# Review of Linear Algebra

Size: px
Start display at page:

Transcription

1 Review of Linear Algebra Throughout these notes, F denotes a field (often called the scalars in this context). 1 Definition of a vector space Definition 1.1. A F -vector space or simply a vector space is a triple (V, +, ), where (V, +) is an abelian group (the vectors), and there is a function F V V (scalar multiplication), whose value at (t, v) is just denoted t v or tv, such that 1. For all s, t F and v V, s(tv) = (st)v. (An analogue of the associative law for multiplication.) 2. For all s, t F and v V, (s + t)v = sv + tv. (Scalar multiplication distributes over scalar addition.) 3. For all t F and v, w V, t(v + w) = tv + tw. (Scalar multiplication distributes over vector addition.) 4. For all v V, 1 v = v. (A kind of identity law for scalar multiplication.) It is a straightforward consequence of the axioms that 0v = 0 for all v V, where the first 0 is the element 0 F and the second is 0 V, that, for all t F, t0 = 0 (both 0 s here are the zero vector), and that, for all v V, ( 1)v = v. Example 1.2. (1) The n-fold Cartesian product F n is an F -vector space, where F n is a group under componentwise addition, i.e. (a 1,..., a n ) + (b 1,..., b n ) = (a 1 + b 1,..., a n + b n ), and scalar multiplication is defined by t(a 1,..., a n ) = (ta 1,..., ta n ). (2) If X is a set and F X denotes the group of functions from X to F, then F X is an F -vector space. Here addition is defined pointwise: (f + g)(x) = f(x) + g(x), and similarly for scalar multiplication: (tf)(x) = tf(x). 1

2 (3) The polynomial ring F [x] is an F -vector space under addition of polynomials, where we define tf(x) to be the product of the constant polynomial t F with the polynomial f(x) F [x]. Explicitly, if f(x) = n i=0 a nx n, then tf(x) = n i=0 ta nx n. (4) More generally, if R is any ring containing F as a subring, then R becomes an F -vector space, using the fact that (R, +) is an abelian group, and defining, for t F and r R, scalar multiplication tr to be the usual multiplication in R applied to t and r. In this case, the properties (1) (4) of a vector space become consequences of associativity, distributivity, and the fact that 1 F is the multiplicative identity in R. In particular, if E is a field containing F as a subfield, then E is an F -vector space in this way. Definition 1.3. Let V be a vector space. An F -vector subspace of V is a subset W V is an additive subgroup of V which is closed under scalar multiplication (i.e. such that, for all v W and for all t F, tv W ). It then follows that (W, +, ) is also a vector space. Example 1.4. {0} and V are always vector subspaces of V. The set P n = { n a i x i : a i F } i=0 of polynomials of degree at most n, together with the zero polynomial, is a vector subspace of F [x]. Finally, we have the analogue of homomorphisms and isomorphisms: Definition 1.5. Let V 1 and V 2 be two F -vector spaces and let f : V 1 V 2 be a function (= map). Then f is linear or a linear map if it is a group homomorphism from (V 1, +) to (V 2, +), i.e. is additive, and satisfies: For all t F and v V 1, f(tv) = tf(v). The function f is a linear isomorphism if it is both linear and a bijection; in this case, it is easy to check that f 1 is also linear. There is an analogue of vector spaces for more general rings: Definition 1.6. Let R be a ring (always assumed commutative, with unity). Then an R-module M is a triple (M, +, ), where (M, +) is an abelian group, and there is a function R M M, whose value at (r, m) is denoted rm or r m, such that: 1. For all r, s R and m M, r(sm) = (rs)m. 2

3 2. For all r, s R and m M, (r + s)m = rm + sm. 3. For all r R and m, n M, r(m + n) = rm + rn. 4. For all m M, 1 m = m. Submodules and homomorphisms of R-modules are defined in the obvious way. For example, R n is an R-module. Despite the similarity with the definition of vector spaces, general R-modules can look quite complicated. For example, a Z-module is the same thing as an abelian group, and hence Z/nZ is a Z-module. For a general ring R, if I is an ideal of R, then both I and R/I are R-modules. 2 Linear independence and span Let us introduce some terminology: Definition 2.1. Let V be an F -vector space and let v 1,..., v d V be a sequence of vectors. A linear combination of v 1,..., v d is a vector of the form t 1 v t d v d, where the t i F. The span of {v 1,..., v d } is the set of all linear combinations of v 1,..., v d. Thus span{v 1,..., v d } = {t 1 v t d v d : t i F for all i}. By definition (or logic), span = {0}. We have the following properties of span: Proposition 2.2. Let v 1,..., v d V be a sequence of vectors. Then: (i) span{v 1,..., v d } is a vector subspace of V containing v i for every i. (ii) If W is a vector subspace of V containing v 1,..., v d, then span{v 1,..., v d } W. In other words, span{v 1,..., v d } is the smallest vector subspace of V containing v 1,..., v d. (iii) For every v V, span{v 1,..., v d } span{v 1,..., v d, v}, and equality holds if and only if v span{v 1,..., v d }. 3

4 Proof. (i) Taking t j = 0, j i, and t i = 1, we see that v i span{v 1,..., v d }. Next, given v = t 1 v 1 + +t d v d and w = s 1 w 1 + +s d w d span{v 1,..., v d }, v + w = (s 1 + t 1 )v (s d + t d )w d span{v 1,..., v d }. Hence span{v 1,..., v d } is closed under addition. Also, 0 = 0 v v d span{v 1,..., v d }, and, given v = t 1 v t d v d span{v 1,..., v d }, v = ( t 1 )v ( t d )v d span{v 1,..., v d } as well. Hence span{v 1,..., v d } is an additive subgroup of V. Finally, for all v = t 1 v t d v d span{v 1,..., v d } and t F, tv = (tt 1 )v (tt d )v d span{v 1,..., v d }. Hence span{v 1,..., v d } is a vector subspace of V containing v i for every i. (ii) If W is a vector subspace of V containing v 1,..., v d, then, for all t 1,..., t d F, t i v i W. Hence t 1 v t d v d W. It follows that span{v 1,..., v d } W. (iii) We always have span{v 1,..., v d } span{v 1,..., v d, v}, since we can write t 1 v t d v d = t 1 v t d v d + 0 v. Now suppose that span{v 1,..., v d } = span{v 1,..., v d, v}. Then in particular v span{v 1,..., v d, v} = span{v 1,..., v d }, by (i), and hence v span{v 1,..., v d }, i.e. v is a linear combination of v 1,..., v d. Conversely, suppose that v span{v 1,..., v d }. Then by (i) span{v 1,..., v d } is a vector subspace of V containing v 1,..., v d and v and hence span{v 1,..., v d, v} span{v 1,..., v d }. Thus span{v 1,..., v d, v} = span{v 1,..., v d }. Definition 2.3. A sequence of vectors w 1,..., w l such that will be said to span V. V = span{w 1,..., w l } Definition 2.4. An F -vector space V is a finite dimensional vector space if there exist v 1,..., v d V such that V = span{v 1,..., v d }. The vector space V is infinite dimensional if it is not finite dimensional. For example, F n is finite-dimensional. But F [x] is not a finite-dimensional vector space. On the other hand, the subspace P n of F [x] defined in Example 1.4 is a finite dimensional vector space, as it is spanned by 1, x,..., x n. The next piece of terminology is there to deal with the fact that we might have chosen a highly redundant set of vectors to span V. 4

5 Definition 2.5. A sequence w 1,..., w r V is linearly independent if the following holds: if there exist real numbers t i such that t 1 w t r w r = 0, then t i = 0 for all i. The sequence w 1,..., w r is linearly dependent if it is not linearly independent. Note that the definition of linear independence does not depend only on the set {w 1,..., w r } if there are any repeated vectors w i = w j, then we can express 0 as the nontrivial linear combination w i w j. Likewise if one of the w i is zero then the set is linearly dependent. By definition or by logic, the empty set is linearly independent. For a less obscure example, e 1,..., e n F n are linearly independent since if t 1 e t n e n = 0, then (t 1,..., t n ) is the zero vector and thus t i = 0 for all i. More generally, for all j n, the vectors e 1,..., e j are linearly independent. Lemma 2.6. The vectors w 1,..., w r are linearly independent if and only if, given t i, s i F, 1 i r, such that then t i = s i for all i. t 1 w t r w r = s 1 w s r w r, Proof. If w 1,..., w r are linearly independent and if t 1 w t r w r = s 1 w s r w r, then after subtracting and rearranging we have (t 1 s 1 )w (t r s r )w r 0. Thus by the definition of linear independence t i s i = 0 for every i, i.e. s i = t i. Conversely, if the last statement of the lemma holds and if t 1 w t r w r = 0, then it follows from t 1 w t r w r = 0 = 0 w w r that t i = 0 for all i. Hence w 1,..., w r are linearly independent. Clearly, if w 1,..., w r are linearly independent, then so is any reordering of the vectors w 1,..., w r, and likewise any smaller sequence (not allowing repeats), for example w 1,..., w s with s r. A related argument shows: Lemma 2.7. The vectors w 1,..., w r are not linearly independent if and only if we can write at least one of the w i as a linear combination of the w j, j i. The proof is left as an exercise. 5

6 Definition 2.8. Let V be an F -vector space. The vectors v 1,..., v d are a basis of V if they are linearly independent and V = span{v 1,..., v d }. For example, the standard basis vectors e 1,..., e n are a basis for F n. The elements 1, x,..., x n are a basis for P n. Thus to say that the vectors v 1,..., v r are a basis of V is to say that every x V can be uniquely written as x = t 1 v t r v r for t i F. Lemma 2.9 (Main counting argument). Suppose that w 1,..., w b are linearly independent vectors contained in span{v 1,..., v a }. Then b a. Proof. We shall show that, possibly after relabeling the v i, span{v 1,..., v a } = span{w 1, v 2,..., v a } = span{w 1, w 2, v 3,..., v a } = = span{w 1, w 2,..., w b, v b+1,..., v a }. From this we will be able to conclude that b a. To begin, we may suppose that {w 1,..., w b }. (If {w 1,..., w b } =, then b = 0 and the conclusion b a is automatic.) Moreover none of the w i is zero. Given w 1, we can write it as a linear combination of the v i : w 1 = a t i v i. i=1 Since w 1 0, {v 1,..., v a }, i.e. a 1, and at least one of the t i 0. After relabeling the v i, we can assume that t 1 0. Thus we can solve for v 1 in terms of w 1 and the v i, i > 1: v 1 = 1 t 1 w 1 + a i=2 ( t ) i v i. t 1 It follows that v 1 span{w 1, v 2,..., v a }. Now using some of the properties of span listed above, we have span{w 1, v 2,..., v a } = span{v 1, w 1, v 2,..., v a } = span{v 1, v 2,..., v a }, where the second equality holds since w 1 span{v 1, v 2,..., v a }. Continuing in this way, write w 2 as a vector in span{w 1, v 2,..., v a }: w 2 = t 1 w 1 + a t i v i. i=2 6

7 For some i 2, we must have t i 0, for otherwise we would have w 2 = t 1 w 1 and thus there would exist a nontrivial linear combination t 1 w 1 + ( 1)w 2 = 0, contradicting the linear independence of the w i. After relabeling, we can assume that t 2 0; notice that in particular we must have a 2. Arguing as before, we may write v 2 = t 1 t 2 w t 2 w 2 + a i=3 ( t ) i v i, t 2 and thus we can solve for v 2 in terms of w 1, w 2, and v i, i 3 and so span{w 1, v 2,..., v a } = span{w 1, w 2, v 3,..., v a }. By induction, for a fixed i < b, suppose that we have showed that i a and that after some relabeling of the v i we have span{v 1,..., v a } = span{w 1, v 2,..., v a } = = span{w 1, w 2, v 3,..., v a } = = span{w 1, w 2,..., w i, v i+1,..., v a }. We claim that the same is true for i + 1. Write which is possible as w i+1 = t 1 w t i w i + t i+1 v i t a v a, w i+1 span{v 1,..., v a } = span{w 1, w 2,..., w i, v i+1,..., v a }. At least one of the numbers t i+1,..., t a is nonzero, for otherwise w i+1 = t 1 w t i w i, which would say that the vectors w 1,..., w i+1 are not linearly independent. In particular this says that i + 1 a. After relabeling, we may assume that t i+1 0. Then as before we can solve for v i+1 in terms of the vectors w 1,..., w i+1, v i+2,..., v a. It follows that span{w 1, w 2,..., w i, v i+1,..., v a } = span{w 1, w 2,..., w i, w i+1, v i+2,..., v a } and we have completed the inductive step. So for all i b, i a, and in particular b a. This rather complicated argument has the following consequences: Corollary (i) Suppose that V = span{v 1,..., v n } and that w 1,..., w l are linearly independent vectors in V. Then l n. 7

8 (ii) If V is a finite-dimensional F -vector space, then every two bases for V have the same number of elements call this number the dimension of V which we write as dim V or dim F V if we want to emphasize the field F. Thus for example dim F n = n and dim P n = n + 1. (iii) If V = span{v 1,..., v d } then some subsequence of v 1,..., v d is a basis for V. Hence dim V d, and if dim V = d then v 1,..., v d is a basis for V. (iv) If V is a finite-dimensional F -vector space and w 1,..., w l are linearly independent vectors in V, then there exist vectors w l+1,..., w r V such that w 1,..., w l, w l+1,..., w r is a basis for V. Hence dim V l, and if dim V = l then w 1,..., w l is a basis for V. (v) If V is a finite-dimensional F -vector space and W is a vector subspace of V, then dim W dim V. Moreover dim W = dim V if and only if W = V. (vi) If v 1,..., v d is a basis of V, then the function f : F d V defined by f(t 1,..., t d ) = d t i v i i=1 is a linear isomorphism from F d to V. Proof. (i) Apply the lemma to the subspace V itself, which is the span of v 1,..., v n, and to the linearly independent vectors w 1,..., w l V, to conclude that l n. (ii) If w 1,..., w b and v 1,..., v a are two bases of V, then by definition w 1,..., w b are linearly independent vectors, and V = span{v 1,..., v a }. Thus b a. But symmetrically v 1,..., v a is a sequence of linearly independent vectors contained in V = span{w 1,..., w b }, so a b. Thus a = b. (iii) If v 1,..., v d are not linearly independent, then, by Lemma 2.7, one of the vectors v i is expressed as a linear combination of the others. After relabeling we may assume that v d is a linear combination of v 1,..., v d 1. By (iii) of Proposition 2.2, span{v 1,..., v d 1 } = span{v 1,..., v d }. Continue in this way until we find v 1,..., v k with k d such that span{v 1,..., v k } = span{v 1,..., v d } and such that v 1,..., v k are linearly independent. Then by definition k = dim V and k d. Moreover k = d exactly when v 1,..., v d are linearly independent, in which case span{v 1,..., v d } is a basis. (iv) If span{w 1,..., w l } V, then there exists a vector, call it w l+1 V with w l+1 / span{w 1,..., w l }. It follows that the sequence w 1,..., w l, w l+1 8

9 is still linearly independent: if there exist t i F, not all 0, such that 0 = l+1 i=1 t iw i, then we must have t l+1 0 since w 1,..., w l are linearly independent. But that would say that w l+1 span{w 1,..., w l }, contradicting our choice. So w 1,..., w l, w l+1 are still linearly independent. We continue in this way. Since the number of elements in a linearly independent sequence of vectors in V is at most dim V, this procedure has to stop after at most dim V l stages. At this point we have found a linearly independent sequence which spans V and thus is a basis. To see the last statement, note that if span{w 1,..., w l } V, then there exist l + 1 linearly independent vectors in V, and hence dim V l + 1. (v) Choosing a basis of W and applying (iv) (since the elements of a basis are linearly independent) we see that it can be completed to a basis of V. Thus dim W dim V. Moreover dim W = dim V if and only if the basis we chose for W was already a basis for V, i.e. W = V. (vi) A straightforward calculation shows that f is linear. It is a bijection by definition of a basis: it is surjective since the v i span V, and it is injective since the v i are linearly independent and by Lemma 2.6. Proposition If V is a finite dimensional F -vector space and W is a subset of V, then W is a vector subspace of V if and only if it is of the form span{v 1,..., v d } for some v 1,..., v d V. Proof. We have already noted that a set of the form span{v 1,..., v d } is a vector subspace of V. Conversely let W be a vector subspace of V. The proof of (iv) of the above corollary shows how to find a basis of W. In particular W is of the form span{v 1,..., v d }. 3 Linear functions and matrices Let f : F n F k be a linear function. Then f(t 1,..., t n ) = f( i t i e i ) = i t i f(e i ). We can write f(e i ) in terms of the basis e 1,..., e k of F k : suppose that f(e i ) = k j=1 a jie j = (a 1i,..., a ki ). We can then associate to f an k n matrix with coefficients in F as follows: Define a 11 a a 1n A = a k1 a k2... a kn 9

10 Then the columns of A are the vectors f(e i ). More generally, suppose that V 1 and V 2 are two finite dimensional vector spaces, and choose bases v 1,..., v n of V 1 and w 1,..., w k of V 2, so that n = dim V 1 and k = dim V 2. If f : V 1 V 2 is linear, then, given the bases v 1,..., v n and w 1,..., w k we again get a k n matrix A by the formula: A = (a ij ), where a ij is defined by f(v i ) = k a ji w j. j=1 We say that A is the matrix associated to the linear map f and the bases v 1,..., v n and w 1,..., w k. With the understanding that we have to choose bases to define the matrix associated to a linear map, composition of linear maps corresponds to multiplication of matrices. In particular, let f : V V be a linear map from a finite dimensional vector space V to itself. In this case, it is simplest to fix one basis v 1,..., v n for V, viewing V as both the domain and range of f, and write A = (a ij ) where f(v i ) = k j=1 a jiv j. Again, having fixed the basis v 1,..., v n once and for all, linear maps from V to itself correspond to n n matrices and composition to matrix multiplication. Finally, we can define the determinant det A of an n n matrix with coefficients in F by the usual formulas (for example, expansion by minors). Moreover, the n n matrix A is invertible det A 0, and in this case there is an explicit formula for A 1 (Cramer s rule). 10

### Some notes on linear algebra

Some notes on linear algebra Throughout these notes, k denotes a field (often called the scalars in this context). Recall that this means that there are two binary operations on k, denoted + and, that

### Chapter 2: Linear Independence and Bases

MATH20300: Linear Algebra 2 (2016 Chapter 2: Linear Independence and Bases 1 Linear Combinations and Spans Example 11 Consider the vector v (1, 1 R 2 What is the smallest subspace of (the real vector space

### Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra

Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra D. R. Wilkins Contents 3 Topics in Commutative Algebra 2 3.1 Rings and Fields......................... 2 3.2 Ideals...............................

### Note that a unit is unique: 1 = 11 = 1. Examples: Nonnegative integers under addition; all integers under multiplication.

Algebra fact sheet An algebraic structure (such as group, ring, field, etc.) is a set with some operations and distinguished elements (such as 0, 1) satisfying some axioms. This is a fact sheet with definitions

### 1 Invariant subspaces

MATH 2040 Linear Algebra II Lecture Notes by Martin Li Lecture 8 Eigenvalues, eigenvectors and invariant subspaces 1 In previous lectures we have studied linear maps T : V W from a vector space V to another

### Homework 5 M 373K Mark Lindberg and Travis Schedler

Homework 5 M 373K Mark Lindberg and Travis Schedler 1. Artin, Chapter 3, Exercise.1. Prove that the numbers of the form a + b, where a and b are rational numbers, form a subfield of C. Let F be the numbers

### and this makes M into an R-module by (1.2). 2

1. Modules Definition 1.1. Let R be a commutative ring. A module over R is set M together with a binary operation, denoted +, which makes M into an abelian group, with 0 as the identity element, together

### Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35

Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35 1. Let R be a commutative ring with 1 0. (a) Prove that the nilradical of R is equal to the intersection of the prime

### Commutative Algebra and Algebraic Geometry. Robert Friedman

Commutative Algebra and Algebraic Geometry Robert Friedman August 1, 2006 2 Disclaimer: These are rough notes for a course on commutative algebra and algebraic geometry. I would appreciate all suggestions

### A linear algebra proof of the fundamental theorem of algebra

A linear algebra proof of the fundamental theorem of algebra Andrés E. Caicedo May 18, 2010 Abstract We present a recent proof due to Harm Derksen, that any linear operator in a complex finite dimensional

### Rings and Fields Theorems

Rings and Fields Theorems Rajesh Kumar PMATH 334 Intro to Rings and Fields Fall 2009 October 25, 2009 12 Rings and Fields 12.1 Definition Groups and Abelian Groups Let R be a non-empty set. Let + and (multiplication)

### Review of linear algebra

Review of linear algebra 1 Vectors and matrices We will just touch very briefly on certain aspects of linear algebra, most of which should be familiar. Recall that we deal with vectors, i.e. elements of

### A linear algebra proof of the fundamental theorem of algebra

A linear algebra proof of the fundamental theorem of algebra Andrés E. Caicedo May 18, 2010 Abstract We present a recent proof due to Harm Derksen, that any linear operator in a complex finite dimensional

Algebra Qualifying Exam August 2001 Do all 5 problems. 1. Let G be afinite group of order 504 = 23 32 7. a. Show that G cannot be isomorphic to a subgroup of the alternating group Alt 7. (5 points) b.

### Linear Algebra. Paul Yiu. Department of Mathematics Florida Atlantic University. Fall 2011

Linear Algebra Paul Yiu Department of Mathematics Florida Atlantic University Fall 2011 Linear Algebra Paul Yiu Department of Mathematics Florida Atlantic University Fall 2011 1A: Vector spaces Fields

### 0.2 Vector spaces. J.A.Beachy 1

J.A.Beachy 1 0.2 Vector spaces I m going to begin this section at a rather basic level, giving the definitions of a field and of a vector space in much that same detail as you would have met them in a

### SUPPLEMENT TO CHAPTER 3

SUPPLEMENT TO CHAPTER 3 1.1 Linear combinations and spanning sets Consider the vector space R 3 with the unit vectors e 1 = (1, 0, 0), e 2 = (0, 1, 0), e 3 = (0, 0, 1). Every vector v = (a, b, c) R 3 can

### Linear Vector Spaces

CHAPTER 1 Linear Vector Spaces Definition 1.0.1. A linear vector space over a field F is a triple (V, +, ), where V is a set, + : V V V and : F V V are maps with the properties : (i) ( x, y V ), x + y

### Definitions. Notations. Injective, Surjective and Bijective. Divides. Cartesian Product. Relations. Equivalence Relations

Page 1 Definitions Tuesday, May 8, 2018 12:23 AM Notations " " means "equals, by definition" the set of all real numbers the set of integers Denote a function from a set to a set by Denote the image of

### Vector Spaces and Linear Maps

Vector Spaces and Linear Maps Garrett Thomas August 14, 2018 1 About This document is part of a series of notes about math and machine learning. You are free to distribute it as you wish. The latest version

### Linear Algebra. Chapter 5

Chapter 5 Linear Algebra The guiding theme in linear algebra is the interplay between algebraic manipulations and geometric interpretations. This dual representation is what makes linear algebra a fruitful

### Mathematics Department Stanford University Math 61CM/DM Vector spaces and linear maps

Mathematics Department Stanford University Math 61CM/DM Vector spaces and linear maps We start with the definition of a vector space; you can find this in Section A.8 of the text (over R, but it works

### Elementary linear algebra

Chapter 1 Elementary linear algebra 1.1 Vector spaces Vector spaces owe their importance to the fact that so many models arising in the solutions of specific problems turn out to be vector spaces. The

### Math 121 Homework 5: Notes on Selected Problems

Math 121 Homework 5: Notes on Selected Problems 12.1.2. Let M be a module over the integral domain R. (a) Assume that M has rank n and that x 1,..., x n is any maximal set of linearly independent elements

### Abstract Vector Spaces

CHAPTER 1 Abstract Vector Spaces 1.1 Vector Spaces Let K be a field, i.e. a number system where you can add, subtract, multiply and divide. In this course we will take K to be R, C or Q. Definition 1.1.

### MATH 326: RINGS AND MODULES STEFAN GILLE

MATH 326: RINGS AND MODULES STEFAN GILLE 1 2 STEFAN GILLE 1. Rings We recall first the definition of a group. 1.1. Definition. Let G be a non empty set. The set G is called a group if there is a map called

### NOTES ON FINITE FIELDS

NOTES ON FINITE FIELDS AARON LANDESMAN CONTENTS 1. Introduction to finite fields 2 2. Definition and constructions of fields 3 2.1. The definition of a field 3 2.2. Constructing field extensions by adjoining

### CHAPTER I. Rings. Definition A ring R is a set with two binary operations, addition + and

CHAPTER I Rings 1.1 Definitions and Examples Definition 1.1.1. A ring R is a set with two binary operations, addition + and multiplication satisfying the following conditions for all a, b, c in R : (i)

### Linear Algebra (part 1) : Vector Spaces (by Evan Dummit, 2017, v. 1.07) 1.1 The Formal Denition of a Vector Space

Linear Algebra (part 1) : Vector Spaces (by Evan Dummit, 2017, v. 1.07) Contents 1 Vector Spaces 1 1.1 The Formal Denition of a Vector Space.................................. 1 1.2 Subspaces...................................................

### 18.312: Algebraic Combinatorics Lionel Levine. Lecture 22. Smith normal form of an integer matrix (linear algebra over Z).

18.312: Algebraic Combinatorics Lionel Levine Lecture date: May 3, 2011 Lecture 22 Notes by: Lou Odette This lecture: Smith normal form of an integer matrix (linear algebra over Z). 1 Review of Abelian

### Math 120 HW 9 Solutions

Math 120 HW 9 Solutions June 8, 2018 Question 1 Write down a ring homomorphism (no proof required) f from R = Z[ 11] = {a + b 11 a, b Z} to S = Z/35Z. The main difficulty is to find an element x Z/35Z

### Linear Algebra. Preliminary Lecture Notes

Linear Algebra Preliminary Lecture Notes Adolfo J. Rumbos c Draft date April 29, 23 2 Contents Motivation for the course 5 2 Euclidean n dimensional Space 7 2. Definition of n Dimensional Euclidean Space...........

### Lecture 28: Fields, Modules, and vector spaces

Lecture 28: Fields, Modules, and vector spaces 1. Modules Just as groups act on sets, rings act on abelian groups. When a ring acts on an abelian group, that abelian group is called a module over that

### Math 550 Notes. Chapter 2. Jesse Crawford. Department of Mathematics Tarleton State University. Fall 2010

Math 550 Notes Chapter 2 Jesse Crawford Department of Mathematics Tarleton State University Fall 2010 (Tarleton State University) Math 550 Chapter 2 Fall 2010 1 / 20 Linear algebra deals with finite dimensional

### ALGEBRA II: RINGS AND MODULES OVER LITTLE RINGS.

ALGEBRA II: RINGS AND MODULES OVER LITTLE RINGS. KEVIN MCGERTY. 1. RINGS The central characters of this course are algebraic objects known as rings. A ring is any mathematical structure where you can add

### Math 115A: Linear Algebra

Math 115A: Linear Algebra Michael Andrews UCLA Mathematics Department February 9, 218 Contents 1 January 8: a little about sets 4 2 January 9 (discussion) 5 2.1 Some definitions: union, intersection, set

### Representations. 1 Basic definitions

Representations 1 Basic definitions If V is a k-vector space, we denote by Aut V the group of k-linear isomorphisms F : V V and by End V the k-vector space of k-linear maps F : V V. Thus, if V = k n, then

### A Do It Yourself Guide to Linear Algebra

A Do It Yourself Guide to Linear Algebra Lecture Notes based on REUs, 2001-2010 Instructor: László Babai Notes compiled by Howard Liu 6-30-2010 1 Vector Spaces 1.1 Basics Definition 1.1.1. A vector space

### Math Introduction to Modern Algebra

Math 343 - Introduction to Modern Algebra Notes Field Theory Basics Let R be a ring. M is called a maximal ideal of R if M is a proper ideal of R and there is no proper ideal of R that properly contains

### Rings If R is a commutative ring, a zero divisor is a nonzero element x such that xy = 0 for some nonzero element y R.

Rings 10-26-2008 A ring is an abelian group R with binary operation + ( addition ), together with a second binary operation ( multiplication ). Multiplication must be associative, and must distribute over

### NOTES ON LINEAR ALGEBRA OVER INTEGRAL DOMAINS. Contents. 1. Introduction 1 2. Rank and basis 1 3. The set of linear maps 4. 1.

NOTES ON LINEAR ALGEBRA OVER INTEGRAL DOMAINS Contents 1. Introduction 1 2. Rank and basis 1 3. The set of linear maps 4 1. Introduction These notes establish some basic results about linear algebra over

### Topics in Module Theory

Chapter 7 Topics in Module Theory This chapter will be concerned with collecting a number of results and constructions concerning modules over (primarily) noncommutative rings that will be needed to study

### We simply compute: for v = x i e i, bilinearity of B implies that Q B (v) = B(v, v) is given by xi x j B(e i, e j ) =

Math 395. Quadratic spaces over R 1. Algebraic preliminaries Let V be a vector space over a field F. Recall that a quadratic form on V is a map Q : V F such that Q(cv) = c 2 Q(v) for all v V and c F, and

### Formal power series rings, inverse limits, and I-adic completions of rings

Formal power series rings, inverse limits, and I-adic completions of rings Formal semigroup rings and formal power series rings We next want to explore the notion of a (formal) power series ring in finitely

### Abstract Vector Spaces and Concrete Examples

LECTURE 18 Abstract Vector Spaces and Concrete Examples Our discussion of linear algebra so far has been devoted to discussing the relations between systems of linear equations, matrices, and vectors.

### Rings and groups. Ya. Sysak

Rings and groups. Ya. Sysak 1 Noetherian rings Let R be a ring. A (right) R -module M is called noetherian if it satisfies the maximum condition for its submodules. In other words, if M 1... M i M i+1...

### Written Homework # 5 Solution

Math 516 Fall 2006 Radford Written Homework # 5 Solution 12/12/06 Throughout R is a ring with unity. Comment: It will become apparent that the module properties 0 m = 0, (r m) = ( r) m, and (r r ) m =

### MATRIX LIE GROUPS AND LIE GROUPS

MATRIX LIE GROUPS AND LIE GROUPS Steven Sy December 7, 2005 I MATRIX LIE GROUPS Definition: A matrix Lie group is a closed subgroup of Thus if is any sequence of matrices in, and for some, then either

### 1 Fields and vector spaces

1 Fields and vector spaces In this section we revise some algebraic preliminaries and establish notation. 1.1 Division rings and fields A division ring, or skew field, is a structure F with two binary

### Introduction to abstract algebra: definitions, examples, and exercises

Introduction to abstract algebra: definitions, examples, and exercises Travis Schedler January 21, 2015 1 Definitions and some exercises Definition 1. A binary operation on a set X is a map X X X, (x,

### Honors Algebra II MATH251 Course Notes by Dr. Eyal Goren McGill University Winter 2007

Honors Algebra II MATH251 Course Notes by Dr Eyal Goren McGill University Winter 2007 Last updated: April 4, 2014 c All rights reserved to the author, Eyal Goren, Department of Mathematics and Statistics,

### MTH 309 Supplemental Lecture Notes Based on Robert Messer, Linear Algebra Gateway to Mathematics

MTH 309 Supplemental Lecture Notes Based on Robert Messer, Linear Algebra Gateway to Mathematics Ulrich Meierfrankenfeld Department of Mathematics Michigan State University East Lansing MI 48824 meier@math.msu.edu

### Chapter 5. Linear Algebra

Chapter 5 Linear Algebra The exalted position held by linear algebra is based upon the subject s ubiquitous utility and ease of application. The basic theory is developed here in full generality, i.e.,

### A Little Beyond: Linear Algebra

A Little Beyond: Linear Algebra Akshay Tiwary March 6, 2016 Any suggestions, questions and remarks are welcome! 1 A little extra Linear Algebra 1. Show that any set of non-zero polynomials in [x], no two

### Linear Algebra. Preliminary Lecture Notes

Linear Algebra Preliminary Lecture Notes Adolfo J. Rumbos c Draft date May 9, 29 2 Contents 1 Motivation for the course 5 2 Euclidean n dimensional Space 7 2.1 Definition of n Dimensional Euclidean Space...........

### Solution to Homework 1

Solution to Homework Sec 2 (a) Yes It is condition (VS 3) (b) No If x, y are both zero vectors Then by condition (VS 3) x = x + y = y (c) No Let e be the zero vector We have e = 2e (d) No It will be false

### Math 530 Lecture Notes. Xi Chen

Math 530 Lecture Notes Xi Chen 632 Central Academic Building, University of Alberta, Edmonton, Alberta T6G 2G1, CANADA E-mail address: xichen@math.ualberta.ca 1991 Mathematics Subject Classification. Primary

### ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ.

ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ. ANDREW SALCH 1. Hilbert s Nullstellensatz. The last lecture left off with the claim that, if J k[x 1,..., x n ] is an ideal, then

### Lecture 2. (1) Every P L A (M) has a maximal element, (2) Every ascending chain of submodules stabilizes (ACC).

Lecture 2 1. Noetherian and Artinian rings and modules Let A be a commutative ring with identity, A M a module, and φ : M N an A-linear map. Then ker φ = {m M : φ(m) = 0} is a submodule of M and im φ is

### NOTES (1) FOR MATH 375, FALL 2012

NOTES 1) FOR MATH 375, FALL 2012 1 Vector Spaces 11 Axioms Linear algebra grows out of the problem of solving simultaneous systems of linear equations such as 3x + 2y = 5, 111) x 3y = 9, or 2x + 3y z =

### Worksheet for Lecture 15 (due October 23) Section 4.3 Linearly Independent Sets; Bases

Worksheet for Lecture 5 (due October 23) Name: Section 4.3 Linearly Independent Sets; Bases Definition An indexed set {v,..., v n } in a vector space V is linearly dependent if there is a linear relation

### Chapter 3. Rings. The basic commutative rings in mathematics are the integers Z, the. Examples

Chapter 3 Rings Rings are additive abelian groups with a second operation called multiplication. The connection between the two operations is provided by the distributive law. Assuming the results of Chapter

### Solutions to Assignment 3

Solutions to Assignment 3 Question 1. [Exercises 3.1 # 2] Let R = {0 e b c} with addition multiplication defined by the following tables. Assume associativity distributivity show that R is a ring with

### ALGEBRA QUALIFYING EXAM PROBLEMS LINEAR ALGEBRA

ALGEBRA QUALIFYING EXAM PROBLEMS LINEAR ALGEBRA Kent State University Department of Mathematical Sciences Compiled and Maintained by Donald L. White Version: August 29, 2017 CONTENTS LINEAR ALGEBRA AND

### NOTES FOR COMMUTATIVE ALGEBRA M5P55

NOTES FOR COMMUTATIVE ALGEBRA M5P55 AMBRUS PÁL 1. Rings and ideals Definition 1.1. A quintuple (A, +,, 0, 1) is a commutative ring with identity, if A is a set, equipped with two binary operations; addition

### Subrings of Finite Commutative Rings

Subrings of Finite Commutative Rings Francisco Franco Munoz arxiv:1712.02025v1 [math.ac] 6 Dec 2017 Abstract We record some basic results on subrings of finite commutative rings. Among them we establish

### ALGEBRA QUALIFYING EXAM SPRING 2012

ALGEBRA QUALIFYING EXAM SPRING 2012 Work all of the problems. Justify the statements in your solutions by reference to specific results, as appropriate. Partial credit is awarded for partial solutions.

### Vector Spaces ปร ภ ม เวกเตอร

Vector Spaces ปร ภ ม เวกเตอร 1 5.1 Real Vector Spaces ปร ภ ม เวกเตอร ของจ านวนจร ง Vector Space Axioms (1/2) Let V be an arbitrary nonempty set of objects on which two operations are defined, addition

### Bulletin of the Iranian Mathematical Society

ISSN: 1017-060X (Print) ISSN: 1735-8515 (Online) Special Issue of the Bulletin of the Iranian Mathematical Society in Honor of Professor Heydar Radjavi s 80th Birthday Vol 41 (2015), No 7, pp 155 173 Title:

### 2. Modules. Definition 2.1. Let R be a ring. Then a (unital) left R-module M is an additive abelian group together with an operation

2. Modules. The concept of a module includes: (1) any left ideal I of a ring R; (2) any abelian group (which will become a Z-module); (3) an n-dimensional C-vector space (which will become both a module

### RINGS: SUMMARY OF MATERIAL

RINGS: SUMMARY OF MATERIAL BRIAN OSSERMAN This is a summary of terms used and main results proved in the subject of rings, from Chapters 11-13 of Artin. Definitions not included here may be considered

### LINEAR ALGEBRA BOOT CAMP WEEK 1: THE BASICS

LINEAR ALGEBRA BOOT CAMP WEEK 1: THE BASICS Unless otherwise stated, all vector spaces in this worksheet are finite dimensional and the scalar field F has characteristic zero. The following are facts (in

### Factorization in Polynomial Rings

Factorization in Polynomial Rings Throughout these notes, F denotes a field. 1 Long division with remainder We begin with some basic definitions. Definition 1.1. Let f, g F [x]. We say that f divides g,

### GEOMETRIC CONSTRUCTIONS AND ALGEBRAIC FIELD EXTENSIONS

GEOMETRIC CONSTRUCTIONS AND ALGEBRAIC FIELD EXTENSIONS JENNY WANG Abstract. In this paper, we study field extensions obtained by polynomial rings and maximal ideals in order to determine whether solutions

### Finite Fields. [Parts from Chapter 16. Also applications of FTGT]

Finite Fields [Parts from Chapter 16. Also applications of FTGT] Lemma [Ch 16, 4.6] Assume F is a finite field. Then the multiplicative group F := F \ {0} is cyclic. Proof Recall from basic group theory

### Math 110, Spring 2015: Midterm Solutions

Math 11, Spring 215: Midterm Solutions These are not intended as model answers ; in many cases far more explanation is provided than would be necessary to receive full credit. The goal here is to make

### Math 429/581 (Advanced) Group Theory. Summary of Definitions, Examples, and Theorems by Stefan Gille

Math 429/581 (Advanced) Group Theory Summary of Definitions, Examples, and Theorems by Stefan Gille 1 2 0. Group Operations 0.1. Definition. Let G be a group and X a set. A (left) operation of G on X is

### Mathematics Department Stanford University Math 61CM/DM Inner products

Mathematics Department Stanford University Math 61CM/DM Inner products Recall the definition of an inner product space; see Appendix A.8 of the textbook. Definition 1 An inner product space V is a vector

### EXERCISE SET 5.1. = (kx + kx + k, ky + ky + k ) = (kx + kx + 1, ky + ky + 1) = ((k + )x + 1, (k + )y + 1)

EXERCISE SET 5. 6. The pair (, 2) is in the set but the pair ( )(, 2) = (, 2) is not because the first component is negative; hence Axiom 6 fails. Axiom 5 also fails. 8. Axioms, 2, 3, 6, 9, and are easily

### Topics in linear algebra

Chapter 6 Topics in linear algebra 6.1 Change of basis I want to remind you of one of the basic ideas in linear algebra: change of basis. Let F be a field, V and W be finite dimensional vector spaces over

### Structure of rings. Chapter Algebras

Chapter 5 Structure of rings 5.1 Algebras It is time to introduce the notion of an algebra over a commutative ring. So let R be a commutative ring. An R-algebra is a ring A (unital as always) together

### 38 Irreducibility criteria in rings of polynomials

38 Irreducibility criteria in rings of polynomials 38.1 Theorem. Let p(x), q(x) R[x] be polynomials such that p(x) = a 0 + a 1 x +... + a n x n, q(x) = b 0 + b 1 x +... + b m x m and a n, b m 0. If b m

### Lecture notes - Math 110 Lec 002, Summer The reference [LADR] stands for Axler s Linear Algebra Done Right, 3rd edition.

Lecture notes - Math 110 Lec 002, Summer 2016 BW The reference [LADR] stands for Axler s Linear Algebra Done Right, 3rd edition. 1 Contents 1 Sets and fields - 6/20 5 1.1 Set notation.................................

### Linear Algebra Lecture Notes-I

Linear Algebra Lecture Notes-I Vikas Bist Department of Mathematics Panjab University, Chandigarh-6004 email: bistvikas@gmail.com Last revised on February 9, 208 This text is based on the lectures delivered

### Vector Spaces. (1) Every vector space V has a zero vector 0 V

Vector Spaces 1. Vector Spaces A (real) vector space V is a set which has two operations: 1. An association of x, y V to an element x+y V. This operation is called vector addition. 2. The association of

### Online Exercises for Linear Algebra XM511

This document lists the online exercises for XM511. The section ( ) numbers refer to the textbook. TYPE I are True/False. Lecture 02 ( 1.1) Online Exercises for Linear Algebra XM511 1) The matrix [3 2

### MATH 304 Linear Algebra Lecture 10: Linear independence. Wronskian.

MATH 304 Linear Algebra Lecture 10: Linear independence. Wronskian. Spanning set Let S be a subset of a vector space V. Definition. The span of the set S is the smallest subspace W V that contains S. If

### Math 4400, Spring 08, Sample problems Final Exam.

Math 4400, Spring 08, Sample problems Final Exam. 1. Groups (1) (a) Let a be an element of a group G. Define the notions of exponent of a and period of a. (b) Suppose a has a finite period. Prove that

### Apprentice Linear Algebra, 1st day, 6/27/05

Apprentice Linear Algebra, 1st day, 6/7/05 REU 005 Instructor: László Babai Scribe: Eric Patterson Definitions 1.1. An abelian group is a set G with the following properties: (i) ( a, b G)(!a + b G) (ii)

### can only hit 3 points in the codomain. Hence, f is not surjective. For another example, if n = 4

.. Conditions for Injectivity and Surjectivity In this section, we discuss what we can say about linear maps T : R n R m given only m and n. We motivate this problem by looking at maps f : {,..., n} {,...,

### (II.B) Basis and dimension

(II.B) Basis and dimension How would you explain that a plane has two dimensions? Well, you can go in two independent directions, and no more. To make this idea precise, we formulate the DEFINITION 1.

### Monoids. Definition: A binary operation on a set M is a function : M M M. Examples:

Monoids Definition: A binary operation on a set M is a function : M M M. If : M M M, we say that is well defined on M or equivalently, that M is closed under the operation. Examples: Definition: A monoid

### Vector spaces. EE 387, Notes 8, Handout #12

Vector spaces EE 387, Notes 8, Handout #12 A vector space V of vectors over a field F of scalars is a set with a binary operator + on V and a scalar-vector product satisfying these axioms: 1. (V, +) is

### (a + b)c = ac + bc and a(b + c) = ab + ac.

2. R I N G S A N D P O LY N O M I A L S The study of vector spaces and linear maps between them naturally leads us to the study of rings, in particular the ring of polynomials F[x] and the ring of (n n)-matrices

### Lecture Notes for Math 414: Linear Algebra II Fall 2015, Michigan State University

Lecture Notes for Fall 2015, Michigan State University Matthew Hirn December 11, 2015 Beginning of Lecture 1 1 Vector Spaces What is this course about? 1. Understanding the structural properties of a wide

### Worksheet for Lecture 15 (due October 23) Section 4.3 Linearly Independent Sets; Bases

Worksheet for Lecture 5 (due October 23) Name: Section 4.3 Linearly Independent Sets; Bases Definition An indexed set {v,..., v n } in a vector space V is linearly dependent if there is a linear relation

### Introduction to modules

Chapter 3 Introduction to modules 3.1 Modules, submodules and homomorphisms The problem of classifying all rings is much too general to ever hope for an answer. But one of the most important tools available