Linear Álgebra Textbook, Apuntes de Álgebra Lineal

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Linear Algebra

x · 1 2 x · 3 1

Jim Hefferon

Notation

R real numbers N natural numbers: { 0 , 1 , 2 ,... } C complex numbers {...

∣ (^)... } set of... such that... 〈... 〉 sequence; like a set but order matters V, W, U vector spaces ~v, ~w vectors ~0, ~ (^0) V zero vector, zero vector of V B, D bases En = 〈~e 1 ,... , ~en〉 standard basis for Rn β,^ ~~δ basis vectors RepB (~v) matrix representing the vector Pn set of n-th degree polynomials Mn×m set of n×m matrices [S] span of the set S M ⊕ N direct sum of subspaces V ∼= W isomorphic spaces h, g homomorphisms H, G matrices t, s transformations; maps from a space to itself T, S square matrices RepB,D(h) matrix representing the map h hi,j matrix entry from row i, column j |T | determinant of the matrix T R(h), N (h) rangespace and nullspace of the map h R∞(h), N∞(h) generalized rangespace and nullspace

Lower case Greek alphabet

name symbol name symbol name symbol alpha α iota ι rho ρ beta β kappa κ sigma σ gamma γ lambda λ tau τ delta δ mu μ upsilon υ epsilon ≤ nu ν phi φ zeta ζ xi ξ chi χ eta η omicron o psi ψ theta θ pi π omega ω

Cover. This is Cramer’s Rule applied to the system x + 2y = 6, 3x + y = 8. The area of the first box is the determinant shown. The area of the second box is x times that, and equals the area of the final box. Hence, x is the final determinant divided by the first determinant.

everything (really these proofs are just verifications), all the way through the uniqueness of reduced echelon form. In particular, in this first chapter, the opportunity is taken to present a few induction proofs, where the arguments just go over bookkeeping details, so that when induction is needed later (e.g., to prove that all bases of a finite dimensional vector space have the same number of members), it will be familiar. Still another consequence is that the second chapter immediately uses this background as motivation for the definition of a real vector space. This typically occurs by the end of the third week. We do not stop to introduce matrix multiplication and determinants as rote computations. Instead, those topics appear naturally in the development, after the definition of linear maps. To help students make the transition from earlier courses, the presentation here stresses motivation and naturalness. An example is the third chapter, on linear maps. It does not start with the definition of homomorphism, as is the case in other books, but with the definition of isomorphism. That’s because this definition is easily motivated by the observation that some spaces are just like each other. After that, the next section takes the reasonable step of defining homomorphisms by isolating the operation-preservation idea. A little mathematical slickness is lost, but it is in return for a large gain in sensibility to students. Having extensive motivation in the text helps with time pressures. I ask students to, before each class, look ahead in the book, and they follow the classwork better because they have some prior exposure to the material. For example, I can start the linear independence class with the definition because I know students have some idea of what it is about. No book can take the place of an instructor, but a helpful book gives the instructor more class time for examples and questions. Much of a student’s progress takes place while doing the exercises; the exer- cises here work with the rest of the text. Besides computations, there are many proofs. These are spread over an approachability range, from simple checks to some much more involved arguments. There are even a few exercises that are reasonably challenging puzzles taken, with citation, from various journals, competitions, or problems collections (as part of the fun of these, the original wording has been retained as much as possible). In total, the questions are aimed to both build an ability at, and help students experience the pleasure of, doing mathematics.

Applications, and Computers. The point of view taken here, that linear algebra is about vector spaces and linear maps, is not taken to the exclusion of all other ideas. Applications, and the emerging role of the computer, are interesting, important, and vital aspects of the subject. Consequently, every chapter closes with a few application or computer-related topics. Some of the topics are: network flows, the speed and accuracy of computer linear reductions, Leontief Input/Output analysis, dimensional analysis, Markov chains, voting paradoxes, analytic projective geometry, and solving difference equations. These are brief enough to be done in a day’s class or to be given as indepen-

ii

dent projects for individuals or small groups. Most simply give a reader a feel for the subject, discuss how linear algebra comes in, point to some accessible further reading, and give a few exercises. I have kept the exposition lively and given an overall sense of breadth of application. In short, these topics invite readers to see for themselves that linear algebra is a tool that a professional must have.

For people reading this book on their own. The emphasis on motivation and development make this book a good choice for self-study. While a pro- fessional mathematician knows what pace and topics suit a class, perhaps an independent student would find some advice helpful. Here are two timetables for a semester. The first focuses on core material.

week Mon. Wed. Fri. 1 1.I.1 1.I.1, 2 1.I.2, 3 2 1.I.3 1.II.1 1.II. 3 1.III.1, 2 1.III.2 2.I. 4 2.I.2 2.II 2.III. 5 2.III.1, 2 2.III.2 exam 6 2.III.2, 3 2.III.3 3.I. 7 3.I.2 3.II.1 3.II. 8 3.II.2 3.II.2 3.III. 9 3.III.1 3.III.2 3.IV.1, 2 10 3.IV.2, 3, 4 3.IV.4 exam 11 3.IV.4, 3.V.1 3.V.1, 2 4.I.1, 2 12 4.I.3 4.II 4.II 13 4.III.1 5.I 5.II. 14 5.II.2 5.II.3 review

The second timetable is more ambitious (it presupposes 1.II, the elements of vectors, usually covered in third semester calculus).

week Mon. Wed. Fri. 1 1.I.1 1.I.2 1.I. 2 1.I.3 1.III.1, 2 1.III. 3 2.I.1 2.I.2 2.II 4 2.III.1 2.III.2 2.III. 5 2.III.4 3.I.1 exam 6 3.I.2 3.II.1 3.II. 7 3.III.1 3.III.2 3.IV.1, 2 8 3.IV.2 3.IV.3 3.IV. 9 3.V.1 3.V.2 3.VI. 10 3.VI.2 4.I.1 exam 11 4.I.2 4.I.3 4.I. 12 4.II 4.II, 4.III.1 4.III.2, 3 13 5.II.1, 2 5.II.3 5.III. 14 5.III.2 5.IV.1, 2 5.IV.

See the table of contents for the titles of these subsections.

iii

Contents

Chapter 1

Linear Systems

1.I Solving Linear Systems

Systems of linear equations are common in science and mathematics. These two examples from high school science [Onan] give a sense of how they arise. The first example is from Physics. Suppose that we are given three objects, one with a mass of 2 kg, and are asked to find the unknown masses. Suppose further that experimentation with a meter stick produces these two balances.

h c^2 15

40 50 c (^2) h

25 50

25

Now, since the sum of moments on the left of each balance equals the sum of moments on the right (the moment of an object is its mass times its distance from the balance point), the two balances give this system of two equations.

40 h + 15c = 100 25 c = 50 + 50h

The second example of a linear system is from Chemistry. We can mix, under controlled conditions, toluene C 7 H 8 and nitric acid HNO 3 to produce trinitrotoluene C 7 H 5 O 6 N 3 along with the byproduct water (conditions have to be controlled very well, indeed — trinitrotoluene is better known as TNT). In what proportion should those components be mixed? The number of atoms of each element present before the reaction

x C 7 H 8 + y HNO 3 −→ z C 7 H 5 O 6 N 3 + w H 2 O

must equal the number present afterward. Applying that principle to the ele- ments C, H, N, and O in turn gives this system.

7 x = 7z 8 x + 1y = 5z + 2w 1 y = 3z 3 y = 6z + 1w

Section I. Solving Linear Systems 3

we repeatedly transform it until it is in a form that is easy to solve.

swap row 1 with row 3 −→

1 3 x^1 + 2x^2 = 3 x 1 + 5x 2 − 2 x 3 = 2 3 x 3 = 9

multiply row 1 by 3 −→

x 1 + 6x 2 = 9 x 1 + 5x 2 − 2 x 3 = 2 3 x 3 = 9

add −1 times row 1 to row 2 −→

x 1 + 6x 2 = 9 −x 2 − 2 x 3 = − 7 3 x 3 = 9

The third step is the only nontrivial one. We’ve mentally multiplied both sides of the first row by −1, mentally added that to the old second row, and written the result in as the new second row. Now we can find the value of each variable. The bottom equation shows that x 3 = 3. Substituting 3 for x 3 in the middle equation shows that x 2 = 1. Substituting those two into the top equation gives that x 1 = 3 and so the system has a unique solution: the solution set is { (3, 1 , 3) }.

Most of this subsection and the next one consists of examples of solving linear systems by Gauss’ method. We will use it throughout this book. It is fast and easy. But, before we get to those examples, we will first show that this method is also safe in that it never loses solutions or picks up extraneous solutions.

1.4 Theorem (Gauss’ method) If a linear system is changed to another by one of these operations

(1) an equation is swapped with another (2) an equation has both sides multiplied by a nonzero constant (3) an equation is replaced by the sum of itself and a multiple of another

then the two systems have the same set of solutions.

Each of those three operations has a restriction. Multiplying a row by 0 is not allowed because obviously that can change the solution set of the system. Similarly, adding a multiple of a row to itself is not allowed because adding − 1 times the row to itself has the effect of multiplying the row by 0. Finally, swap- ping a row with itself is disallowed to make some results in the fourth chapter easier to state and remember (and besides, self-swapping doesn’t accomplish anything).

Proof. We will cover the equation swap operation here and save the other two cases for Exercise 29.

4 Chapter 1. Linear Systems

Consider this swap of row i with row j. a 1 , 1 x 1 + a 1 , 2 x 2 + · · · a 1 ,nxn = d 1 .. . ai, 1 x 1 + ai, 2 x 2 + · · · ai,nxn = di .. . aj, 1 x 1 + aj, 2 x 2 + · · · aj,nxn = dj .. . am, 1 x 1 + am, 2 x 2 + · · · am,nxn = dm

a 1 , 1 x 1 + a 1 , 2 x 2 + · · · a 1 ,nxn = d 1 .. . aj, 1 x 1 + aj, 2 x 2 + · · · aj,nxn = dj .. . ai, 1 x 1 + ai, 2 x 2 + · · · ai,nxn = di .. . am, 1 x 1 + am, 2 x 2 + · · · am,nxn = dm

The n-tuple (s 1 ,... , sn) satisfies the system before the swap if and only if substituting the values, the s’s, for the variables, the x’s, gives true statements: a 1 , 1 s 1 +a 1 , 2 s 2 +· · ·+a 1 ,nsn = d 1 and... ai, 1 s 1 +ai, 2 s 2 +· · ·+ai,nsn = di and... aj, 1 s 1 + aj, 2 s 2 + · · · + aj,nsn = dj and... am, 1 s 1 + am, 2 s 2 + · · · + am,nsn = dm. In a requirement consisting of statements and-ed together we can rearrange the order of the statements, so that this requirement is met if and only if a 1 , 1 s 1 + a 1 , 2 s 2 + · · · + a 1 ,nsn = d 1 and... aj, 1 s 1 + aj, 2 s 2 + · · · + aj,nsn = dj and... ai, 1 s 1 + ai, 2 s 2 + · · · + ai,nsn = di and... am, 1 s 1 + am, 2 s 2 + · · · + am,nsn = dm. This is exactly the requirement that (s 1 ,... , sn) solves the system after the row swap. QED

1.5 Definition The three operations from Theorem 1.4 are the elementary re- duction operations, or row operations, or Gaussian operations. They are swap- ping, multiplying by a scalar or rescaling, and pivoting.

When writing out the calculations, we will abbreviate ‘row i’ by ‘ρi’. For instance, we will denote a pivot operation by kρi + ρj , with the row that is changed written second. We will also, to save writing, often list pivot steps together when they use the same ρi.

1.6 Example A typical use of Gauss’ method is to solve this system.

x + y = 0 2 x − y + 3z = 3 x − 2 y − z = 3

The first transformation of the system involves using the first row to eliminate the x in the second row and the x in the third. To get rid of the second row’s 2 x, we multiply the entire first row by −2, add that to the second row, and write the result in as the new second row. To get rid of the third row’s x, we multiply the first row by −1, add that to the third row, and write the result in as the new third row.

−ρ 1 +ρ 3 −→ − 2 ρ 1 +ρ 2

x + y = 0 − 3 y + 3z = 3 − 3 y − z = 3

(Note that the two ρ 1 steps − 2 ρ 1 + ρ 2 and −ρ 1 + ρ 3 are written as one opera- tion.) In this second system, the last two equations involve only two unknowns.

6 Chapter 1. Linear Systems

the second equation has no leading y. To get one, we look lower down in the system for a row that has a leading y and swap it in.

ρ 2 ↔ρ 3 −→

x − y = 0 y + w = 0 z + 2w = 4 2 z + w = 5

(Had there been more than one row below the second with a leading y then we could have swapped in any one.) The rest of Gauss’ method goes as before.

− 2 ρ 3 +ρ 4 −→

x − y = 0 y + w = 0 z + 2 w = 4 − 3 w = − 3

Back-substitution gives w = 1, z = 2 , y = −1, and x = −1.

Strictly speaking, the operation of rescaling rows is not needed to solve linear systems. We have included it because we will use it later in this chapter as part of a variation on Gauss’ method, the Gauss-Jordan method. All of the systems seen so far have the same number of equations as un- knowns. All of them have a solution, and for all of them there is only one solution. We finish this subsection by seeing for contrast some other things that can happen.

1.11 Example Linear systems need not have the same number of equations as unknowns. This system

x + 3y = 1 2 x + y = − 3 2 x + 2y = − 2

has more equations than variables. Gauss’ method helps us understand this system also, since this

− 2 ρ 1 +ρ 2 −→ − 2 ρ 1 +ρ 3

x + 3 y = 1 − 5 y = − 5 − 4 y = − 4

shows that one of the equations is redundant. Echelon form

−(4/5)ρ 2 +ρ 3 −→

x + 3 y = 1 − 5 y = − 5 0 = 0

gives y = 1 and x = −2. The ‘0 = 0’ is derived from the redundancy.

Section I. Solving Linear Systems 7

That example’s system has more equations than variables. Gauss’ method is also useful on systems with more variables than equations. Many examples are in the next subsection. Another way that linear systems can differ from the examples shown earlier is that some linear systems do not have a unique solution. This can happen in two ways. The first is that it can fail to have any solution at all.

1.12 Example Contrast the system in the last example with this one.

x + 3y = 1 2 x + y = − 3 2 x + 2y = 0

− 2 ρ 1 +ρ 2 −→ − 2 ρ 1 +ρ 3

x + 3 y = 1 − 5 y = − 5 − 4 y = − 2

Here the system is inconsistent: no pair of numbers satisfies all of the equations simultaneously. Echelon form makes this inconsistency obvious.

−(4/5)ρ 2 +ρ 3 −→

x + 3 y = 1 − 5 y = − 5 0 = 2

The solution set is empty.

1.13 Example The prior system has more equations than unknowns, but that is not what causes the inconsistency — Example 1.11 has more equations than unknowns and yet is consistent. Nor is having more equations than unknowns necessary for inconsistency, as is illustrated by this inconsistent system with the same number of equations as unknowns.

x + 2y = 8 2 x + 4y = 8

− 2 ρ 1 +ρ 2 −→

x + 2y = 8 0 = − 8

The other way that a linear system can fail to have a unique solution is to have many solutions.

1.14 Example In this system

x + y = 4 2 x + 2y = 8

any pair of numbers satisfying the first equation automatically satisfies the sec- ond. The solution set {(x, y)

∣ (^) x + y = 4} is infinite — some of its members

are (0, 4), (− 1 , 5), and (2. 5 , 1 .5). The result of applying Gauss’ method here contrasts with the prior example because we do not get a contradictory equa- tion.

− 2 ρ 1 +ρ 2 −→ x + y = 4 0 = 0

Section I. Solving Linear Systems 9

(a) 2 x + 2y = 5 x − 4 y = 0

(b) −x + y = 1 x + y = 2

(c) x − 3 y + z = 1 x + y + 2z = 14 (d) −x − y = 1 − 3 x − 3 y = 2

(e) 4 y + z = 20 2 x − 2 y + z = 0 x + z = 5 x + y − z = 10

(f ) 2 x + z + w = 5 y − w = − 1 3 x − z − w = 0 4 x + y + 2z + w = 9 X 1.18 There are methods for solving linear systems other than Gauss’ method. One often taught in high school is to solve one of the equations for a variable, then substitute the resulting expression into other equations. That step is repeated until there is an equation with only one variable. From that, the first number in the solution is derived, and then back-substitution can be done. This method both takes longer than Gauss’ method, since it involves more arithmetic operations and is more likely to lead to errors. To illustrate how it can lead to wrong conclusions, we will use the system x + 3y = 1 2 x + y = − 3 2 x + 2y = 0 from Example 1.12. (a) Solve the first equation for x and substitute that expression into the second equation. Find the resulting y. (b) Again solve the first equation for x, but this time substitute that expression into the third equation. Find this y. What extra step must a user of this method take to avoid erroneously concluding a system has a solution? X 1.19 For which values of k are there no solutions, many solutions, or a unique solution to this system? x − y = 1 3 x − 3 y = k

X 1.20 This system is not linear: 2 sin α − cos β + 3 tan γ = 3 4 sin α + 2 cos β − 2 tan γ = 10 6 sin α − 3 cos β + tan γ = 9 but we can nonetheless apply Gauss’ method. Do so. Does the system have a solution? X 1.21 What conditions must the constants, the b’s, satisfy so that each of these systems has a solution? Hint. Apply Gauss’ method and see what happens to the right side. (a) x − 3 y = b 1 3 x + y = b 2 x + 7y = b 3 2 x + 4y = b 4

(b) x 1 + 2x 2 + 3x 3 = b 1 2 x 1 + 5x 2 + 3x 3 = b 2 x 1 + 8x 3 = b 3

1.22 True or false: a system with more unknowns than equations has at least one solution. (As always, to say ‘true’ you must prove it, while to say ‘false’ you must produce a counterexample.) 1.23 Must any Chemistry problem like the one that starts this subsection — a balance the reaction problem — have infinitely many solutions? X 1.24 Find the coefficients a, b, and c so that the graph of f (x) = ax^2 + bx + c passes through the points (1, 2), (− 1 , 6), and (2, 3).

10 Chapter 1. Linear Systems

1.25 Gauss’ method works by combining the equations in a system to make new equations. (a) Can the equation 3x− 2 y = 5 be derived, by a sequence of Gaussian reduction steps, from the equations in this system? x + y = 1 4 x − y = 6

(b) Can the equation 5x− 3 y = 2 be derived, by a sequence of Gaussian reduction steps, from the equations in this system? 2 x + 2y = 5 3 x + y = 4

(c) Can the equation 6x − 9 y + 5z = −2 be derived, by a sequence of Gaussian reduction steps, from the equations in the system? 2 x + y − z = 4 6 x − 3 y + z = 5 1.26 Prove that, where a, b,... , e are real numbers and a 6 = 0, if ax + by = c has the same solution set as ax + dy = e then they are the same equation. What if a = 0?

X 1.27 Show that if ad − bc 6 = 0 then

ax + by = j cx + dy = k has a unique solution.

X 1.28 In the system

ax + by = c dx + ey = f each of the equations describes a line in the xy-plane. By geometrical reasoning, show that there are three possibilities: there is a unique solution, there is no solution, and there are infinitely many solutions. 1.29 Finish the proof of Theorem 1.4. 1.30 Is there a two-unknowns linear system whose solution set is all of R^2?

X 1.31 Are any of the operations used in Gauss’ method redundant? That is, can any of the operations be synthesized from the others? 1.32 Prove that each operation of Gauss’ method is reversible. That is, show that if two systems are related by a row operation S 1 ↔ S 2 then there is a row operation to go back S 2 ↔ S 1. 1.33 A box holding pennies, nickels and dimes contains thirteen coins with a total value of 83 cents. How many coins of each type are in the box? 1.34 [Con. Prob. 1955] Four positive integers are given. Select any three of the integers, find their arithmetic average, and add this result to the fourth integer. Thus the numbers 29, 23, 21, and 17 are obtained. One of the original integers is: