null{A}.

Example: Lambda Spaces and Eigenspaces of a Square Matrix

LetA be asquarennmatrix whose entries come from some fieldF, and let be any scalar in F. The-space ofA corresponding to the scalar , denoted

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by {A}, is the subspace ofFn consisting of all n 1 matrix solutions to the

equation AnnXn1 = Xn1. Note that for = 0, we have 0{A}= Rnull{A}.In addition, note that when {A} is not just the trivial subspace consisting ofonly the zero vector in Fn, then is called an eigenvalue of A, and {A} iscalled the eigenspace ofA corresponding to the eigenvalue , and is denoted byEigen{A,}. Any non-zero vectors in Eigen{A,} are called eigenvectors ofAcorresponding to the eigenvalue .

Example: Differential Equations

Consider the vector space D[x, R; a, b]consisting of all infinitely differentiablefunctions on the interval a < x < b. Let U be the set of all functions y in

D[x, R; a, b]that satisfy the second order differential equation

y00(x) + P(x)y0(x) + Q(x)y(x) = 0

for functions P and Q in D[x, R; a, b]. Then we see that U D[x, R; a, b].This follows from the fact that U is not empty since y(x) = 0 is in U, and ify1andy2 are inUso that

y001(x) + P(x)y0

1(x) + Q(x)y1(x) = 0

and

y00

2(x) + P(x)y0

2(x) + Q(x)y2(x) = 0,then

(1y1+2y2)00(x) + P(x)(1y1+2y2)

0(x) + Q(x)(1y1+2y2)(x) = 0

which means that 1y1+ 2y2 is also in U. Since U is a non-empty subset ofD[x, R; a, b]and Uis closed under scalar multiplication and vector addition, wemay conclude thatUis a subspace ofD[x, R; a, b]. We shall see below that thedimension ofU is2.

2.7 Spanning Sets and Linear Independence

LetVbe any vector space over a fieldFand let

M={x1, x2, x3, . . . , xm}

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be a set ofn vectors inV. The set of all linear combinations ofM, i.e.,

Span(M) =( mX

k=1

kxk

k F fork = 1, 2, 3, . . . , m

)is a subspace ofVover the same fieldFand is called the spanning space of theset M. It is the smallest subspace of V that contains the vectors in M. Thismeans any other subspace ofV that contains the vectors in Mmust also containthe vectors in Span(M).

Examples: Column, Row and Null Spaces of a Matrix

LetA= [aij]nm be an n m matrix whose entries come from some fieldF.

Thecolumn spaceofA, denoted by Col{A}, is the subspace ofFn spannedby the columns ofA.

Therow space ofA, denoted by Row{A}, is the subspace ofFm spannedby the rows ofA.

If Span(M) = V, then M is called a spanning set forVand note that ifMis a spanning set for V, then M

A for any other set of vectors A inV is also a

spanning set for V. However, ifMis a spanning set for V, then M Cfor anyproper subset CofMmay, or may not, be a spanning set for V.

This says we may add any vectors to a spanning set of Vand not destroy itsability to span all of V, but if we remove any vectors from a spanning set of V,we may (or may not) destroy its ability to span all of V.

In addition, ifM={x1, x2, x3, . . . , xm} is a spanning set for V, then for anyx V, we can find scalars 1,2,3, . . . ,m F such that

x=

mXk=1

kxk

but the choice of scalars 1,2,3, . . . ,m Fneed not be unique. To lead to aunique choice of scalars, we must define the concept of linear independence.

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A set M={x1, x2, x3, . . . , xm} of a vector space V is calledlinearly indepen-dent ofmXk=1

kxk =0

leads only to 1 = 2 = 3 = = m = 0. If the setM in V is not linearlyindependent, it is called linearly dependent.

Note that ifMis linearly independent inV, thenMCfor any proper subsetC ofMis also linearly independent in V. However, ifMis linearly independentinV, thenM Afor any other subsetA in Vmay not be linearly independent.

This says we may remove any vectors from a linearly independent set inV andnot destroy its linear independence in V, but if we add any vectors to a linearlyindependent set inV, we may (or may not) destroy its linear independence inV.

2.8 The Dimension of a Vector Space and Basis Sets

A vector space V is called finite dimensional if there exist a positive integern such that V contains a linearly independent set ofn vectors whereas any setofn+ 1 or more vectors in V is linearly dependent. In this casen is called thedimension ofV, written as dim(V). Note thatdim(V) is the largestnumber of

vectors inVthat can form a linearly independent set and its the smallestnumberof vectors in Vthat can span V.

Example: The Space Pn[x, R]

Let us show that the set ofn + 1 vectors

Bn= {1, x , x2, x3, . . . , xn}

inPn[x; R]forms a linearly independent set. Toward this end, consider the equa-tion

0+1x +2x2

+3x3

+ +nxn

= 0which must be true for all real values oft. By differentiating this equationntimeswe get the set of equations

1+ 22x + 33x2 + + nnx

n1 = 0

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22+ 63x + + n(n 1)nxn2 = 063+ + n(n 1)(n 2)nxn

3

= 0

and so on until

(n 1)(n 2) (2)(1)n1+ n(n 1)(n 2) (2)(1)nx = 0n(n 1)(n 2) (2)(1)n = 0.

The last of these gives n = 0. Putting this into the next-to-the-last equationgives n1= 0. By continuing this process we conclude that

0= 1= 2= = n= 0

which means that Bn is a linearly independent set. Note that the set of vectors

Bm= {1, x , x2, x3, . . . , xm}

formn+ 1 is also a linearly independent set in Pn[x; R]. Now any set of theform

Sn= Bn {p(x)}where p(x) is any element in Pn[x; R] must be a linearly dependent set since wemay always writep(x)(by definition ofPn[x; R]) as

p(x) =a0+ a1x + a2x2 + + anx

n

and hence0+1x +2x

2 +3x3 + +nx

n +p(x) = 0

will be zero for = 1 andk =ak fork = 0, 1, 2, . . . , n , thereby showing thatthe set Sn is linearly dependent for any choice of p(x) in Pn[x; R]. This showsthatdim(Pn[x; R]) =n + 1.

Example: The Space P[x, R]

Let have already shown that the set ofn + 1 vectors

Bn= {1, x , x2

, x3

, . . . , xn}

in Pn[x; R] forms a linearly independent set. This says that the same set ofvectors forms a linearly independent set in P[x; R]for any choice ofn and hencethe dimension ofP[x; R] is infinite.

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A set of vectors B={e1, e2, e3, . . . , en}

inVis called abasis set forV if (a)B is linearly independent and (b) Span(B) =V, and now we may say for any vector x V, there is a unique set of scalars1,2,3, . . . ,n F such that

x=nX

k=1

kek.

This follows because if1, 2, 3, . . . ,n Fis another set of scalars such that

x=

nXk=1

kek,

thennX

k=1

kek nX

k=1

kek = x x

ornX

k=1

(k k)ek=0 which leads to k k = 0

ork =k fork = 1, 2, 3, . . . , n.

Every non-trivial vector space (finite or infinite) has a basis set, and ifB1andB2 are two basis sets in V, then there must exist a one-to-one correspondencebetweenB1 andB2. For the case offinite dimensional vector spaces, this meansthat the number of vectors in B1 must be the same as the number of vectors inB2, and this number is equal to the dimension ofV, i.e., |B1|= |B2|= dim(V).If B1 is any basis set for V and if |B1| is infinite in size, then V is called aninfinitely dimensional vector space. In this course, we will be mostly concernedwith infinitely dimensional vector spaces. Finally, note that ifU is a subspace ofV (i.e.,U V) thendim(U) dim(V).

2.9 Coordinate Matrix Representations

LetVbe a vector space with coefficient fieldF, and let

B= {e1, e2, e3, . . . , en}

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be a ordered basis set for V, where n could be infinity. Then we know that for

anyx V, we may expressed xas

x=nX

k=1

xkek

where x1, x2, x3, . . ., xn are scalars in F. This can be used to define an n1matrix of coefficients

[x]B =

x1x2x3...

xn

Fn

known as the coefficient (or coordinate) matrix representation ofx with respectto the ordered basis set B .

It can easily be shown that ifVis a vector space with vector additionandscalar multiplication over a fieldF, and having a basis

B={e1, e2, e3, . . . , en},

then for any x and y inV and any in F, the coordinate matrices satisfy,

[x y]B = [x]B+ [y]B (9.1)

and[ x]B =[x]B (9.2)

where the operations on the right hand side of these equation (9.1) and (9.2) arevector addition and scalar multiplication in the vector space Fn.

These two equations imply that all finite dimensional vector spaces (with di-mension n) over a field Fcan be made to behave like Fn under the usual operationsof vector addition and scalar multiplication. This is accomplished by starting withthe vector spaceVand an ordered basis setB inV, transforming each of the vec-tors x in V to their coordinate matrix representations [x]B in F

n, perform alloperations on the coordinate vectors in Fn, and then transform back into V. In

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symbols, Equation (9.1) can be written as

x y x y Fn

[x]B [y]B [x]B+ [y]B

while Equation (9.2) says

x x Fn [x]B [x]B

Let us look at two examples of how this is done.

Example: An Exotic Vector Space

LetVbe the set,

V =

xy

x R andy R

with vector addition and scalar multiplication defined as

xy x0

y0

= x + x0 2

y+ y0 1

and

xy

=

x + 2(1 )y+ (1 )

,

respectively. It can be shown that Vis a vector space over the field of real numbersR. Note that the zero of this vector space is

0= 0 xy =

0x + 2(1 0)0y+ (1

0) =

21

since xy

21

=

21

xy

=

x + 2 2y+ 1 1

=

xy

.

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Now suppose we have the set

B= 1

0

, 1

1

(9.3)

inV, then we can show that this is a basis for V. Starting with

10

11

=

xy

(9.4)

we get

1 + 2(1 )0 + (1

)

1 + 2(1 )1 + (1

)

=

xy or

2 1

2 1

=

xy

which reduces to

(2 ) + (2 ) 2(1 ) + (1) 1

=

xy

or

2 1

=

xy

.

Solving this for and gives

2 = x1 = y

which leads to= 1 y and = 1 x + y.

Putting these into Equation (9.4) gives,xy

=

(1 y)

10

(1 x + y)

11

(9.5)

for any vector xy

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in V showing that the set B spansthe vector space V. Next we observe that if

the zero vector0=

21

is placed into Equation (9.3) we get

= 1 1 = 0 and = 1 2 + 1 = 0(which are both the additive identity in R) showing that the set B is a linearlyindependent set in V. ThereforeB is a basis set for V. In addition, Equation(9.5) shows that the coordinate matrixof an arbitrary vector in V with respectto the basis B is given as,

xy

B

= = 1 y

1 x + y . (9.6)

Note that given xy

,

we can compute xy

B

using Equation (9.6). Conversely, given xy

B

=

we can compute

xy

using Equation (9.4).So for example, suppose we check out the results in Equations(9.1) and (9.2) using

x= 52 and y=

3

1 .

If we directly calculate x y, we get

x y=

52

31

=

(5) + (3) 2(2) + (1) 1

=

60

.

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If we instead, calculate [x]B+ [y]B, we get

[x]B+ [y]B =

52

B

+

31

B

=

1 (2)

1 (5) + (2)

+

1 (1)

1 (3) + (1)

=

12

+

23

=

(1) + (2)(2) + (3)

=

15

so that (from Equation (9.4) with = 1and= 5),x y =

1

10

(5)

11

=

1(1) + 2(1 1)

1(0) + (1 1)

(5)(1) + 2(1 (5))(5)(1) + (1 (5))

=

10

71

=

(1) + (7) 2(0) + (1) 1

=

60

which agrees with the direct calculation and so

[x y]B = [x]B+ [y]B.

Suppose, now we want to calculate 4 x. A direct calculation yields

4 x= 4

52

=

4(5) + 2(1 4)

4(2) + (1 4)

=

14

5

.

An indirect calculation using the coordinate matrices and Equation (9.2) yields

[4 x]B = 4[x]B = 4 5

2 B = 4 1 (2)

1 (5) + (2) = 4

12

=

4(1)4(2)

=

48

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thereby leading to (Equation (9.4) with = 4and= 8)

4 x =

(4) 1

0

(8) 1

1

=

4(1) + 2(1 (4))4(0) + (1 (4))

8(1) + 2(1 (8))

8(1) + (1 (8))

=

65

101

=

(6) + (10) 2

(5) + (1) 1

=

14

5

which agrees with the direct calculation, and so

[4 x]B = 4[x]B.

Example: Application To The Common Logarithm

LetWbe the set ofpositivereal numbers and define vector addition and scalarmultiplication as

x y= xy and x= x

for R. ThenWis a vector space over the field of real numbersR. Note thatthezero for this vector space is the number 1 since

x 1 = 1 x= 1x=x

for allx W. The setB = {10}

inWis a basis set forW since

10 =x gives 10 =x

so that= log10(x) = log(x)

showing that B spansall ofW, while

10 = 1 (thezerovector inW)

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leads to = log(1) = 0 showing that B is a linearly independent set in W. In

addition we note that the coordinate matrixof an element in W can be writtenas[x]B = [log(x)].

In the days before calculators and slide rules, people used Equations (9.1) and(9.2) and common logarithms to obtain approximations to complex arithmeticcalculations. For example, suppose we wanted to calculate

(2856)(497)3/2350

then, (in the days before calculators and slide rules), we would say the above

calculation (as viewed from the vector space W), would read(2856)(497)3/2

350= (2856)(497)3/2(350)1/2

= (2856) (497)3/2 (350)1/2

= (2856)

3

2 (497)

1

2 (350)

so that

(2856)(497)3/2350 B

=

(2856)

3

2 (497)

12 (350)

B= [2856]B+

3

2[497]B+

12

[350]B

= log(2856) +3

2log(497) +

12

log(350)

' 3.4558 +3

2(2.6964) 1

2(2.5441) = 6.2284

(where each common log was looked up in a table of logarithms). Therefore, wehave,

(2856)(497)3/2

350' 6.228410 = 106.2284 = 100.2284 106

= antilog(0.2284)106 '1.69106,

where, the antilog(0.2284)was also looked up in a table of antilogarithms. Notethat if you had chosen the basis set, B0 = {2}, then you would be dealing with

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the log to the base 2. Other than that, the calculations would be essentially the

same. Note that old-time slides rules were designed around this basis principle.

2.10 The Change-Of-Basis Matrix

Suppose thatV is ann-dimensional vector space with bases

A={a1, a2, a3, . . . , an}

andB={b1, b2, b3, . . . , bn}

and suppose that xis a vector in V. Then we may expression x as

x= 1a1+2a2+3a3+ +nan

or asx= 1b1+2b2+3b3+ +nbn.

so thatnX

i=1

ibi=nX

j=1

jaj.

Suppose also that we express each aj as

aj =nX

i=1

ijbi so that [aj]B =

1j2j3j

...nj

for each j = 1, 2, 3, . . . , n. Placing these expressions foraj in the above equationleads to

n

Xi=1ibi =

n

Xj=1jaj =

n

Xj=1j

n

Xi=1ijbi=

n

Xi=1n

Xj=1jijbi

ornX

i=1

i

nXj=1

jij

!bi= 0.

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The linear independence of the set of vectors in B allows us to conclude that

i nX

j=1

jij = 0 or i=nX

j=1

ijj

for eachi = 1, 2, 3, . . . , n. In matrix form, this reads

123...n

=

11 12 13 in21 22 23 2n31 32 33 3n

... ...

... . . .

...n1 n2 n3 nn

123...n

or[x]B = (B[T]A)[x]A

where

B[T]A =

11 12 13 in21 22 23 2n31 32 33 3n

... ...

... . . .

...n1 n2 n3 nn

= [[a1]B|[a2]B|[a3]B| |[an]B]

is called the transformation matrix from the basis set A to the basis set B . In asimilar way, we may write

[x]A = (A[T]B)[x]B

where

A[T]B = [[b1]A|[b2]A|[b3]A| |[bn]A]

is the transformation matrix from the basis set B to the basis setA. It is easy toshow that

(A[T]B)(B[T]A) = (A[T]A) = [I]

for any two basis sets A and Bin V, and hence both A [T]Band B [T]Aare invertibleand equals each others inverse, i.e.,

A[T]B = (B[T]A)1 and B[T]A = (A[T]B)

1.

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Note that ifA, B andCare three basis sets in a vector space V, then it is easy

to show thatC[T]A = (C[T]B)(B[T]A).

This is very convenient when B is a basis set (such as the natural or standardbasis set) in which each of B[T]A and B[T]C are easy to compute. For then wemay write

C[T]A = (C[T]B)(B[T]A) = (B[T]C)1(B[T]A)

thereby helping in the computation of C[T]A.

Example: The Vector Space P2[x; R]

LetV =P2[x; R] with the basis sets

A={a1, a2, a3}= {1, 1 + x, 1 + x + x2}

andB= {b1, b2, b3}={x + x

2, 1 + x2, x}.

To find[a + bx + cx2]A, we must solve for 1, 2, and 3 so that

1(1) +2(1 + x) +3(1 + x + x2) =a + bx + cx2.

This leads to

(1+2+3) + (2+3)x +3x2

=a + bx + cx2

which leads to the system

1+2+3 = a

2+3 = b

3 = c

which has the solution: 1= a b, 2= b c and 3= c, so that

[a + bx + cx2]A =

a bb

c

c

.

To find[a + bx + cx2]B, we must solve for 1, 2, and 3 so that

1(x + x2) +2(1 + x

2) +3(x) =a + bx + cx2.

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This leads to

(2) + (1+3)x + (1+2)x2 =a + bx + cx2

which leads to the system

2 = a

1+3 = b

1+2 = c

which has the solution: 1= c a, 2= a and 3= a + b c, so that

[a + bx + cx2]B = c aa

a + b c

.

To determine

A[T]B = [[b1]A|[b2]A|[b3]A]

we use

[b1]A = [0 + 1x + 1x2]A =

0 11 1

1

=

10

1

and

[b2]A = [1 + 0x + 1x2]A =

1 00 1

1

=

11

1

and

[b3]A = [0 + 1x + 0x2]A =

0 11 0

0

=

11

0

and so

A[T]B = [[b1]A|[b2]A|[b3]A| |[bn]A] =

1 1 10

1 1

1 1 0

.

As a check we should find that

[a + bx + cx2]A = (A[T]B)[a + bx + cx2]B

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which gives

a bb c

c

=

1 1 10 1 1

1 1 0

c aa

a + b c

=

a bb c

c

which checks out. Note that by using the standard (or natural) basis set ofP2[x; R],

S={s1, s2, s3}= {1, x , x2} for which [a + bx + cx2]S =

ab

c

,

we can get A[T]B via

A[T]B = (A[T]S)(S[T]B) = (S[T]A)1(S[T]B)

= [[a1]S|[a2]S|[a3]S]1 [[b1]S|[b2]S|[b3]S]

=

1 1 10 1 1

0 0 1

1 0 1 01 0 1

1 1 0

=

1 1 10 1 1

1 1 0

,

which agrees with our earlier results. Note also that

B[T]A= (A[T]B)1 =

1 1 10 1 11 1 0

1

= 1 1 01 1 1

1 2 1

,

which agrees with our results from class.

2.11 The Vector Space of Geometric Arrows

LetG be the set of all geometric arrows in space. Each of these arrows havea length (called its magnitude) and a direction in space (called its direction),including the zero arrow (0) which has zero length and (by definition) any

direction in space. The beginning of an arrow is called its tailand the end of thearrow is called itstip. Note that an arrow in G need not be fixed in space since thatthe arrow can be translatedanywhere in space without changing its magnitudeand direction, and hence not changing the arrow itself. The magnitude of an arrowx in G is denoted by the symbol ||x||, and the direction of an arrow x in G is

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denoted by the symbol

bx. A direction opposite that ofx is denoted by the symbol

bx. Note that the direction of an arrow x inG (i.e.,bx) is also considered as anarrow inG which has the same direction as that ofx, but havingunit magnitude,so that||bx|| 1.

We now define arrow addition by the tip-to-tail rule. This says if you want toadd the arrows x and yin G, then first you translate yuntil its tail is placed on thetip ofx. This forms a two-arrow tip-to-tail chain. The sum ofxand y is definedas that arrow which connects the tail ofx to the tip of y, i.e., its connects thebeginning of the two-arrow tip-to-tail chain to the end of the two-arrow tip-to-tailchain.

More generally, if you want to add the n arrows x1, x2, x3,. . ., xn inG, thenfirst translate x2 until its tail is placed on the tip of x1, then you translate x3until its tail is placed on the tip ofx2, then you translate x4until its tail is placedon the tip ofx3, etc., until you translate xn until its tail is placed on the tip ofxn1. This forms ann-arrow tip-to-tail chain. The sum ofx1, x2, x3, . . ., xn inG is defined as that arrow which connects the tail ofx1 to the tip ofxn, i.e., itsconnects the beginning of the n-arrow tip-to-tail chain to the end of the n-arrowtip-to-tail chain.

Using the real numbers R as the field of scalars, we next define scalar multi-

plication by the rule which says when an arrow x is multiplied by a scalar , theresult is a arrow whose length is multiplied by|| (i.e.,||x||= ||||x||) and thedirection of the result is the same as that ofxwhen > 0, or opposite that ofxwhen < 0. That is,

cx=

+bx, when > 0bx, when < 0 .

Note that for = 0, we definex= 0.

It can easily be shown that G with the above definitions of arrow addition

and scalar multiplication forms a 3-dimensional real vector space. Using whatyou already know about the vector space G, i.e., the dot product, cross product,triple scalar product, etc., you should show that a set of three vectors

B = {u1, u2, u3}

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in G is a basis set in G if and only if the triple scalar product between u1, u2,

andu3 is non-zero, i.e., if and only if

[u1, u2, u3] u1(u2 u3)6= 0.

Recall that the dot productbetween two vectors is defined by

u1 u2= ||u1||||u2|| cos()

and the cross productbetween two vectors is defined by

u1 u2= ||u1||||u2|| sin()

bn

wherebnis a unit vector perpendicular to both u1and u2and satisfying the right-hand rule where you place the fingers of your right hand along u1 and then curlthese fingers through the angle until your fingers are aligned with u2 at whichpoint your thumb will be pointing in the direction ofbn.Properties Of The Dot Product

Some important properties of the dot product are as follows.

(1) u1(u2+ u3) =u1 u2+ u1 u3 for all u1, u2, and u3 inG.

(2) u1(u2) = (u1) u2= (u1 u2)for all u1 andu2 inG and all in R.

(3) u1 u2= u2 u1 for all u1 andu2 inG.

(4) u1 u1= ||u1||2 0 for all u1 inG, and u1 u1= 0 if and only ifu1= 0.

(5) u1 u2= 0 if and only ifu1u2.(6) u1 0= 0 for all u1 inG.

Properties Of The Cross ProductSome important properties of the cross product are as follows.

(1) u1(u2+ u3) =u1 u2+ u1 u3 for all u1, u2, and u3 inG.

(2) u1 (u2) = (u1)u2= (u1u2)for allu1and u2 in G and all in R.

(3) u1 u2= (u2 u1)for all u1 andu2 inG.

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(4) u1 u2= 0 if and only ifu1ku2.

(5) u1 0= 0 for all u1 inG.

Properties Of The Triple-Scalar ProductSome important properties of the triple-scalar product are as follows.

(1) [u1, u2, u3+ u4] = [u1, u2, u3] + [u1, u2, u4]for all u1, u2, u3 andu4 inG.

(2) [u1, u2,u3] = [u1,u2, u3] = [u1, u2, u3] =[u1, u2, u3]for all u1, u2 andu3 inG and all inR.

(3) [u1, u2, u3] = [u3, u1, u2] = [u2, u3, u1] =

[u2, u1, u3] =

[u1, u3, u2] =

[u3, u2, u1] for all u1, u2 andu3 inG.(4) [u1, u2, u3] = 0 if and only ifu1, u2 andu3 all lie in the same plane.

(5) [u1, u2, 0] = 0for all u1 andu2 inG.

Suppose now that we wanted to write any xin G as

x= 1u1+2u2+3u3

which is possible provided that [u1, u2, u3] 6= 0. Then we could start with thisresult and write

x u1= 1u1 u1+2u2 u1+3u3 u1

andx u2= 1u1 u2+2u2 u2+3u3 u2

andx u3= 1u1 u3+2u2 u3+3u3 u3

which, in matrix form, leads to

u1 u1 u2 u1 u3 u1u1 u2 u2 u2 u3 u2

u1 u3 u2 u3 u3 u3

12

3

=

x u1x u2

x u3

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so that 12

3

= u1 u1 u2 u1 u3 u1u1 u2 u2 u2 u3 u2u1 u3 u2 u3 u3 u3

1 x u1x u2

x u3

.If the basis set

B = {u1, u2, u3}

satisfies u1 u1 u2 u1 u3 u1u1 u2 u2 u2 u3 u2

u1 u3 u2 u3 u3 u3

=

||u1||

2 0 00 ||u2||

2 00 0 ||u3||

2

,

it is called anorthogonalbasis set and the above equation leads to

12

3

=

||u1||

2 0 00 ||u2||

2 00 0 ||u3||

2

1 x u1x u2

x u3

=

1/||u1||

2 0 00 1/||u2||

2 00 0 1/||u3||

2

x u1x u2

x u3

or

123

= (x u1)/||u1||

2

(x u2)/||u2||2

(x u3)/||u3||2

so that

x=

x u1||u1||2

u1+

x u2||u2||2

u2+

x u3||u3||2

u3

for eachxin G. In addition, if the basis set

B = {u1, u2, u3}

satisfies u1 u1 u2 u1 u3 u1u1 u2 u2 u2 u3 u2u1 u3 u2 u3 u3 u3

= 1 0 00 1 00 0 1

,

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it is called anorthonormalbasis set and the above equation leads to

12

3

=

x u1x u2

x u3

so thatx= (x u1)u1+ (x u2)u2+ (x u3)u3

for each x in G. IfB is not orthonormal or orthogonal, then system of equations

u1 u1 u2 u1 u3 u1u1 u2 u2 u2 u3 u2

u1 u3 u2 u3 u3 u3

12

3

=

x u1x u2

x u3

can be solved using Cramers rule from linear algebra, resulting in the formulas

1=

det

x u1 u2 u1 u3 u1x u2 u2 u2 u3 u2

x u3 u2 u3 u3 u3

det

u1 u1 u2 u1 u3 u1u1 u2 u2 u2 u3 u2

u1 u3 u2 u3 u3 u3

, 2=

det

u1 u1 x u1 u3 u1u1 u2 x u2 u3 u2

u1 u3 x u3 u3 u3

det

u1 u1 u2 u1 u3 u1u1 u2 u2 u2 u3 u2

u1 u3 u2 u3 u3 u3

and

3=

det

u1 u1 u2 u1 x u1u1 u2 u2 u2 x u2

u1 u3 u2 u3 x u3

det

u1 u1 u2 u1 u3 u1u1 u2 u2 u2 u3 u2

u1 u3 u2 u3 u3 u3

.

In the case ofG, since we have the additional operations of dot product, crossproduct and triple scalar product and we may determine1,2, and3by startingwith

x= 1u1+2u2+3u3and writing

x (u2 u3) =1u1(u2 u3) +2u2(u2 u3) +3u3(u2 u3).

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Then using the fact that

u2(u2 u3) = 0 and u3(u2 u3) = 0

we may conclude that

x (u2 u3) =1u1(u2 u3)

or

1= x (u2 u3)

u1(u2 u3)=

[x, u2, u3]

[u1, u2, u3].

Similarly we get

2= [u1, x, u3][u1, u2, u3]

and 3= [u1, u2, x][u1, u2, u3]

,

and so

[x]B = 1

[u1, u2, u3]

[x, u2, u3][u1, x, u3]

[u1, u2, x]

so that

x= [x, u2, u3]u1+ [u1, x, u3]u2+ [u1, u2, x]u3

[u1, u2, u3]

for each vector xin G.

2.12 A Generalization to Higher Dimensions

Suppose thatV is ann-dimensional vector space with basis set

B={u1, u2, u3, . . . , un}

and suppose that xis a vector in Vwhich is to be expressed as

x= 1u1+2u2+3u3+ +nun.

Then we would like to compute the coefficients 1, 2, 3, . . ., n by using anidea similar to that in the previous section. In order to do this, we need to definethe notion of a dot product in an n-dimensional vector space. Toward this end,we shall study the ideas behind the inner productfor vector spaces.

Groups, Fields and Vector Spaces

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