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Algebra I lecture notes

version

Imperial College LondonMathematics 2005/2006

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CONTENTS Algebra I lecture notes

Contents

1 Groups 51.1 Definition and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.1.1 Group table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.2 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.2.1 Criterion for subgroups . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.3 Cyclic subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.3.1 Order of an element . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.4 More on the symetric groups Sn . . . . . . . . . . . . . . . . . . . . . . . . 181.4.1 Order of permutation . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.5 Lagranges Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.5.1 Consequences of Lagranges Theorem . . . . . . . . . . . . . . . . . 22

1.6 Applications to number theory . . . . . . . . . . . . . . . . . . . . . . . . . 221.6.1 Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1.7 Applications of the group Zp . . . . . . . . . . . . . . . . . . . . . . . . . . 261.7.1 Mersenne Primes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.7.2 How to find Meresenne Primes . . . . . . . . . . . . . . . . . . . . . 28

1.8 Proof of Lagranges Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 30

2 Vector Spaces and Linear Algebra 342.1 Definition of a vector space . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.2 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.3 Solution spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.4 Linear Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

2.5 Span . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.6 Spanning sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432.7 Linear dependence and independence . . . . . . . . . . . . . . . . . . . . . 442.8 Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 482.9 Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512.10 Further Deductions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

3 More on Subspaces 573.1 Sums and Intersections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573.2 The rank of a matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

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Algebra I lecture notes CONTENTS

3.2.1 How to find row-rank(A) . . . . . . . . . . . . . . . . . . . . . . . . 613.2.2 How to find column-rank(A)? . . . . . . . . . . . . . . . . . . . . . 63

4 Linear Transformations 684.1 Basic properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 704.2 Constructing linear transformations . . . . . . . . . . . . . . . . . . . . . . 714.3 Kernel and Image . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 734.4 Composition of linear transformations . . . . . . . . . . . . . . . . . . . . . 784.5 The matrix of a linear transformation . . . . . . . . . . . . . . . . . . . . . 794.6 Eigenvalues and eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . 82

4.6.1 How to find evals / evecs ofT? . . . . . . . . . . . . . . . . . . . . 834.7 Diagonalisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 844.8 Change of basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

5 Error-correcting codes 885.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 885.2 Theory of Codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

5.2.1 Error Correction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 915.3 Linear Codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

5.3.1 Minimum distance of linear code . . . . . . . . . . . . . . . . . . . 925.4 The Check Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 935.5 Decoding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

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CONTENTS Algebra I lecture notes

Introduction

(1) Groups used throughout maths and science to describe symmetry e.g. everyphysical object, algebraic equation or system of differential equations, . . . , has agroup associated with it.

(2) Vector spaces have seen and studied some of these already, e.g. Rn.

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Chapter 1

Groups

1.1 Definition and examples

Definition 1.1. Let S be a set. A binary operation on S is a rule which assigns to anyordered pair (a, b) (a, b S) an element a b S.In other words, its a function from S S S.Eg 1.1.

1) S = Z, a b = a + b

2) S = C, a b ab3) S = R, a b = a b4) S = R, a b = min(a, b)5) S = {1, 2, 3}, a b = a (eg. 1 1 = 1, 2 3 = 2)

Given a binary operation on a set S and a,b,c S, can form a b c in two ways

(a b) ca

(b

c)

These may or may not be equal.

Eg 1.2. In 1), (a b) c = a (b c).In 3), (3 5) 4 = (3 5) 4 = 6. Whereas 3 (5 4) = 3 (5 4) = 2Definition 1.2. A binary operation is associative if for all a,b,c S

(a b) c = a (b c)

Associativity is important.

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1.1. DEFINITION AND EXAMPLES Algebra I lecture notes

Eg 1.3. Solve 5 + x = 2.We add

5 to get

5 + (5 + x) =

5 + 2. Now we use associativity! We rebracket to

(5 + 5) + x = 5 + 2. Thus 0 + x = 5 + 2, so x = 3.To do this, we needed

1) associativity of +

2) the existence of 0 (with 0 + x = x)

3) existence of5 (with 5 + 5 = 0)

Generally, suppose we have a binary operation and an equation

a x = b(a, b S constants, x S unknown) To be able to solve, we need

1) associativity

2) existence ofe S such that e x = x for x S

3) existence ofa S such that a a = e

Then can solve

a x = ba (a x) = a b(a a) x = a b

e x = a bx = a b

Group will be a structure in which we can solve equations like this.

Definition 1.3. A group (G, ) is a set G with a binary operation satisfying the followingaxioms (for all a,b,c S)

(1) ifa, b S then a b S (closure)

(2) (a b) c = a (b c) (associativity)

(3) there exists e S such that

e x = x e = x

(identity axiom)

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1) Suppose e, e are identity elements. So

e x = x e = xe x = x e = x

Then

e = e e = e

2) Let x G and suppose x, x are inverses of x. That means

x

x = x x

= ex x = x x = e

Then

x = x e= x (x x)= (x x) x= e x= x

2

Notation 1.1.

e is the identity element of G x1 is the inverse of x Instead of (G, ) is a group, we write G is a group under .

Often drop the in a b, and write just ab.Eg 1.5.In (Z, +), x1 = x. Z is a group under addition.In (Q {0}, x), x1 = 1x .

Definition 1.4. We say (G, ) is a finite group if |G| is finite; (G, ) is an infinite group if|G| is infinite.Eg 1.6. All groups in example 2.4 are infinite except the last which has size (order) 4.

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Algebra I lecture notes 1.1. DEFINITION AND EXAMPLES

Eg 1.7. Let F = R or C. Say a matrix (aij) is a matrix over F if all aij F.Set of all n

n matrices over F under matrix multiplication is not a group (problem with

inverse axiom). But lets define GL(n, F) to be the set of all n n invertible matrices overF.

Definition 1.5. Denote the set of all invertible matrices over field F as GL(n, F)

GL(n, F) = {(aij) | 1 < i, j n, aij F}

Claim GL(n, F) is a group under matrix multiplication.

Proof. Write G = GL(n, F).

Closure Let A, B G. So A, B are invertible. Now

(AB)1 = B1A1

since

(AB)(B1A1) = A(BB1B) = AIA1 = I

(B1A1)(AB) = B1(A1A)B = B1IB = I

So AB G.Associativity proved in M1GLA.

Identity is identity matrix In.

Inverse of A is A1 (since AA1 = A1A = I). Note A1 G as it has an inverse,A.

2

1) GL(1,R) is the set of all (a) with a R, a = 0. This is just the group (R {0}, ).

2) GL(2,C) is the set of

a bc d

, a,b,c,d C, ad bc = 0.

Note 1.1. Usually, in a group (G, ) a b is not same as b a.

Definition 1.6. Let (G, ) be a group. If for all a, b G, a b = b a we call G an abeliangroup.Eg 1.8. (Z, +) is abelian as a + b = b + a for all a, b Z. So are the other groups in 2.4.So is GL(1, F).But GL(2,R) is not abelian, since

1 10 2

0 11 0

=

1 12 0

0 11 0

1 10 2

=

0 21 1

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Groups of permutations

Definition 1.7. S a set. A permutation of S is a function f : S

S which is a bijection(both injection and surjection).

Eg 1.9. S = {1, 2, 3, 4}, f : 1 2, 2 3, 3 4, 4 1 is a permutation.Notation 1.2.

f =

1 2 3 42 4 3 1

is a permutation 1 2, 2 4, 3 3, 4 1.Let

g = 1 2 3 43 1 2 4

The composition f q is defined byf g = f(g(s))

Here

f g =

1 2 3 43 2 4 1

Recall the inverse function f1 is the inverse of f. Here

f1 = 1 2 3 4

4 1 3 2Notice f1 f =

1 2 3 41 2 3 4

= e, the identity function.

Proposition 1.2. Let S = {1, 2, 3, . . . , n} and let G be the set of all permutations ofS. Then (G, ), being the function composition, is a group, i.e. G is a group undercomposition.

Proof. Notation for f G is

f = 1 2 n

f(1) f(2) f(n) Closure By M1F, if f, g are bijections S S then f g is a bijection. Associativity Let f , g , h G and apply s S Then

f (g h)(s) = f(g h(s))= f(g(h(s)))

(f g) h(s) = (f g)(h(s)) = f(g(h(s)))So f (g h) = (f g) h

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Algebra I lecture notes 1.1. DEFINITION AND EXAMPLES

Identity is e =

1 2 n1 2 n

, since e f = f e = f.

Inverse of f is f1 =

f(1) f(2) f(n)1 2 n

and f1 f = f f1 = e

2

Definition 1.8. The group of all permutations of{1, 2, . . . , n} is written Sn and called thesymmetric group of degree n.

Eg 1.10. S2 =

e,

1 22 1

. So |S2| = 2.

Eg 1.11. S3 = 1 2 31 2 3 ,1 2 32 1 3 ,1 2 33 2 1 ,1 2 31 3 2 ,1 2 32 3 1 ,1 2 33 1 2|S3| = 6Proposition 1.3. Sn is a finite group of size n!.

Proof. For f =

1 2 n

f(1) f(2) f(n)

, number of choices for f(1) is n, for f(2) is n1,f(3) is n2, . . . , f(n) is 1. The total number of permutations is n (n1) (n2) 1 = n!.2

Notation 1.3. (Multiplicative notation for groups)

If (G, ) is a group, well usually write just ab instead of a b. We can define powersa2 = a aa3 = a a a

an = a a a

n

When we write Let G be a group, we mean the binary operation is understood, andwere writting ab instead of a b, etc.

Eg 1.12. In (Z, +), ab = a b = a + b and a2

= a a = 2a, i.e. an

= na.

1.1.1 Group table

Definition 1.9. Let G be a group, with elements a , b , c, . . . . Form a group table

a b c a a2 ab ac b ba b2 bc ...

...

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1.2. SUBGROUPS Algebra I lecture notes

Eg 1.13. S3 =

e = 1 2 31 2 3a =

1 2 32 3 1

a2 =

1 2 33 1 2

b =

1 2 32 1 3

ab = 1 2 3

3 2 1a2b =

1 2 31 3 2

e a a2 b ab a2be e a a2 b ab a2ba a a2 e ab a2b b

a2 a2 e a a2b b abb b a2b ab e a2 a

ab ab b a2b a e a2

a2

b a2

b ab b a2

a e

a3 = e ba = a2b b2 = e

1.2 Subgroups

Definition 1.10. Let (G,

) be a group. A subgroup of (G,

) is a subset of G which isitself a group under .Eg 1.14.

(Z, +) is a subgroup of (R, +). (Q {0}, ) is not a subgroup of (R, +) ({1, 1}, ) is a subgroup of ({1, 1, i, i}, ) ({1, i}, ) is not a subgroup of ({1, 1, i, i}, ) (closure fails i i = 1)

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Algebra I lecture notes 1.2. SUBGROUPS

1.2.1 Criterion for subgroups

Proposition 1.4. Let G be a group.1

Let H G. Then H is a subgroup of G if thefollowing conditions are true

(1) e H (where e is the identity of G)

(2) ifh, k H then hk H (H is cloesed)

(3) ifh H then h1 H

Proof. Assume (1)-(3). Check the group axioms for H:

Closure true by (2)

Associativity true by associativity for G

Identity by (1)

Identity by (3)

2

Eg 1.15. Let G = GL(2,R) (22 invertible matrices overR). Let H =

1 n0 1

| n Z

.

Claim H is a subgroup of G.

Proof. Check (1)-(3) of previous proposition

(1) e =

1 00 1

H

(2) Let h =

1 n0 1

, k =

1 p0 1

. Then nk =

1 p + n0 1

H.

(3) Let h =

1 n0 1

. Then h1 =

1 n0 1

H

2

1so is understood and we write ab instead ofa b

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1.3. CYCLIC SUBGROUPS Algebra I lecture notes

1.3 Cyclic subgroups

Let G be a group. Let a G. Recalla1 = a, a2 = aa, . . .

Negative powers

a0 = e

a2 = a1a1

an = a1 a1

n

Note 1.2. All the powers an

(n Z) lie in G (by closure).Lemma 1.1. For any m, n Z

aman = am+n

Proof. For m, n > 0

aman =

m+n a a

n

a a m

= am+n

For m 0, n < 0aman = a a

n

a1 a1 n

= am(n)

Similarly for m < 0, n 0. Finally, when m, n < 0

aman =

mn

a1 a1

na1 a1

m= am+n

2

Proposition 1.5. Let G be a group. Let a G. DefineA = {an | n Z} = . . . , a2, a1, e , a , a2, . . .

Then A is a subgroup of G.

Proof. Check (1)-(3) of 2.4

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Algebra I lecture notes 1.3. CYCLIC SUBGROUPS

(1) e = a0 A

(2) am

, an

A then am

an

= am+n

A(3) an A then (an)1 = an A

2

Definition 1.11. Write A = a, called the cyclic subgroup of G generated by a. So foreach element a G we get a cyclic subgroup a of G.Eg 1.16. (1) G = (Z, +). What is the cyclic subgroup 3?

Well, 31 = 3, 32 = 3 + 3 = 6, 3n = 3n, 31 = 3, 3n = 3n. So 3 = {3n | n Z}.Similarly 1 = {n | n Z} = Z.

(2) G = S3, a =1 2 3

2 3 1

. What is a?

Well

a0 = e

a1 = a

a2 =

1 2 33 1 2

a3 = e

a4 = a

a5 = a2...

a1 = a3a1 = a2

a2 = a...

Hence a = {an | n Z} = {e,a,a2} .Now consider b, b =

1 2 32 1 3

. Here b0 = e, b1 = b, b2 = e, . . . .

So b = {e, b}.(3) All the cyclic subgroups ofS3 = {e,a,a2,b,ab,a2b}

e = {e}a = {e,a,a2}a2

= {e,a,a2}b = {e, b}

ab = {e,ab}

a2b

= {e, a2b}

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Definition 1.12. Say a group G is a cyclic group, if there exists an element a G suchthat

G = a = {an | n Z}Call a a generator for G.

Eg 1.17.

(1) (Z, +) = 1, So (Z, +) is cyclic with generator 1(2) ({1, 1, i, i}, ) is cyclic, generator i, since

i = {i0, i1, i2, i3} = {1, i, 1, i}Another generator is i, but 1 and 1 are not generators.

(3) S3 is not cyclic, as non of its 5 cyclic subgroups is the whole of S3.For any n N there exists a cyclic group of size n (having n elements) Cn

Cn = {x C | xn = 1}the complex n-th roots of unity, under multiplication. By M1F, we know Cn = {1, , 2, . . . , n1},where = e2

i

n , SoCn =

a cyclic subgroup of (C {0}, ).

1.3.1 Order of an element

Definition 1.13. Let G be a group, let a G. The order of a, written o(a), is thesmallest positive integer k such that ak = e. So o(a) = k means ak = e and ai = e fori = 1, . . . , k 1.If no such k exists, we say a has infinite order and we write o(a) = .Eg 1.18.

(1) e has order 1, and is the only such element.

(2) G = S3, a =

1 2 32 3 1

. Then a1 = a, a2 =

1 2 33 1 2

, a3 =

1 2 31 2 3

. So

o(a) = 3.For b =

1 2 32 1 3

, b1 = e, b2 = e, so o(b) = 2. Full list:

o(e) = 1

o(a) = 3

o(a2) = 3

o(b) = 2

o(ab) = 2

o(a2b) = 2

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Algebra I lecture notes 1.3. CYCLIC SUBGROUPS

(3) G = (Z, +). What is o(3)? In G, e = 0, 3n = n 3. So 3n = e for any n N, soo(3) =

.

(4) G = GL(2,C), A =

i 00 e2i/3

. Then

Ak =

ik 00 e2ik/3

So smallest k for which this is the identity is 12 o(A) = 12.

Proposition 1.6. G a group, a G. The number of elements in the cyclic subgroupgenerated by a is equal to o(a).

| a

|= o(a)

Proof.

(1) Suppose o(a) = k, finite. This means ak = e, but ai = e for 1 i k 1.Write A = a = {an | n Z}. Then A contains

e,a,a2, . . . , ak1

These are all different elements of G since for 1 y < j k 1,ai = aj

a1ai = aiaj

e = a

ji

o(a) = j i < k contradiction.Hence A contains e , a , . . . , ak1, all distinct, so

|A| KWe now show that every element of A is one of e , a , . . . , ak1. Let an A. Write

n = qk + r, 0 r k 1Then

an

= aqk+r

= aqkar

= (ak)qar

= eqar

= ar

So an = ar {e,a,a2, . . . , ak1}. Weve shownA = {e,a,a2, . . . , ak1}

so |A| = k = o(a).

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(2) Suppose o(a) = . This means

ai

= e for i 1If i < j then ai = aj since

ai = aj

e = aji

contradiction.Then

A = {. . . , a2, a1, e , a , a2, . . . }and all these elements are different elements of G. So

|A

|=

= o(a).

2

Eg 1.19.

(1) G = S3, a =

1 2 32 3 1

. Then a = {e,a,a2}, size 3, o(a) = 3.

(2) G = (Z, +). Then 3 = {3n | n Z}, is infinite and o(3) = .(3) Cn = = {1, , . . . , n1}, size n, and o() = n.

1.4 More on the symetric groups Sn

Eg 1.20. Let f =

1 2 3 4 5 6 7 84 5 6 3 2 7 1 8

S8. What is f2, f5?

We need better notation to see answers quicklky. Observe the numbers 143671 arein a cycle, as well as numbers 25 and 8.We will write f = (1 4 3 6 7)(2 5)(8). These are the cycles of f. Each symbol in the firstcycle goes to the next except for the last 7 which goes back to the first 1. The cycles aredisjoint they have no symbols in common. Call this the cycle notation for f.

Definition 1.14. In general, an r cycle is a permutationa1 a2 . . . ar

which sends a1 a2 ar a1 . . . .Eg 1.21. Can easily go from cycle notation to original, e.g.

g = (1 5 3)(2 4)(6 7) S7g =

1 2 3 4 5 6 75 4 1 2 3 7 6

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Algebra I lecture notes 1.4. MORE ON THE SYMETRIC GROUPS SN

Proposition 1.7. Every permutation f in Sn can be expressed in the cycle notation, i.e.as a product of disjoint cycles.

Proof. Following procedure works:Start with 1, and write down sequence

1, f(1), f2(1), . . . , f r1(1)

until the first repeat fr(1). Then in fact fr(1) = 1, since

fr(1) = fi(1)

fi(1)fr(1) = 1

fri(1) = 1

which is a contradiction as fr(1) is the first repeat. So have the r-cycle

(1 f(1) fr1(1))first cycle of f.Second cycle: Pick a symbol i not in the first cycle, and write

i, f(i), f2(i), . . . , f s1(i)

where fs(i) = i. Then this is the second cycle of f. This cycle is disjoint from the first

since if not, say fj

(i) = k in first cycle, then fsj(k)

= fs

(i) = i would be in the first cycle.Now carry on: pick j not in first two cycles and repeat to get third cycle and carry on untilwe have used all the symbols 1, . . . , n. so

f = (1 f(1) fr1(1)(i f(i) fs1(i)) . . .a product of disjoint cycles. 2

Note 1.3. Cycle notation is not quite unique e.g. (1 2 3 4) can be written as (2 3 4 1)AND (1 2)(3 4 5) = (3 4 5)(1 2). Notation is unique appart from such changes.

Eg 1.22.

1. The elements of S3 in cycle notation

e = (1)(2)(3)

a = (1 2 3)

a2 = (1 3 2)

b = (1 2)(3)

ab = (1 3)(2)

a2b = (2 3)(1)

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2. For disjoint cycles, order of multiplication does not matter, e.g.

(1 2)(3 4 5) = (3 4 5)(1 2)

For non-disjoint cycles it does matter, e.g.

(1 2)(1 3) = (1 3)(1 2)

3. Multiplication is easy using cycle notation, e.g.

f = (1 2 3 5 4) S5g = (2 4)(1 5 3) S5

then

f g = (1 4 3 2)(5)

Definition 1.15. Let g = (1 2 3) (4 5)(5 7)(8)(9) S9. The cycle shape of g is

(3, 2, 2, 1, 1)

i.e. the sequence of numbers giving the cycle-length of g in descending order. Abbreviate:

(3, 22, 12)

Eg 1.23. How many permutation of each cycle-shape in S4?

cycle-shape e.g number in S4(14) e 1

(2, 12) (1 2)(3)(4) 42 = 6(3, 1) (1 2 3)(4)

43

2 = 8(4) (1 2 3 4) 3! = 6

(22) (1 2)(3 4) 3

Total 24 = 4!.

1.4.1 Order of permutation

Recall the order o(f) of f Sn is the smallest positive integer k such that fk = e.

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Eg 1.24. f = (1 2 3 4), 4-cycle then

f1 = f

f2 = (1 3)(2 4)

f3 = (1 4 3 2)

f4 = e

So o(f) = 4. Similarly, if f = (1 2 . . . r) then o(f) = r.

Eg 1.25. g = (1 2 3)(4 5 6 7). What is o(g)?

g2 = (1 2 3)(4 5 6 7) (1 2 3)(4 5 6 7)(disjoint) = (1 2 3)2(4 5 6 7)2

Similarly

gi = (1 2 3)i(4 5 6 7)i

To make gi = e, need i to be divisible by 3 (to get rid of (1 2 3)i) and by 4 (to get rid of(4 5 6 7)i). So o(g) = lcm(3, 4) = 12.

Same argument gives

Proposition 1.8. The order of a permutation in cycle notation is the least commonmultiple of the cycle lengths

Eg 1.26. Order of (1 2)(3 4 5 6) is lcm(2, 4) = 4. The order of (1 3)(3 4 5 6) is not 4 (not

disjoint)Eg 1.27. Pack of 8 cards. Shuffle by dividing into two halves and interlacing, so if originalorder is 1, 2, . . . , 8 then the order is 1, 5, 2, 6, 3, 7, 4, 8. How many shuffles bring cards backto original order?This is the permutation s in S8

s =

1 2 3 4 5 6 7 81 5 2 6 3 7 4 8

In cycle notation s = (1)(2 5 3)(4 6 7)(8). So order of s o(s) = lcm(3, 3, 1, 1) = 3, so 3shuffles are required.

1.5 Lagranges Theorem

Recall G a finite group means G has a finite number of elements. Size of G is |G|, e.g.|S3| = 6.Theorem 1.1. Let G be a finite group. If H is any subgroup of G, then |H| divides |G|.Eg 1.28. Subgroups of S3 have size 1, 2, 3 or 6.

Note 1.4. It does not work the other wat round, i.e. ifa is a number dividing |G|, thenthere may well not exist a subgroup of G of size a.

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1.6. APPLICATIONS TO NUMBER THEORY Algebra I lecture notes

1.5.1 Consequences of Lagranges Theorem

Corollary 1. If G is a finite group and a G then o(a) divides G.Proof. Let H = a, cyclic subgroup of G generated by a. By 1.6, |H| = o(a) so byLagrange, o(a) divides |G|. 2

Corollary 2. Let G be a finite group and let n = |G|. If a G, then an = e.

Proof. Let k = o(a). By 1, k divides n. Say n = kr. So an =

akr

= er = e. 2

Corollary 3. If |G| is a prime number, then G is cyclic.

Proof. Let|G

|= p, prime. Pick a

G with a

= e. By Lagrange, the cyclic subgroup

ahas size dividing p. It contains e, a, so has size 2, therefore has size p. As |G| = p, this

implies G = a, cyclic. 2

Eg 1.29. Subgroups of S3. These have size 1 e, 2, 3 cyclic by 3. So we know all thesubgroups of S3.

1.6 Applications to number theory

Definition 1.16. Fix a positive integer m N. For any integer r, the residue class of rmodulo m denoted [r]m is

[r]m = {km + r | k Z}

Eg 1.30.

[0]5 = {5k | k Z}[1]5 = {. . . , 9, 4, 1, 6, 11, . . . } = [1]5

[2]5 = [3]5 = [8]5Since every integer is congruent to 0, 1, 2, . . . , m 1 modulo m,

[0]m [1]m [m 1]m = Z

and every integer is in exactly one of these residue classes.

Proposition 1.9.

[a]m = [b]m a b mod m

Proof.

Suppose [a]m = [b]m. As a [a]m this implies a [b]m, so a b mod m.

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Suppose a b mod m. Now

x a mod m x b mod m(as is an equivalence relation). So

x [a]m x [b]mTherefore [a]m = [b]m.

2

Eg 1.31.[17]9 = [

19]9

Definition 1.17. Write Zm for the set of all the residue classes

[0]m , [1]m , . . . , [m 1]mFrom now on well usually drop the subscript m and write

[r] = [r]m

Definition 1.18. Define +, on Zm by[a] + [b] = [a + b]

[a] [b] = [ab]This is OK, as

[a] = [a] [b] = [b]

[a + b] = [a + b] [ab] = [ab]

Eg 1.32.

[2] + [4] = [1][3] + [3] = [1]

[3] [3] = [4]

1.6.1 Groups

Eg 1.33. (Zm, +) is a group. What about (Zm, )? Identity will be [1]. So [0] will haveno inverse (as [0] [a] = [0]). So let

Zm = Zm {[0]}

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1.6. APPLICATIONS TO NUMBER THEORY Algebra I lecture notes

For which m is (Zm, ) a group?

Eg 1.34.

Z2 = {[1]}. This is a group.Z3 = {[1] , [2]}

[1] [2][1] [1] [2][2] [2] [1]

Compare with S2 to see it is a group.

Z4

[1] [2] [3][1] [1] [2] [3][2] [2] [0]

Here [2] Z4, but [2] [2] = [0] / Z4.Theorem 1.2. (Zm,

) is a group iff m is a prime number.

Proof.

Suppose Zm is a group. If m is not a prime, then

m = ab, 1 < a, b < m

so [a], [b] Zm (neither is [0]). but

[a] [b] = [ab] = [m] = [0]

This contradicts closure. So m is prime.

Suppose m is a prime, write m = p.We show that Zp is a group.

Closure Let [a] , [b] Zp. Then [a] , [b] = [0], so p |a and p |b. Then p |ab (asp is prime result from M1F). So

[a] [b] = [ab] = [0]

Thus [a] [b] Zp.

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Algebra I lecture notes 1.6. APPLICATIONS TO NUMBER THEORY

Associativity

([a] [b]) [c] = [ab] c = [(ab)c][a] ([b] [c]) = [a] [bc] = [a(bc)]

These are equal as (ab)c = a(bc) for a,b,c Z. Identity is [1] as [a] [1] = [1] [a] = [a].

Inverses Let [a] Zp. We want to find [a] such that [a] [a] = [a] [a] = [1], i.e.

[aa] = [1]

aa 1 mod p

Heres how. Well, [a] = [0] so p |a. As p is prime, hcf(p, a) = 1. By M1F,there exist integers s, t Z with

sp + ta = 1

Then

ta = 1 sp 1 mod p

So

[t] [a] = [1]Then [t] Zp ([t] = [0]) and [t] = [a]1.

2

So, Zp (p prime)

(1) is abelian

(2) has p 1 elements

Eg 1.35. Z

5 = {[1] , [2] , [3] , [4]}. Is Z

5 cyclic?Well

[2]2 = 4 [2]3 = [3] [2]4 = [1]

So Z5 = [2].Eg 1.36. In the group Z31 what is [7]

1?From the proof above, want to find s, t with

7s + 31t = 1

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1.7. APPLICATIONS OF THE GROUP ZP Algebra I lecture notes

Use Euclidean algrithm

31 = 4 7 + 37 = 2 3 + 1

So

1 = 7 2 3= 7 2(31 4 7)= 9 2 31

So [7]1 = [9] .

1.7 Applications of the group ZpTheorem 1.3. (Fermats Little Theorem) Let p be a prime, and let n be an integer notdivisible by p. Then

np1 1 mod p

Proof. Work in the groupZp =

{[1] , . . . , [p

1]

}As p |n, [n] = [0],so[n] Zp

Now Cor.?? says: if |G| = k then ak = e a G.Hence

[n]p1 = identity ofZp= [1]

Since

[n]p1 = [n] [n] = np1so

np1

= [1] np1 1 mod p(from prop. 1.9). 2

Corollary 4. Let p be prime. Then for all integers n

np n mod p

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Algebra I lecture notes 1.7. APPLICATIONS OF THE GROUP ZP

Proof. If p |n then by FLT

np1

1 mod pnp n mod p

If p|n then both np and n are congruent to 0 mod p. 2Eg 1.37. p = 5, then 1314 1 mod 5p = 17, then 6216 1 mod 17.Eg 1.38. Find remainder when divide 682 by 17.

616 1 mod 17 (FLT)(616)5 = 680 1 mod 17

682 = 680 66 62 2 mod 17

Second application.

1.7.1 Mersenne Primes

Definition 1.19. A prime number p is called a Mersenne prime if p = 2n 1 for somen N.

Eg 1.39.

22 1 = 323 1 = 724 1 = 1525 1 = 3127 1 = 127

The largest known primes are Mersenne primes. Largest known 2/2/06

230402457

1Connection with perfect numbers

Definition 1.20. A positive integer N is perfect if N is equal to the sum of its positivedivisors (including 1, not N).

Eg 1.40.

6 = 1 + 2 + 3

28 = 1 + 2 + 4 + 7 + 14

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1.7. APPLICATIONS OF THE GROUP ZP Algebra I lecture notes

Theorem 1.4. (Euler)

(1) If 2n

1 is prime then 2n1

(2n

1) is perfect.(2) Every even perfect number is of this form

Proof.

Sheet 4.

Harder - look it up.

2

It is still unsolved is there an odd perfect number?

1.7.2 How to find Meresenne Primes

Proposition 1.10. If 2n 1 is prime, then n must be prime.

Proof. Suppose n is not prime. So

n = ab, 1 < a, b < n

Then

2n 1 = 2ab 1= (2a 1)(2a(b1) + 2a(b2) + + 2a + 1)

(using xb 1 = (x 1)(xb1 + ) with x = 2a)So 2n 1 has factor 2a 1 > 1, so is not prime. Hence 2n 1 implies n prime. 2

Eg 1.41. Know22 1, 23 1, 25 1, 27 1

are prime. Next cases

211 1, 213 1, 217 1Are these prime?

We will answer this using the group Zp. We will need

Proposition 1.11. Let G be a group, and let a G. Suppose an = e. Then o(a)|n.

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Algebra I lecture notes 1.7. APPLICATIONS OF THE GROUP ZP

Proof. Let K = o(a). Write

n = qK + r, 0 r < KThen

e = an = aqK+r

= aqKar = (aK)qar

= eqar

= ar

So ar = e. This K is smallest positive integer such that aK = e and 0 r < K, this forcesr = 0. Hence K = o(a) divides n. 2

Proposition 1.12. Let N = 2p 1, p prime. Let q be prime, and suppose q|N. Thenq 1 mod p.Proof. q|N means N 0 mod q, i.e.

2p 1 mod qThis means that

[2]p = [1] Zq

We know that Zq is a group of order q 1. We also know that o([2]) in Zq divides p, so is1 or p as p is prime.If o([2]) = 1, then

[2] = [1] in Zq

that is

2 1 mod q1 0 mod q

so q|1, a contradiction.Hence we must have o([2]) = p

By Corollary 1,o([2]) divides |Zq|

That is, p divides q 1q 1 0 mod p

q 1 mod p2

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1.8. PROOF OF LAGRANGES THEOREM Algebra I lecture notes

Test for a Mersenne primeN = 2p

1

List all the primes q with q 1 mod p and q < N and check, one by one, to see if anydivide N. If none of them divide N, we have a prime.

Eg 1.42. p = 11. N = 2p 1 = 2047, N < 50. Which primes q less than 50 have q 1mod 11? We check through all numbers congruent to 1 mod 11.

12, 23, 34, 45

The only prime less than 50 that can possibly divide 2047 is 23. Now we check to see if23|211 1, i.e., if 211 = 1 mod 23.

25 32 9 mod 23210 (25)2 92 mod 23

12 mod 23211 23 mod 23

1 mod 23Conclusion 211 1 is not a prime it has a factor of 23.Eg 1.43. 213 1 is prime Exercise sheet.

1.8 Proof of Lagranges Theorem

Now we have to prove the Lagranges Theorem

Theorem 1.5. Let G be a finite group of order |G|, with a subgroup H of order |H| = m.Then m divides |G|.Note 1.5. The idea write H = {h1, . . . , hm}. Then we divide G into blocks.

H Hx Hyh1 h1x h1y . . .

h2 h2x h2y . . .... . . .1 2 3 . . . r

We want the blocks to have the following three properties

(1) Each block has m distinct elements

(2) No element of G belongs to two blocks

(3) Every element ofG belongs to (exactly) one block

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Algebra I lecture notes 1.8. PROOF OF LAGRANGES THEOREM

Then |G| is the total number of elements listed in the blocks, i.e. rm, so m||G|.

Definition 1.21. For x G, H subgroup of G, define the right cosetHx = {hx | h H} = {hx | h H}

= {h1x, h2x , . . . , hmx}The official name for a block is a right coset.

Note 1.6. Hx GEg 1.44. G = S3, H = a = {e,a,a2}, a = (1 2 3).

H = He = Ha = Ha2 = ea2, aa2, a2a2a2, e , a

Take b = (1 2), so b2 = e,

Hb =

eb,ab,a2b

=

b,ab,a2b

e ba aba2 a2n

Lemma 1.2. For any x in G

|Hx| = mProof. By definition, we have

Hx = {h1, x , . . . , hmx}These elements are all different, as

hix = hjx

hixx1 = hjxx

1

hi = hj

So |Hx| = m. 2Lemma 1.3. If x, y G then either Hx = Hy or Hx Hy = .Proof. Suppose

Hx Hy = We will show this implies Hx = Hy.We can choose an element a Hx Hy. Then

a = hix a = hjy

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1.8. PROOF OF LAGRANGES THEOREM Algebra I lecture notes

for some hi, hj H.

a =ix = hiyx = h1i hjy

Then for any h Hhx = hh1i hjy

As H is a subgroup, hh1i hj H. Hencehx Hy

This shows Hx Hy.Similarly

hix = hjy

y = h1j hix

so for any h Hhy = hh1j hix Hx

So Hy Hx.We conclude Hx = Hy. 2

Lemma 1.4. Let x

G. Then x lies in the right coset Hx.

Proof. As H is a subgroup, e H. So x = ex Hx. 2Theorem 1.6. Let G be a finite group of order |G|, with a subgroup H of order |H| = m.Then m divides |G|.Proof. By 1.4, G is equal to the union all the right cosets of H, i.e.

G =xG

Hx

Some of these right cosets will be equal (eg. G = S3, H =

a

, then H = He = Ha = Ha2).

Let the list of different right cosets be

Hx1, . . . , H xr

ThenG = Hx1 Hx2 Hxr

and Hxi = Hxj if i = j (eg. in G = S3, G = H Hb).By 1.3, Hxi Hxj = if i = j. Picture

G = Hx1 Hx2 Hxr (1.1)

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Algebra I lecture notes 1.8. PROOF OF LAGRANGES THEOREM

So |G| = |Hx1| + + |Hxr|. By 1.2

|Hxi| = m = |H|So

|G| = rm = r|H|Therefore |H| divides |G|. 2Proposition 1.13. Let G be a finite group, and H a subgroup of G. Let

r =|G||H|

Then there are exactly r different right cosets of H in G, say

Hx1, . . . H xr

They are disjoint, andG = Hx1 Hxr

Definition 1.22. The integer r = |G||H|

is called the index of H in G, written

r = |G : H|

Eg 1.45.

(1) G = S3, H = a = {e,a,a2}. Index |G : H| = 63 = 2. There are 2 right cosets H,Hb and G = H Hb.

(2) G = S3, K = b = {e, b} where b = (1 2)(3). Index |G : K| = 62 = 3. So there are 3right cosets they are

Ke = K = {e, b}Ka = {a,ba}

= a, a2bKa2 = a2, ba2 = a2, ab

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Chapter 2

Vector Spaces and Linear Algebra

Recall

Rn = {(x1, x2, . . . , xn) | xi R}Basic operations on Rn:

addition (x1, . . . , xn) + (y1, . . . , yn) = (x1 + y1, . . . , xn + yn)

scalar multiplication (x1, . . . , xn) = (x1, . . . , xn), R

These operations satisfy the following rules:

Addition rulesA1 u + (v + w) = (u + v) + w associativity

A2 v + 0 = 0 + v = v identity

A3 v + (v) = 0 inversesA4 u + v = v + u abelian

(These say (Rn, +) is an abelian group)

Scalar multiplication rules

S1 (v + w) = v + w

S2 ( + )v = v + v

S3 (v) = ()v

S4 1v = v

These are easily proved for Rn:

Eg 2.1.

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Algebra I lecture notes 2.1. DEFINITION OF A VECTOR SPACE

A1

u + (v + w) = (u1, . . . , un) + ((v1, . . . , vn) + (w1, . . . , wn))= (u1, . . . , un) + (v1 + w1, . . . , vn + wn)

= (u1 + (v1 + w1), . . . )

= ((u1 + v1) + w1, . . . )

= ((u1, . . . ) + (v1, . . . )) + (w1, . . . )

= (u + v) + w

S3

(v) = ((v1, . . . , vn))= ((v1), . . . , (vn))

= (()v1, . . . , ()vn) (assoc. of (R, ))= ()v

2.1 Definition of a vector space

A vector space will be a set of objects with addition and scalar multiplication definedsatisfying the above axioms. Want to let the scalars be either R or C (or a lot of otherthings). So let

F = either R or C

Definition 2.1. A vector space over F is a set V of objects called vectors together witha set of scalars F and with

a rule for adding any two vectors v, w V to get a vector v + w V a rule for multiplying any vector v V by any scalar F to get a vector v V. a zero vector 0 V

for any v V, a vector v VSuch that the axioms A1-A4 and S1-S4 are satisfied.

There are many different types of vectors spaces

Eg 2.2.

(1) Rn is a vector space over R

(2) Cn = {(z1, . . . , z n) | zi C} with addition u + v and scalar multiplication v ( C)is a vector space over C.

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2.1. DEFINITION OF A VECTOR SPACE Algebra I lecture notes

(3) Let m, n N. Define

Mm,n = set of all m n matrices with real entries(So in this example, vectors are matrices.) Adopt the usual rules for addition andscalar multiplication of matrices: A = (aij), B = (bij), R

A + B = (aij + bij)

A = (aij)

Zero vector is the matrix 0 (m n zero matrix). And A = (aij). Then Mm,nbecomes a vector space over R (check axioms).

(4) A non-example: Let V = R2, with usual addition defined and new scalar multiplica-tion: (x1, x2) = (x1, 0). Lets check axioms

A1-A4 hold

S1 (v + w) = v + w holds S2 ( + ) v = v + v holds S3 ( v) = ()v holds S4 1 v = v fails. To show this, need to produce just one v for which it fails,

eg. 1

(17, 259) = (17, 0)

= (17, 259)

(5) Functions. LetV = set of all functions f : R R

So vectors are functions.

Addition f + g is a function x f(x) + g(x) Scalar multiplication is a function x f(x) ( R) Zero vector is the function 0 : x 0. Inverses f is a function x f(x)

Check the axioms

A1 using associativity ofR

(f + (g + h))(x) = f(x) + (g + h)(h)

= f(x) + (g(x) + h(x))

= (f(x) + g(x)) + h(x)

= (f + g)(x) + h(x)

= ((f + g) + h) (x)

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Algebra I lecture notes 2.1. DEFINITION OF A VECTOR SPACE

Conclude V is a vector space over R.

(6) Polynomials. Recall a polynomial overR

is an expressionp(x) = a0 + a1x + + anx

with all ai R. LetV = set of all polynomials over R

Addition If p(x) =

aixi, q(x) =

bix

i then

p(x) + q(x) =

(ai + bi)xi

Scalar multiplication If p(x) =

aix

i, ai R then

p(x) = aixi Zero vector is 0 the poly with all coefficients 0

Negative of p(x) =

aixi is

p(x) =

aixi

Now check A1-A4, S1-S4. So V is a vector space over R.

Consequence of axioms

Proposition 2.1. Let V be a vector space over F and let v V, F(1) 0v = 0

(2) 0 = 0

(3) ifv = 0 then = 0 or v = 0

(4) ()v = v = (v)Proof.

(1) Observe

0v = (0 + 0)v

= 0v + 0v by S2

0v + (0v) = (0v + 0v) + (0v)0 = 0v

(2)

0 = (0 + 0)

= 0 + 0 by S1

0 = 0

Parts (3), (4) Ex. sheet 5 2

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2.2. SUBSPACES Algebra I lecture notes

2.2 Subspaces

Definition 2.2. Let V be a vector space over F, and let W V. Say W is a subspace ofV if W is itself a vector space, with the same addition and scalar multiplication as V.

Criterion for subspaces

Proposition 2.2. W is a subspace of vector space V if the following hold:

(1) 0 W(2) ifv, w W then v + w W(3) ifw W, F then w W

Proof. Assume (1), (2), (3). We show W is a vector space. Addition and scalar multiplication on W are defined by (2), (3). Zero vector 0 W by (1) Negative w = (1)w W by (3).

Finally, A1-A4, S1-S4 hold for W since they hold for V. 2

Eg 2.3.

1. V is a subspace of itself.

2. {0} is a subspace of any vector space.3. Let V = R2 and

W = {(x1, x2) | x1 + 2x2 = 0}Claim W is a subspace ofR2.

Proof. Check (1)-(3) from the proposition

(1) 0 W since 0 + 2 0 = 0(2) Let v = (v1, v2)

W, w = (w1, w2)

W. So

v1 + 2v2 = w1 + 2w2 = 0

v1 + w1 + 2(v2 + w2) = 0

v + w = (v1 + w1, v2 + w2) W(3) Let v = (v1, v2) W, R. Then

v1 + 2v2 = 0

v1 + 2v2 = 0

v = (v1, v2) W

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Algebra I lecture notes 2.3. SOLUTION SPACES

So W is a subspace by 2.2. 2

4. Same proof shows that any line through 0 (ie. px1 + qx2 = 0) is a subspace ofR2.

Note 2.1. A line not through the origin is not a subspace (no zero vector).The only subspace ofR2 are: lines through 0, R2 itself, {0}.

5. Let V = vector space of polynomials over R. Define

W = polynomials of degree at most 3

(recall deg(p(x)) = highest power ofx appearing in p(x)).Claim W is a subspace of V.

Proof.

(1) 0 W(2) ifp(x), q(x) W then deg(p), deg(q) 3, hence deg(p + q) 3, so p + q W.(3) ifp(x) W, R, then p(x) has degree of most 3, so p(x) W.

2

2.3 Solution spaces

Vast collection of subspaces ofRn is provided by the following

Proposition 2.3. Let A be an m n matrix with real entries and letW = {x Rn | Ax = 0}

(The set of solutions of the system of linear equations Ax = 0)Then W is a subspace ofRn.

Proof. We check 3 conditions of 2.2.

(1) 0 W (as A0 = 0)(2) ifv, w W then Av = Aw = 0. Hence A(v + w) = 0, so v + w W(3) ifv W, R (Av = 0), then A(v) = (Av) = 0 = 0, so v W

2

Definition 2.3. The system Ax = 0 is a homogeneous system of linear equations, and Wis called the solution space

Eg 2.4.

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2.4. LINEAR COMBINATIONS Algebra I lecture notes

1. m = 1, n = 2, A =

a b

. Then

W = x R2 | ax1 + bx2 = 0which is a line through 0.

2. m = 1, n = 3, A =

a b c

. Then

W =

x R3 | ax1 + bx2 + cx3 = 0

a plane through 0.

3. m = 2, n = 4, A =

1 2 1 0

1 0 1 2

Here

W = x R4 | x1 + 2x2 + x3 = 0, x1 + x3 + 2x4 = 04. Try a non-linear equation:

W =

(x1, x2) R2 | x1x2 = 0

Answer is no. To show this, need a single counterexample to one of the conditionsof 2.2, eg:(1, 0), (0, 1) W, but (1, 0) + (0, 1) = (1, 1) / W.

2.4 Linear Combinations

Definition 2.4. Let V be a vector space over F and let v1, v2, . . . , vk be vectors in V. Avector v V of the form

v = 1v1 + 2v2 + + kvkis called a linear combination of v1, . . . , vk.

Eg 2.5.

1. V = R2. Let v1 = (1, 1). The linear combinations of v1 are the vectors

v = v1 ( R) = (, )These form the line through origin and v1, ie. x1 x2 = 0.

2. V = R2. Let

v1 = (1, 0)

v2 = (0, 1)

The linear combinations of v1, v2 are

1v1 + 2v2 = (1, 2)

So every vector in R2 is a linear combination of v1, v2.

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3. V = R3. Let

v1 = (1, 1, 1)v2 = (2, 2, 1)Typical linear combination is

1v1 + 2v2 = (1 + 22, 1 + 22, 1 2)This gives all vectors in the plane containing origin, v1, v2, which is x1 x2 = 0. So eg.(1, 0, 0) is not a linear combination of v1, v2.

2.5 Span

Definition 2.5. Let V be a vector space over F, and let v1, . . . , vk be vectors in V. Definethe span of v1, . . . , vk, written

Sp(v1, . . . , vk)

to be the set of all linear combinations of v1, . . . , vk. In other words

Sp(v1, . . . , vk) = {1v1 + + kvk | i F} VEg 2.6.

1. V = R2, any v1 V. Then

Sp(v1) = all vectors v1 ( R)= line through 0, v1

2. In R2,Sp((1, 0), (0, 1)) = R2

3. In R3, v1 = (1, 1, 1), v2 = (2, 2, 1)Sp(v1, v2) = plane containing 0, v1, v2

= plane x1 = x2

4. In R3

Sp(v1 = (1, 0, 0), v2 = (0, 1, 0), v3 = (0, 0, 1)) = whole of R3

5. V = R3. Let

w1 = (1, 0, 0)

w2 = (1, 1, 0)

w3 = (1, 1, 1)

Claim: Sp(w1, w2, w3) = R3.

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Proof. Observe

v1 = w1v2 = w2 w1v3 = w3 w2

Hence any linear combination ofv1, v2, v3 is also a linear combination ofw1, w2, w3 (i.e.(1, 2, 3) = 1v1+2v2 +3v3 = 1w1+2(w2w1)+3(w3w2) Sp(w1, w2, w3))2

6. V = vector space of polynomials over R. Let

v1 = 1

v2 = x

v3 = x2

Then

Sp(v1, v2, v3) = {1v1 + 2v2 + 3v3 | i R}=

1 + 2x + 3x

2 | i R

= set of all polynomials of degree 2

Eg 2.7. In general, If v1, v2 are vectors in R3, not on same line through 0 (i.e. v1

= v1),

thenSp(v1, v2) = plane through 0, v1, v2

Proposition 2.4. V vector space, v1, . . . , vk V. ThenSp(v1, . . . , vk)

is a subspace of V.

Proof. Check the conditions of 2.2

(1) Taking all i = 0 (using 2.1)

0v1 + 0v2 + + 0vk = 0 + + 0 = 0So 0 is a linear combination of v1, . . . , vk, so 0 Sp(v1, . . . , vk)

(2) Let v, w Sp(v1, . . . , vk), sov = 1v1 + + kvk

w = 1v1 + + kvkThen v + w = (1 + 1)v1 + + (k + k)vk Sp(v1, . . . , vk).

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(3) Let v Sp(v1, . . . , vk), F, so

v = 1v1 + + kvkso

v = (1)v1 + + (kvk) Sp(v1, . . . , vk)

2

2.6 Spanning sets

Definition 2.6. V vector space, W a subspace of V. We say vectors v1, . . . , vk span W if

(1) v1, . . . , vk W and

(2) W = Sp(v1, . . . , v2)

Call the set {v1, . . . , vk} a spanning set of W.

Eg 2.8.

{(1, 0, 0) , (1, 1, 0) , (1, 1, 1)} is a spannig set for R3.

(1, 1, 1) , (2, 2, 1) span the plane x1 x2 = 0. Let

W =

x R4

1 1 3 12 3 1 1

1 0 8 2

x = 0

Find a (finite) spanning set for W.Solve system

1 1 3 1 02 3 1 1 01 0 8 2 0

1 1 3 1 0

0 1 5 1 00 1 5 1 0

1 1 3 1 00 1 5 1 0

0 0 0 0 0

Echelon form:

x1 + x2 + 3x3 + x4 = 0

x2 5x3 x4 = 0

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General solution

x4 = ax3 = b

x2 = a + 5b

x1 = a 3b (a + 5b)= 2a 8b

i.e. x = (2a 8b, a + 5b,b,a). So W = {(2a 8b, a + 5b,b,a) | a, b R} Definetwo vectors (take a = 1 and b = 0 and vice versa)

w1 = (

2, 1, 0, 1) a = 1, b = 0

w2 = (8, 5, 1, 0) a = 0, b = 1Claim W = Sp(w1, w2)

Proof. Observe

(2a 8b, a + 5b,b,a) = a(2, 1, 0, 1) + b(8, 5, 1, 0)= aw1 + bw2

This gives a general method of finding spanning sets of solution spaces. 2

2.7 Linear dependence and independence

Definition 2.7. V vector space over F. We say a set of vectors v1, . . . , vk in V is a linearlyindependent set if the following condition holds

1v1 + + kvk = 0 all i = 0

Usually just say the vectors v1, . . . , vk are linearly independent vectors.We say the set {v1, . . . , vk} is linearly dependent if the oposite true, i.e. if we can findscalars i such that

(1) 1v1 + + kvk = 0(2) at least one i = 0

Eg 2.9.

1. V = R2, v1 = (1, 1). Then {v1} is a linearly independent set, as

v1 = 0 (, ) = (0, 0) = 0

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2. V = R2, the set {0} is linearly dependent, e.g.

20 = 0

3. In R2, let v1 = (1, 1), v2 = (2, 1). Is {v1, v2} linearly independent?Consider equation

1v1 + 2v2 = 0

i.e.(1, 1) + (22, 2) = (0, 0)

i.e.

1 + 22 = 01 + 2 = 0

1 = 2 = 0

4. In R3, let

v1 = (1, 0, 1)

v2 = (2, 2, 1)v3 = (1, 4, 5)

Are v1, v2, v3 linearly independent?Consider system

x1v1 + x2v2 + x3v3 = 0 (2.1)

This is the system of linear equations 1 2 10 2 4

1 1 5

x = 0

(i.e.

v1 v2 v3

x = 0)Solve

1 2 1 00 2 4 01

1

5 0

1 2 1 00 2 4 00

3

6 0

1 2 1 00 2 4 00 0 0 0

Solution x = (3a, 2a, a) (any a). So

3v1 2v2 + v3 = 0So v1, v2, v3 are linearly dependent. Geometrically, v1, v2 span a plane in R

3 and v3 =3v1 + 2v2 Sp(v1, v2) is in this plane.In general: in R3, three vectors are linearly dependent iff they are coplanar.

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2.7. LINEAR DEPENDENCE AND INDEPENDENCE Algebra I lecture notes

4. V = vector space of polynomials over R. Let

p1(x) = 1 + x

2

p2(x) = 2 + 2x x2p3(x) = 1 + 4x 5x2

Are p1, p2, p3 linearly dependent?Consider equation

1p1 + 2p2 + 3p3 = 0

Equating coefficients

1 + 22 + 3 = 0

22 + 43 = 0

1 2 53 = 0Showed in previous example that a solution is

1 = 3, 2 = 2, 3 = 1So

3p1 2p2 +p1 = 0So linearly dependent.

5. V = vector space of functions R R. Letf1(x) = sin x, f2(x) = cos x

So f1, f2 V. Are f1, f2 linearly independent? Sheet 6.

Two basic results about linearly independent sets.

Proposition 2.5. Any subset of a linearly independent set of vectors is linearly indepen-dent.

Proof. Let S be a lin. indep. set of vectors, and T

S. Label vectors in S, T

T = {v1, . . . , vt}S = {v1, . . . , vt, vt+1, . . . , vs}

Suppose

1v1 + + tvt = 0Then

1v1 + + tvt + 0vt+1 + + 0vs = 0As S is lin. indep., all coeffs must be 0, so all i = 0. Thus T is lin. indep. 2

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Proposition 2.6. V vector space, v1, . . . , vk V. Then the following two statements areequivalent (i.e. (1)

(2)).

(1) v1, . . . , vk are lin. dependent

(2) there exists i such that vi is a linear combination of v1, . . . , vi1.

Proof.

(1) (2) Suppose v1, . . . , vk is lin. dep., so there exist i such that1v1 + + kvk = 0

and j = 0 for some j. Choose the largest j for which j = 0. So1v1 + + jvj = 0

Then

jvj = 1v1 j1vj1So

vj = 1j

j1j

which is a linear combination of v1, . . . , vj1.

(1) (2) Assume vi is a linear combination of v1, . . . , vi1, sayv1 = 1v1 + + i1vi1

Then

1v1 + + i1vi1 vi + 0vi+1 + + 0vk = 0Not all the coefficients in this equation are zero (coef of vi is 1). So v1, . . . , vk arelin. dependent.

2

Eg 2.10. v1 = (1, 0, 1), v2 = (2, 2, 1), v3 = (1, 4, 5) in R3. These are linearly dependent:3v1 2v2 + v3 = 0. And v3 = 3v1 + 2v2 a linear combination of previous ones.Proposition 2.7. V vector space, v1, . . . , vk V. Suppose vi is a linear combination ofv1, . . . , vi1. Then

Sp(v1, . . . , vk) = Sp(v1, . . . , vi1, vi+1, . . . , vk)

(i.e. throwing out vi does not change Sp(v1, . . . , vk))

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2.8. BASES Algebra I lecture notes

Proof. Letvi = 1v1 +

+ i1vi1 (j

F)

Now considerv = 1v1 + + kvk Sp(v1, . . . , vk)

Then

v = 1v1 + + i1vi1 ++i(1v1 + + i1vi1) ++i+1vi+1 + + kvk

So v is a lin. comb. of

v1, . . . , vi1, vi+1, . . . , vk

Therefore Sp(v1, . . . , vk) Sp(v1, . . . , vi1, vi+1, . . . vk). 2Eg 2.11. v1 = (1, 0, 1), v2 = (2, 2, 1), v3 = (1, 4, 5). Here

v3 = 3v1 + 2v2So Sp(v1, v2, v3) = Sp(v1, v2).

2.8 Bases

Definition 2.8. V a vector space. We say a set of vectors {v1, . . . , vk} in V, is basis of Vif

(1) V = Sp(v1, . . . , vk)

(2) {v1, . . . , vk} is a linearly independent set.Informally, a basis is a spanning set for which we cannot throw any of the vectors away.

Eg 2.12.

1. {(1, 0)v1

, (0, 1)v2

} is a basis ofR2.

Proof.

(1) (x1, x2) = x1v1 + x2v2 so R2 = Sp(v1, v2)

(2) v1, v2 are linearly independent as

1v1 + 2v2 = 0 (1, 2) = (0, 0) 1 = 2 = 0

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2

2. (1, 0, 0), (1, 1, 0), (1, 1, 1) is a basis ofR3

Proof.

(1) They span R3 previous example.

(2)

x1v1 + x2v2 + x3v3 = 0

leads to the system 1 1 1 00 1 1 0

0 0 1 0 with the only solution x1 = x2 = x3 = x4 = 0 v1, v2, v3 are lin. indep.

2

Theorem 2.1. Let V be a vector space with a spanning set v1, . . . , vk (i.e. V = Sp(v1, . . . , vk)).Then there is a subset of{v1, . . . , vk} which is a basis of V.Proof. Consider the set

v1, . . . , vk

We throw away vectors in this list which are linear combinations of the previous vectors

in the list. End up with a basis. Process:Casting out Process

First, throw away any zero vectors in the list.

Start at v2: if it is a linear combination v1, (i.e. v2 = v1), then delete it; if not,leave it there.

Now consider v3: if it is a linear combination of the remaining previous vectors, deleteit; if not, leave it there.

Continue, moving from left to right, deleting any vi, which is a linear combination of

previous vectors in the list.

End up with a subset {w1, . . . , wm} of{v1, . . . , vk} such that(1) V = Sp(w1, . . . , wm) (by 2.7)

(2) no wi is a linear combination of previous ones.

Then {w1, . . . , wm} form a linearly independent set by 2.6. Therefore {w1, . . . , wm} is abasis of V. 2

Eg 2.13.

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2.8. BASES Algebra I lecture notes

1. V = R3, v1 = (1, 0, 1), v2 = (2, 2, 1), v3 = (1, 4, 5). Let W = Sp(v1, v2, v3). Finda basis of W.

1) Is v2 a linear combination of v1? No: leave it in.

2) Is v3 a linear combination of v1, v2? Yes: v3 = 3v1 + 2v2.So cast out v3: basis for W is {v1, v2}.

2. Heres a meatier example of The Casting out Process. Let V = R4 and

v1 = (1, 2, 3, 1)v2 = (2, 2, 2, 1)

v3 = (5, 2, 1, 3)v4 = (11, 2, 1, 7)v5 = (2, 8, 2, 3)

Let W = Sp(v1, . . . , v5), subspace ofR4. Find a basis of W.

The all-in-one-go method: Form 5 4 matrix

v1...

v5

and reduce it to echelon form:

1 2 3 12 2 2 15 2 1 3

11 2 1 72 8 2 3

v1

v2v3v4v5

1 2 3 10 6 8 10 12 16 20 24 32 40 12 4 1

v1

v2 2v1v3 5v1v4 11v1v5 2v1

1 2 3 10 6 8 10 0 0 00 0 0 00 0 12 3

v1v2 2v1v3 5v1 2(v2 2v1)v4 11v1 4(v2 2v1)v5 2v1 2(v2 2v1)

So v3 is a linear combination of v1 and v2: cast it out. And v4 is linear combinationof previous ones: cast it out.Row vectors in echelon form are linearly independent: So last row v5 + 2v1 2v2 isnot a linear combination of the first two rows. So v5 is not a linear combination ofv1, v2.Conclude: Basis of W is {v1, v2, v5}.

To help with spanning calculations:

Eg 2.14. Let v1 = (1, 2, 1), v2 = (2, 0, 1), v3 = (0, 1, 3), v4 = (1, 2, 3). Do v1, v2, v3, v4span the whole ofR3?

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Algebra I lecture notes 2.9. DIMENSION

Let b R3. Then b Sp(v1, v2, v3, v4) iff system x1v1+x2v2+x3v3+x4v4 = b has a solutionfor x1, x2, x3, x4

R. This system is

1 2 0 1 b2 0 1 2 b21 1 3 3 b3

1 2 0 1 b10 4 1 0 b2 2b1

0 3 3 4 b3 + b1

1 2 0 1 b10 4 1 0 b2 2b1

0 0 9 12

This system has a solution for any b R3. Hence Sp(v1, . . . , v4) R3.

2.9 DimensionDefinition 2.9. A vector space V is finite-dimensional if it has a finite spanning set, i.e.there is a finite set of vectors v1, . . . , vk such that V = Sp(v1, . . . , vk).

Eg 2.15. Rn is finite dimensional. To show this, let

e1 = (1, 0, 0, . . . , 0)

e2 = (0, 1, 0, . . . , 0)...

en = (0, 0, 0, . . . , 1)

Then for any x = (x1, . . . , xn) Rn

x = x1e1 + x2e2 + + xnenSo Rn = Sp(e1, . . . , en) is finite-dimensional.

Note 2.2. {e1, . . . , en} is linearly independent since 1e1+ +nen = 0 implies (1, . . . , n) =0, so all i = 0. So {e1, . . . , en} is a basis for Rn, called the standard basis.Eg 2.16. Let V be a vector space of polynomials over R.

Claim: V is not finite-dimensional.

Proof. By contradiction. Assume V has a finite spanning set p1, . . . , pk. Let deg (pi) = niand let n = max (n1, . . . , nk. Any linear combination 1p1 + + kpk ( R) has degree n. So the poly xn+1 is not a linear combination of vectors from our assumed spanningset; contradiction. 2

Proposition 2.8. Any finite-dimensional vector space has a basis.

Proof. Let V be a finite-dimensional vector space. Then V has a finite spanning set. Thiscontains a basis of V by Theorem 2.1. 2

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Definition 2.10. The dimension ofV is the number of vectors in any basis of V.1 Writtendim V

Eg 2.17. Rn has basis e1, . . . , en, so dimRn = n.

Eg 2.18. Let v R2, v = 0 and let L be the line through 0 and v. So L is a subspace.L = {v | R}

So L = Sp(v) and {v} is a basis ofL. So dim L = 1.Eg 2.19. Let v1, v2 R3 with v1, v2 = 0 and v2 = v1. Then Sp(v1, v2) = P is a planethrough 0, v1, v2. As v2 = v1, {v1, v2} is linearly indepenednt, so is a basis of P. Sodim P = 2.

Major result:Theorem 2.2. Let V be a finite-dimensional vector space. Then all bases of V have thesame number of vectors.

Proof. Based on:

Lemma 2.1. (Replacement Lemma) V a vector space. Suppose v1, . . . , vk and x1, . . . , xrare vectors in V such that

v1, . . . , vk span V x1, . . . , xr are linearly independent

Then

(1) r k and(2) there is a subset {w1, . . . , wkr} of{v1, . . . , vk} such that x1, . . . , xr, w1, . . . , wkr span

V (i.e. we can replace r of the vs by the xs and still span V)

Eg 2.20. V = R3.

e1, e2, e3 span R3

x1 = (1, 1, 1)According to 2.1(2), we can replace one of the eis by x1 and get a spanning set {x1, ei, ej}.How? Consider spanning set

x1, e1, e2, e3

This set is linearly dependent since x = e1 + e2 e3. By 2.6, one of the vectors is thereforea linear combination of previous ones in this case

e3 = e1 + e2 x1So cast out e3 spanning set is {x1, e1, e2}.

1following theorem shows the uniqueness of this number

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Algebra I lecture notes 2.10. FURTHER DEDUCTIONS

Proof. (of Lemma 2.1)Consider S1 =

{x1, v1, . . . , vk

}. This spans V. It is linearly dependent, as x1 is a linear

combination of the spanning set v1, . . . , vk. So by 2.6, some of the vectors in S1 is alinear combination of previous ones. This vector is not x1, so it is some vi. By 2.7,V = Sp(x1, v1, . . . , vi, . . . , vk). Now let S2 = {x1, x2, v1, . . . , vi, . . . , vk}. This spans V andis linearly dependent, as x2 is a linear combination of others. By 2.6, there exists a vectorin S2 which is linear combination of previous ones. It is not x1 and x2, as x1, x2 are linearlyinedpendent. So it is some vj. By 2.7, V = Sp(x1, x2, v1, . . . , vi, . . . , vj , . . . , vk). Continuelike this, adding xs, deleting vs.If r > k, then eventually, we delete all the vs and get V = Sp(x1, . . . , xk). Then xk+1 is alinear combination of x1, . . . , xk. This cant happed as x, . . . , xk+1 is linearly independentset. Therefore r k.This proces ends when weve used up all the xs, giving

V = Sp(x1, . . . , xr, k r remaining vs)2

(Proof of 2.2 continued)Let {v1, . . . , vk} and {x1, . . . , xr} be the bases ofV. Both are spanning sets for V and bothare linearly independent. Well v1, . . . , vk span, x1, . . . , xr is linearly inedpendent, so by theprevious lemma, r k.Similarly, x1, . . . , xr span. v1, . . . , vk is linearly independent, so by the previous lemmaagain, k

r.

Hence r = k. So all bases of V have the same number of vectors. 2

2.10 Further Deductions

Proposition 2.9. Let dim V = n. Any spanning set for V of size n is a basis ofV.

Proof. Let {v1, . . . , vn} be the spanning set. By 2.1, this set contains a basis of V. By 2.2,all bases of V have the size n. Therefore, {v1, . . . , vn} is a basis of V. 2Eg 2.21. Is (1, 2, 3), (0, 2, 5), (1, 0, 6) a basis ofR3?

1 2 30 2 5

1 0 6

1 2 30 2 5

0 2 9

1 2 30 2 5

0 0 14

The rows of this echelon form are linearly independent, so cant cast out any vectors. Sothey form a basis.

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Proposition 2.10. If{x1, . . . , xr} is a linearly independent set in V, then there is a basisof V containing x1, . . . , xr.

Proof. Let v1, . . . , vn be a basis ofV. By 2.1(2), there exists {w1, . . . , wnr} {v1, . . . , vn}such that

V = Sp(x1, . . . , xr, w1, . . . , wnr)

Then x1, . . . , xr, w1, . . . , wnr is a spanning set of size n, hence is a basis by 2.9. 2

Eg 2.22. Let v1 = (1, 0, 1, 2), v2 = (1, 1, 2, 5) R4. Find a basis ofR4 containing v1, v2.Claim: v1, v2, e1, e2 is a basis ofR

4.

Proof. Clearly can get all standard basis vectors e1, e2, e3, e4 as linear combination ofv1, v2, e1, e2. So v1, v2, e1, e2 span R

4, so they are basis by 2.9. 2

Proposition 2.11. Let W be subspace of V. Then

(1) dim W dim V(2) IfW = V, then dim W < dim V

Proof.

(1) Let w1, . . . , wr be a basis of W. This set is linearly independent. So by Proposition2.10 there is a basis of V containing it. Say w1, . . . , wr, v1, . . . , vs. Then dim V =

r + s r = dim W.(2) If dim W = dim V, then s = 0 and w1, . . . , wr is a basis ofV, so V = Sp(w1, . . . , wr) =

W.

2

Eg 2.23. (The subspaces ofR3) Let W be a subspace ofR3. Then dim W dimR3 = 3.Possibilities:

dim W = 3 Then W = R3

dim W = 2 Then W has a basis {v1, v2} so W = Sp(v1, v2), which is a plane through0, v1, v2.

dim W = 1 Then W has a basis {v1} so W = Sp(v1), which is a line through 0, v1. dim W = 0 Then W = {0}.

Conclude: The subspaces ofR3 are {0}, R3 and lines and planes containing 0.Proposition 2.12. Let dim V = n. Any set of n vectors which is linearly independent isa basis of V.

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Algebra I lecture notes 2.10. FURTHER DEDUCTIONS

Proof. Let v1, . . . , vn be linearly independent. By Proposition 2.10 there is a basis contain-ing v1, . . . , vn. As all bases have n vectors, v1, . . . , vn must be a basis. 2

Eg 2.24. Is the set (1, 1, 1, 0), (2, 0, 1, 2), (0, 3, 1, 1), (2, 2, 1, 0) a basis ofR4?v1v2v3v4

1 1 1 02 0 1 20 3 1 12 2 1 0

1 1 1 00 2 2 20 3 1 10 0 3 0

1 1 1 00 1 1 10 0 4 50 0 3 0

1 1 1 00 1 1 10 0 4 50 0 0 1

w1w2w3w4

The vectors w1, w2, w3, w4 are linearly independent (clear as they are in echelon form). By2.12, w1, . . . , w4 are a basis ofR

4, therefore v1, . . . , v4 span R4 (as ws are linear combina-

tions of vs), therefore v1, . . . , v4 is a basis ofR4, by 2.9.

Proposition 2.13. Let dim V = n. Then any set ofn + 1 vectros or more vectors in V islinearly dependent.

Proof. Let S be a set ofn + 1 or more vectors. IfS is linearly independent, it is containedin a basis by Proposition 2.10, which is impossible as all bases have n vectors. So S islinearly dependent. 2

Eg 2.25. (A fact about matrices)Let V = M2,2, the vector space of all 2 2 matrices over R (usual addition A + B andscalar multiplication A of matrices) Basis: Let

E11 = 1 00 0

E12 =0 1

0 0

E21 =

0 01 0

E22 =

0 00 1

Claim: E11, E12, E21, E22 is a basis of V = M2,2.

Proof.

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2.10. FURTHER DEDUCTIONS Algebra I lecture notes

Span

a bc d = aE11 + bE12 + cE21 + dE22 Linear independence

1E11 + 2E12 + 3E21 + 4E22 = 0 implies1 23 4

=

0 00 0

i = 0

2

So dim V = 4.Now let A V = M2,2. Consider I, A, A2, A3, A4. These are 5 vectors in V, so they arelinearly dependent by 2.13. This means there exist i R (at least one non zero) suchthat

4A4 + 3A

3 + 2A2 + 1A + 0I = 0

This means, if we write

p(x) = 4x4 + 3x

3 + 2x2 + 1x + 0

then p(x) = 0 and p(A) = 0. So weve proved the following:Proposition 2.14. For any 2 2 matrix A there exists a nonzero polynomial p(x) ofdegree 4, such that p(A) = 0.Note 2.3. This generalizes to n

n matrices.

Summary so far

V a finite-dimensional vector space (i.e. V has a finite spanning set)

Basis of V is a linear independent spanning set All bases have the same size called dim V (Theorem 2.2) Every spanning set contains a basis (Theorem 2.1)

Write dim V = n Any spanning set of size n is a basis (Proposition 2.9) Any linearly independent set of size n is a basis (Proposition 2.12) Any linearly independent set is contained in a basis (Proposition 2.10) Any set of n + 1 or more vectors is linearly dependent (Proposition 2.13) Any subspace W of V has dim W n, and dim W = n W = V (Proposition

2.11)

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Chapter 3

More on Subspaces

3.1 Sums and Intersections

Definition 3.1. V a vector space. Let U, W be subspaces of V. The intersection of Vand W is

U W = {v | v U and v W}The sum of U and W

U + W = {u + w | u U and w W}Note 3.1. U + W contains

all the vectors in u U (as u + 0 U + W) all the vectors in w W many more vectors (usually)

Eg 3.1. V = R2 U = Sp(1, 0), W = Sp(0, 1). Then U + W contains all vectors 1(1, 0) +2(0, 1) = (1, 2). So U + W is the whole of R

2.

Proposition 3.1. U W and U + W are subspaces of V.

Proof. Use subspaes criterion, Proposition 2.3

U + W

(1) As U, W are subspaces, both contain 0, so 0 + 0 = 0 U + W(2) Let u1 + w1, u2 + w2 U+ W (where ui U, wi W). Then (u1 + w1) + (u2 +

w2) = (u1 + u2) + (w1 + w2) U + W(3) Let u + w U + W, F. Then

(u + w) = u + w U + W

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3.1. SUMS AND INTERSECTIONS Algebra I lecture notes

U W Sheet 8.2

What about dimensions of U + W, U W?First:

Proposition 3.2. If U = Sp(u1, . . . , ur), W = Sp(w1, . . . , ws). Then

U + W = Sp(u1, . . . , ur, w1, . . . , ws)

Proof. Let u + w U + W. Then (i, i F)

u = 1u1 +

+ rvr

w = 1w1 + + swsSo

u + w = 1u1 + + rvr + 1w1 + + sws Sp(u1, . . . , ur, w1, . . . , wr)

So

U + W Sp(u1, . . . , ur, w1, . . . , ws)

All the ui, wi are in U+W. As U+W is a subspace, it therefore contains Sp(u1, . . . , ur, w1, . . . , ws).HenceU + W = Sp(u1, . . . , ur, w1, . . . , ws)

2

Eg 3.2. In the above example, U + W = Sp((1, 0), (0, 1)) = R2.

Eg 3.3. Let U = {x R3 | x1 + x2 + x3 = 0}, W = {x R3 | x1 + 2x2 + x3 = 0} sub-spaces ofR3. Find bases of U, W, U W, U + W.

For U general solution is (a b,b,a), so basis for U is: {(1, 0, 1), (1, 1, 0)}. For W general solution is (2b + a,b,a), so basis of W is: {(1, 0, 1), (2, 1, 0)}.

U W: this is

x R3

1 1 11 2 1

x = 0. Solve

1 1 1 0

1 2 1 0

1 1 1 00 3 2 0

General solution is (a, 2a, 3a). Basis for U + W is {(1, 2, 3)}.

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Algebra I lecture notes 3.1. SUMS AND INTERSECTIONS

U + W: By Proposition 3.2

U + W = Sp((1, 0, 1), (1, 1, 0), (1, 0, 1), (2, 1, 0))Check that can cast out only 1 vector. So U + W has dimension 3, so

U + W = R3

So

dim U = dim W = 2

dim U W = 1dim U + W = 3

Theorem 3.1. Let V be a finite-dimensional space and let U, W be subspaces ofV. Then

dim(U + W) = dim U + dim W dim(U W)Proof. Let

dim U = m

dim W = n

dim(U W) = r

Aim: to prove dim (U + W) = m + n r. Start with basis {x1, . . . , xr} ofU W. By 2.10,can extend this to bases

Bu = {x1, . . . , xr, u1, . . . , umr} basis of UBw = {x1, . . . , xr, w1, . . . , wnr} bases of W

Let

B = Bu Bw = {x1, . . . , xr, u1, . . . , umr, w1, . . . , wnr}Claim B is a basis ofU + W.

Proof.

(1) Span: B spans U + W by Proposition 3.2.

(2) Linear independence: We show that B is linearly independent. Suppose

1x1 + + rxr + 1u1 + + mrumr + 1w1 + + nrwnri.e.

r

i=1ixi +

mr

i=1iui +

nr

i=1iwi = 0 (3.1)

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Let

v =nr

i=1

iwi

Then v W. Alsov =

ixi

iui U

So v is in U W.As x1, . . . , xr is a basis of U W

v =ri=1

ixi ( F)

As v = iwi, this gives

ri=1

ixi +nri=1

iwi = 0

Since Bw = {x1, . . . , xr, w1, . . . , wnr} is linearly independent, this forces (for all i)i = 0

i = 0

(i.e. v = 0). Then by (3.1)

ri=1

ixi +mri=1

iui = 0

Since Bu = {x1, . . . , xr, u1, . . . , umr} is linearly independent, this forces (for all i)i = 0

i = 0

So weve shown that in (3.1), all coefficients i, i, i are zero, showing that B =

Bu Bw is linearly independent. Hence B is a basis of U + W.2

Then we proved that

dim(U + W) = r + m r + n r = r + m r2

Eg 3.4. V = R4. Suppose U, W are subspaces with dim U = 2, dim W = 3. Thendim U + W 3 (as it contains W) and dim U + W 4 (as U + W R4). Possibilities

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Algebra I lecture notes 3.2. THE RANK OF A MATRIX

dim U + W = 3 Then U + W = W and so U W.

dim U + W = 4 (in other words U + W = R4

). Thendim(U W) = dim U + dim W dim U + W = 1

For example this happens:

U = Sp(e1, e2) = {(x1, x2, 0, 0) | xi R}W = Sp(e1, e3, e4) = {(x1, 0, x2, x3)}

3.2 The rank of a matrix

Definition 3.2. Let A be an m n matrix with real entries. Definerow-space(A) = subspace ofRn spanned by the rows of A

column-space(A) = subspace ofRm spanned by the columns of A

Eg 3.5. A =

3 1 20 1 1

row-space(A) = Sp((3, 1, 2), (0, 1, 1))

column-space(A) = Sp

30

,

1

1

,

21

Definition 3.3. Let A be m n matrix. Definerow-rank(A) = dim row-space(A)

column-rank(A) = dim column-space(A)

Eg 3.6. In above example

row-rank(A) = column-rank(A) = 2

3.2.1 How to find row-rank(A)

Procedure:

(1) Reduce A to echelon form by row operations, say

A =

0 . . . 1 . . . . . . . . . . . . . . .0 . . . 0 . . . 1 . . . . . . . . .

...0 . . . 0 . . . 0 1 . . . . . .

Then (we will prove this)

row-space(A) = row-space(A)

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3.2. THE RANK OF A MATRIX Algebra I lecture notes

(2) Then row-rank(A) = number of nonzero rows in echelon form A and these nonzerorows are a basis for row-space(A).

Proof.

(1) Rows of A are linear combinations of the rows ofA (since they are obtained by rowoperations ri ri + rj, etc.) Therefore

row-space(A) Sp( rows of A)= row-space(A)

By reversing the row operations to go from A to A, we see that rows of A are linearcombinations of rows of A, so

row-space(A) row-space(A)

Therefore row-space(A) = row-space(A)

(2) Let the nonzero rows of A be v1, . . . , vr

A =

0 . . . 1 . . . . . . . . . . . . . . .0 . . . 0 . . . 1 . . . . . . . . .

...

0 . . . 0 . . . 0 1 . . . . . .

v1v2...

vr

Then

row-space(A) = Sp(v1, . . . , vr)

Also v1, . . . , vr are linearly independent, since

1v1 + + rvr = 0

implies

1 = 0 (since 1 is i1 coordinate of LHS)

2 = 0 (since 1 is i2 coordinate of LHS)

j = 0 (since 1 is ij coordinate of LHS)

Therefore v1, . . . , vr is a basis for row-space(A), hence for row-space(A). So

row-rank(A) = r = no. of nonzero rows ofA

2

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Eg 3.7. Find the row-rank of

A = 1 2 52 1 0

1 4 15

Reduce to echelon form:

A 1 2 50 3 10

0 6 20

1 2 5

0 3 100 0 0

Row-rank is 2.

Eg 3.8. Find the dimension of

W = Sp((1, 1, 0, 1), (2, 3, 1, 0), (0, 1, 2, 3)) R4

Observe

W = row-space(

1 1 0 12 3 1 00 1 2 3

) = ASo dim W = row-rank(A).

A 1 1 0 10 5 1 2

0 1 2 3

1 1 0 1

0 5 1 20 0 9 12

So dim W = row-rank(A) = 3.

3.2.2 How to find column-rank(A)?

Clearly

column-rank(A) = row-rank(AT)

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Eg 3.9. column-rank() of A =

1 2 52 1 0

1 4 15

AT =

1 2 12 1 4

5 0 15

1 2 10 3 6

0 0 0

So column-rank(A) = 2.

Theorem 3.2. For any matrix A,

row-rank(A) = column-rank(A)

Proof. Let

A =

a11 a1n... . . . ...

am1 amn

v1...

vm

So vi = (ai1, . . . , ain).Let

k = row-rank(A)

= dim Sp(v1, . . . , vm)

Let w1, . . . , wk be a basis for row-space(A). Say

w1 = (b11, . . . , b1n)...

wk = (bk1, . . . , bkn)

Each vi

Sp(w1, . . . , wk), so (ij

F)

v1 = 11w1 + + 1kwk...

vm = m1w1 + + mkwkEquating coordinates:

ith coord ofv1 : a1i = 11b1i + + 1kbki...

ith coord of vm : ami = m1b1i + + mkbki

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This says

ith column of A =a1i...

ami

= b1i11...m1

+ + bki1k...mk

Hence each column of A is a linear combination of the k column vectors

11...m1

, . . . ,

1k...

mk

= l1, . . . , lk

So column-space(A) is spanned by these k vectors. So

column-rank(A) = dim column-space(A) k = row-rank(A)

So weve shown that

column-rank(A) row-rank(A)

Applying the same to AT:

column-rank(AT) row-rank(AT)

i.e.

row-rank(A) column-rank(A)Hence row-rank(A) = column-rank(A). 2

Eg 3.10. illustrating the proofLet

A =

1 2 1 01 1 0 1

0 3 1 2

v1v2v3

As v3 = v1 + v2, basis of row-space is w1, w2 where

w1 = v1

w2 = v2

Write each vi as a linear combination of w1, w2

v1 = w1 = 1w1 + 0w2

v2 = w2 = 0w1 + 1w2

v3 = w1 + w2 = 1w1 + 1w2

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So the column vectors l1, l2 are

101

l2

, 01

1

l2

These span column-space(A). Check 11

0

= l1 l2

213 = 2l1 + l2

101

= l1

Definition 3.4. The rank of a matrix is its row-rank or its col-rank, written

rank(A) or rk(A)

Proposition 3.3. Let A be n

n. Then the following four statements are equivalent:

(1) rank(A) = n

(2) rows of A are a basis ofRn

(3) columns of A are a basis ofRn

(4) A is invertible

Proof.

(1) (2)

rank(A) = n

dim row-space(A) = n

the n rows of A span R3

the n rows are a basis ofR3 (2.9)

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(1) (3) Similarly

(1) (4)rank(A) = n

A can be reduced to echelon form

A can be reduced to In

A is invertible (M1GLA, 7.5)

2

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Chapter 4

Linear Transformations

Linear transformations are functions from one vector space to another which preserveaddition and scalar multiplication, i.e. for linear transformation T if

v1 w1 = T(v1)v2 w2 = T(v2)

thenv1 + v2 w1 + w2 = T(v1) + T(v2)

andv1 w1 = T(v1)

Definition 4.1. Let V, W be vector spaces. A function T : V W is a linear transfor-mation if

1) T(v1 + v2) = T(v1) + T(v2) for all v1, v2 V2) T(v) = T(v) for all v V, F

Eg 4.1.

(1) Define T : R1 R1 byT(x) = sin x

Then T is not a linear transformation: e.g.

T() = sin = 0

2T

2

= 2 sin

2= 2

So 2T2

= T()(2) T : R2 R1,

T(x1, x2) = x1 + x2

T is a linear transformation:

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1)

T((x1, x2) + (y1, y2)) = T(x1 + y1, x2 + y2)= x1 + x2 + y1 + y2

= T(x1, x2) + T(y1, y2)

2)

T((x1, x2)) = T(x1, x2)

= x1 + x2

= (x1 + x2)

= T(x1, x2)

(3) T : R2 R2T(x1, x2) = x1 + x2 + 1

T is not linear: e.g.

T(2(1, 0)) = T(2, 0) = 3

2T(1, 0) = 4

(4) V vector space of polynomials. Define T : V V by

T(p(x)) = p(x)

e.g.T(x3 3x) = 3x2 3

Then T is linear transformation:

1)

T(p(x) + q(x)) = p(x) + q(x)

= T(p(x)) + T(q(x))

2)

T(p(x)) = p(x) = T(p(x))

Basic examples:

Proposition 4.1. Let A be an m n matrix over R. Define T : Rn Rm by (for allx Rn, column vectors)

T(x) = Ax

Then T is a linear transformation.

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4.1. BASIC PROPERTIES Algebra I lecture notes

Proof.

1)

T(v1 + v2) = A(v1 + v2)

= Av1 + Av2

= T(v1) + T(v2)

2)

T(v) = A(v)

= Av

= T(v)

2

Eg 4.2.

1. Define T : R3 R2

T

x1x2x3

=

x1 3x2 + x3x1 + x2 2x3

Then

T(x) =

1 3 11 1 2

x

So T is a linear transformation.

2. Rotation : R2 R2 is

=

cos sin

sin cos

x

so is a linear transformation.

4.1 Basic properties

Proposition 4.2. Let T : V W be a linear transformation(i) T(0V) = 0W

(ii) T(1v2 + + kvk) = 1T v1 + + kT(vk)Proof.

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Algebra I lecture notes 4.2. CONSTRUCTING LINEAR TRANSFORMATIONS

(i)

T(0V) = T(0v)= 0T(v)

= 0W

(ii)

T(1v1 + + kvk) = T(1v1 + + k1vk1) + T(kvk)= T(1v1 + + k1vk1) + kT(vk)

Repeat to get (ii).

2

4.2 Constructing linear transformations

Eg 4.3. Find a linear transformation T : R2 R3 which sends10

e1

11

2

w1

01e2

0

13

w2

We are forced to define

T

x1x2

= T(x1e1 + x2e2)

= x1T(e1) + x2T(e2)

= x1w1 + x2w2

So the only possible choice for T is

T

x1x2

=

x1x1 + x2

2x1 + 3x2

This is a linear transformation, as it is

T(x) =

1 01 1

2 3

x

And it does send e1 w1, e2 w2.

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4.2. CONSTRUCTING LINEAR TRANSFORMATIONS Algebra I lecture notes

In general:

Proposition 4.3. Let V, W be vector spaces, and let v1, . . . , vn be a basis of V. For anyn vectors w1, . . . , wn in W there is a unique linear transformation T : V W such that

T(v1) = w1...

T(vn) = wn

Proof. Let v V. Write

v = 1v1 + + nvnBy 4.2(ii), the only possible choice for T(v) is

T(v) = T(1v1 + + nvn)= 1T(v1) + + nT(vn)= 1w1 + + nwn

So this is our definition of T : V W if v = 1v1 + + nvn, then

T(v) = 1w1 + + nwn

We show this function T is a linear transformation:

1) Let v =

ivi, w =

ivi. Then v + w =

(i + i)vi, so

T(v + w) =

(i + i)wi

=

iwi +

iwi

= T(v) + T(w)

2) Let v = ivi, F. ThenT(v) = T

ivi

=

iwi

=

iwi

= T(v)

So T is a linear transformation sending v1 w1 for all i and is unique.2

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Algebra I lecture notes 4.3. KERNEL AND IMAGE

Remark 4.1. This shows that once we know what a linear transformation does to thevectors in a basis, we know what it does to all vectors.

Eg 4.4. V =vector space of polynomials over R of degree 2. Basis of V: 1, x, x2. Pick3 vectors in V: w1 = 1 + x, w2 = x x2, w3 = 1 + x2. By 4.3 there exists a unique lineartransformation T : V V sending

1 w1x w2

x2 w3Then

T(a + bx + cx2) 4.2= aT(1) + bT(x) + cT(x2)

= a(1 + x) + b(x x2) + x(1 + x2)= a + c + (a + b)x + (c b)x2

4.3 Kernel and Image

Definition 4.2. T : V V linear transformation. Define the image Im(T) to be

Im(T) = {T(v) | v V} W

The kernel Ker(T) isKer(T) = {v V | T(v) = 0} V

Eg 4.5. T : R3 R2

T

x1x2

x3

= 3 1 21 0 1

A

x1x2

x3

= 3x1 + x2 + 2x3x1 + x3

Then

Ker(T) =

x R3 | T(X) = 0=

x R3 | Ax = 0 = solution space of Ax

Im (T) = set of all vectors

3x1 + x2 + 2x3

x1 + x3

= set of all vectors x1

3

1

+ x2

10

+ x3

21

= col-space ofA

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4.3. KERNEL AND IMAGE Algebra I lecture notes

Proposition 4.4. T : V W linear transformation. Then

i) Ker(T) is a subspace of V

ii) Im(T) is a subspace of W

Proof.

i) Use 2.3:

1) 0 Ker(T) since T(0V) 4.2= 0W2) Let v, w Ker(T). Then T(v) = T(w) = 0, so

T(v + w) = T(v) + T(w)

= 0 + 0 = 0

So v + w Ker(T).3) Let v Ker(T), F. Then

T(v) = T(v)

= 0 = 0

So v Ker(T)ii)

1) 0 Im (T) as 0 = T(0)2) w1, w2 Im (T) so w1 = T(v1), w2 = T(v2)

w1 + w2 = T(v1) + T(v2)

= T(v1 + v2)

so w1 + w2 Im (T)3) w Im (T), F. Then

w = T(v)

w = T(v)

= T(v)

so w Im (T).

2

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Eg 4.6. Let vn = vector space of polynomials of degree n. Define T : Vn Vn1 by

T(p(x)) = p

(x)

Then T is a linear transformation.

Ker(T) = {p(x) | T(p(x)) = 0}= {p(x) | p(x) = 0}= V0 (the constant polynomials)

and Im(T) = Vn1. This has basis 1, x , x2, . . . , xn1, dim n.

Proposition 4.5. Let T : V

W be a linear transformation. If v1, . . . , vn is a basis ofV, then

Im (T) = Sp(T(v1), . . . , T (vn))

Proof. Let T(v) Im (T). Write

v = 1v1 + + nvnThen

T(v)4.2= 1T(v1) + + nT(vn)

Sp(T(v1), . . . , T (vn))This shows

Im (T) Sp(T(v1), . . . , T (vn))All T(v1) Im T, so as Im(T) is a subspace, Sp(T(v1), . . . , T (vn)) Im (T). ThereforeSp(T(v1), . . . , T (vn)) = Im(T). 2

Important class of kernels and images:

Proposition 4.6. Let A be an m n matrix, and define T : Rn Rm by (x Rn)

T(x) = Ax

Then

1) Ker(T) = solution space of the system Ax = 0.

2) Im T = column-space(A)

3) dimIm(T) = rank(A)

Proof.

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4.3. KERNEL AND IMAGE Algebra I lecture notes

1)

Ker(T) = {x | T(x) = 0}= {x Rn | Ax = 0}= solution space ofAx = 0

2) Take a standard basis e1, . . . , en ofRn. By 4.5

Im (T) = Sp(T(e1), . . . , T (en))

Here

T(ei) = Aei = A

0...1...0

= i-th column of A

So Im(T) = Sp(columns of A) = column-space(A).

3) dim(Im(T)) = dim (column-space(A)) = rank(A)

2

Eg 4.7. T : R3 R3

T(x) =

1 2 31 0 1

1 4 7

Find bases for Ker (T) and Im (T).

Ker(T) 1 2 3 01 0 1 0

1 4 7 0

1 2 3 00 2 4 0

0 2 4 0

1 2 3 00 2 4 0

0 0 0 0

General solution (a, 2a, a). Basis for Ker (T) is (1, 2, 1).

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Im (T). Basis for Im (T) = column-space((A)). Dimension is rank((A)) = 2. So basis

is 1

11 ,

2

04.

Theorem 4.1 (Rank-nullity Theorem). 1 Let V, W be vector spaces, and T : V W bea linear transformation. Then

dim(Ker(T)) + dim (Im (T)) = dim (V)

Proof. Let r = dim(Ker(T)). Let u1, . . . , ue be a basis of Ker (T). By 2.10 we can extendthis to

u1, . . . , ur, v1, . . . , vs

basis of V. So dim V = r + s.We want to show that dim (Im (T)) = s. By 4.5

Im (T) = Sp(T(u1), . . . , T (ur), T(v1), . . . , T (vs))

Each T(u1) = 0, as ui Ker T. SoIm T = Sp(T(v1), . . . , T (vs)) ()

Claim: T(v1), . . . , T (vs) is a basis of Im T.

Proof. Span shown by (). Suppose1T(v1) + + sT(vs) = 0

Then

T(1v1 + + svs) 4.2= 0So 1v1 + + svs Ker T. So

1v1 + + svs = 1u1 + + rur(as u1, . . . , ur are basis of Ker T). That is

1u1 +

+ rur

1v1

svs = 0

As u1, . . . , ur, v1, . . . , vs is a basis ofV, it is linearly independent, and so

i = i = 0 iThis shows T(v1), . . . , T (vs) is linearly independent, hence a basis of Im T. 2

So dim (Im T) = s and

dim (Ker T) + dim (Im T) = r + s = dim V.

2

1dim(Ker(T)) is sometimes called the nullity ofT

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4.4. COMPOSITION OF LINEAR TRANSFORMATIONS Algebra I lecture notes

Consequences for linear equations

Proposition 4.7. Let A be m

n matrix, and

W = solution space ofAx = 0

= {x Rn | Ax = 0}Then

dim W = n rank(A)Proof. Define

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