Linear Algebra: Textbook PDF

David Poole A M 0 'D E R N I N T R 0 D U, C T I 0 N 4th edition  Fourth edition David Poole Trent University.. CENGAGE - Learning· Australia Brazil Mexico S...

David Poole A M 0 'D E R N I N T R 0 D U, C T I 0 N 4th edition  Fourth edition David Poole Trent University.. CENGAGE - Learning· Australia Brazil Mexico Singapore United l AB = [ 3, 2 ] = [ 6 - 3, 3 - l] Similarly, CD = r - 1 - ( -4), 1 - ( - 1) J [ 3, 2 ] -----> ----> and thus AB = CD , as expected. A vector such as oP with its tail at the origin is said to be in standard position. The foregoing discussion shows that every vector can be drawn as a vector in stan­ dard position. Conversely, a vector in standard position can be redrawn (by transla­ tion) so that its tail is at any point in the plane. Exa m p l e 1. 1 If A = ( - 1 , 2 ) and B = ( 3, 4), find AB and redraw it (a) in standard position and (b) with its tail at the point C = (2, - 1). -----> -----> Solulion We compute AB = [3 - ( - 1), 4 - 2] = [4, 2]. If AB is then translated to CD, where C = ( 2, - 1 ), then we must have D = ( 2 + 4, - 1 + 2) = (6, 1). (See Figure 1.5. )  Section 1.1 The Geometry and Algebra of Vectors 5 y D(6, 1 ) Figure 1. 5 New Veclors from Old As in the racetrack game, we often want to "follow" one vector by another. This leads to the notion of vector addition, the first basic vector operation. If we follow u by v, we can visualize the total displacement as a third vector, denoted by u + v. In Figure 1.6, u = [ 1, 2] and v = [ 2, 2 ] , so the net effect of follow­ ing u by v is [ 1 + 2, 2 + 2 ] = [3, 4] which gives u + v. In general, if u = [u 1 , u 2 ] and v = [ v 1 , v2 ] , then their s u m u + v is the vector It is helpful to visualize u + v geometrically. The following rule is the geometric version of the foregoing discussion. y (/lj2 / 2_ _ : 2 4 I I U /_ ) _ _ _ _ _ _ _ _ _ _ _n 3 Figure 1. 6 Vector addition 6 Chapter 1 Vectors The Head-to-Tail R u l e Given vectors u and v in IR 2 , translate v s o that its tail coincides with the head of u. The s u m u + v of u and v is the vector from the tail of u to the head of v. (See Figure 1.7. ) Figure 1. 1 The head-to-tail rule Figure 1. 8 By translating u and v parallel to themselves, we obtain a parallelogram, as The parallelogram shown in Figure 1.8. This parallelogram is called the parallelogram determined by u determined by u and v and v. It leads to an equivalent version of the head-to-tail rule for vectors in standard position. The Parallelogram Rule Given vectors u and v in IR 2 (in standard position), their s u m u + v is the vector in standard position along the diagonal of the parallelogram determined by u and v. (See Figure 1.9. ) y Figure 1. 9 The parallelogram rule Exa m p l e 1. 2 If u = [3, - 1 ] and v = [ l , 4], compute and draw u + v. Solulion We compute u + v = [3 + 1, - 1 + 4] = [4, 3]. This vector is drawn using the head-to-tail rule in Figure l. l O(a) and using the parallelogram rule in Figure l. l O(b).  Section 1.1 The Geometry and Algebra of Vectors 1 y y u (a) (b) Figure 1. 1 0 The second basic vector operation is scalar multiplication. Given a vector v and a real number c, the scalar multiple c v is the vector obtained by multiplying each component of v by c. For example, 3 [ - 2, 4] = [ - 6, 12]. In general, Geometrically, cv is a "scaled" version of v. Exa m p l e 1. 3 If v = [ - 2, 4], compute and draw 2v, tv, and - 2v. Solution We calculate as follows: 2v = [ 2 ( - 2 ) , 2 ( 4 ) ] = [ - 4, 8 J tv = [t ( - 2 ) , t ( 4 ) ] = [ - 1 , 2 ] - 2v = [ - 2 ( - 2 ) , - 2 ( 4 ) ] = [ 4, - 8 ] These vectors are shown in Figure 1. 1 1. y 2v - 2v Figure 1. 1 1 8 Chapter 1 Vectors /./ u 2v - 2v u + ( - v) Figure 1. 1 3 Figure 1. 1 2 Vector subtraction Observe that cv has the same direction as v if c > 0 and the opposite direction if The term scalar comes from the c < 0. We also see that cv is le I times as long as v. For this reason, in the context of Latin word scala, meaning "lad­ vectors, constants (i.e., real numbers) are referred to as scalars. As Figure 1. 1 2 shows, dd' The equally spaced rungs on when translation of vectors is taken into account, two vectors are scalar multiples of a ladder suggest a scale, and in vec­ tor arithmetic, multiplication by a each other if and only if they are parallel. constant changes only the scale (or A special case of a scalar multiple is ( - l )v, which is written as -v and is called length) of a vector. Thus, constants the negative ofv. We can use it to define vector subtraction: The difference of u and became known as scalars. v is the vector u - v defined by u - v = u + ( - v) Figure 1. 1 3 shows that u - v corresponds to the "other" diagonal of the parallelo­ gram determined by u and v. Exa m p l e 1. 4 If u = [ 1 , 2] and v = [ - 3, 1 ] , then u - v = [ 1 - ( - 3), 2 - 1 ] = [ 4, l ]. The definition of subtraction in Example 1.4 also agrees with the way we cal­ y culate a vector such as AB.---- If the -> points A and B correspond to the vectors a and b A in standard position, then AB b - a, as shown in Figure 1. 14. [Observe that the = head-to-tail rule applied to this diagram gives the equation a + (b - a ) = b. If we had accidentally drawn b - a with its head at A instead of at B, the diagram would have read b + (b - a) = a, which is clearly wrong! More will be said about algebraic expressions involving vectors later in this section.] Figure 1. 1 4 vec1ors in 3 Everything we have just done extends easily to three dimensions. The set of all or­ dered triples of real numbers is denoted by IR 3. Points and vectors are located using three mutually perpendicular coordinate axes that meet at the origin 0. A point such as A = ( 1 , 2, 3) can be located as follows: First travel 1 unit along the x-axis, then move 2 units parallel to the y-axis, and finally move 3 units parallel to the z-axis. The corresponding vector a = [ l , 2, 3] is then OA, as shown in Figure 1. 15. Another way to visualize vector a in IR 3 is to construct a box whose six sides are de­ termined by the three coordinate planes (the xy-, xz-, and yz-planes) and by three planes through the point ( 1 , 2, 3 ) parallel to the coordinate planes. The vector [ 1 , 2, 3] then corre­ sponds to the diagonal from the origin to the opposite corner of the box (see Figure 1. 16).  Section 1.1 The Geometry and Algebra of Vectors 9 z z A ( l , 2, 3 I I 3) I ,. I :3 I I - I -- 1 2 - x y x Figure 1. 1 5 Figure 1. 1 6 The "componentwise" definitions of vector addition and scalar multiplication are extended to IR 3 in an obvious way. Vectors in n In general, we define !R n as the set of all ordered n-tuples of real numbers written as row or column vectors. Thus, a vector v in !R n is of the form The individual entries of v are its components; V; is called the ith component. We extend the definitions of vector addition and scalar multiplication to !R n in the obvious way: If u = [u 1 , u 2 , , u n l and v = [ v 1 , v2 ,... , vn ] , the ith component of u + v is U ; + V; and the ith component of cv is just C V;. Since in !R n we can no longer draw pictures of vectors, it is important to be able to calculate with vectors. We must be careful not to assume that vector arithmetic will be similar to the arithmetic of real numbers. Often it is, and the algebraic calculations we do with vectors are similar to those we would do with scalars. But, in later sections, we will encounter situations where vector algebra is quite unlike our previous experi­ ence with real numbers. So it is important to verify any algebraic properties before attempting to use them. One such property is commutativity of addition: u + v = v + u for vectors u and v. This is certainly true in IR 2 Geometrically, the head-to-tail rule shows that both u + v and v + u are the main diagonals of the parallelogram determined by u and v. (The parallelogram rule also reflects this symmetry; see Figure 1. 1 7.) Note that Figure 1. 1 7 is simply an illustration of the property u + v = v + u. It is not a proof, since it does not cover every possible case. For example, we must also include the cases where u = v, u = - v, and u = 0. (What would diagrams for these cases look like?) For this reason, an algebraic proof is needed. However, it is just as easy to give a proof that is valid in !R n as to give one that is valid in IR 2. Figure 1. 1 1 The following theorem summarizes the algebraic properties of vector addition u+v=v+u and scalar multiplication in !R n. The proofs follow from the corresponding properties of real numbers. 10 Chapter 1 Vectors Theorem 1. 1 Algebraic Properties o f Vectors i n !R n Let u, v, and w be vectors in !R n and let c and d be scalars. Then a. u+v=v+u Commutativity b. (u + v) + w = u + (v + w) Associativity c. u+0=u d. u + ( -u) = 0 e. c (u + v) = cu + cv Distributivity f. (c + d)u = cu + du Distributivity g. c (du) = (cd)u h. lu = u R e m a rks The word theorem is derived from Properties (c) and (d) together with the commutativity property (a) imply the Greek word theorema, which that 0 + u = u and - u + u = 0 as well. in turn comes from a word mean­ If we read the distributivity properties (e) and (f) from right to left, they say ing "to look af' Thus, a theorem that we can factor a common scalar or a common vector from a sum. is based on the insights we have when we look at examples and Proof We prove properties (a) and (b) and leave the proofs of the remain­ extract from them properties that ing properties as exercises. Let u = [u 1 , u 2 ,... , u n ] , v = [ v 1 , v2 ,... , vn ] , and w = we try to prove hold in general. [ W1 , Wz ,... , Wn ]. Similarly, when we understand something in mathematics-the (a) U + = [U1, U z ,... , U n ] + [V1, Vz ,... , Vn ] V proof of a theorem, for example­ we often say, "I see:' = [U1 + V1 , U z + Vz ,... , U n + Vn ] = [ v 1 + u 1 , v2 + u 2 , , v n + u n ].. = [v 1 , Vz ,... , Vn ] + [u 1 , U z ,... , U n ] =v+u The second and fourth equalities are by the definition of vector addition, and the third equality is by the commutativity of addition of real numbers. (b) Figure 1. 1 8 illustrates associativity in IR 2. Algebraically, we have [ (u 1 + v 1 ) + w 1 , (u 2 + vz ) + w2 ,... , (u n + vn ) + wn ] [ u 1 + (v l + W1 ), U 2 + ( Vz + Wz ),... , U n + (vn + Wn ) J [ u 1 , u 2 ,... , u n ] + ( [ v 1 , v2 ,... , vn ] + [ w 1 , w2 ,... , w n ] ) = u + (v + w) u The fourth equality is by the associativity of addition of real numbers. Note the care­ ful use of parentheses. Figure 1. 1 8  Section 1.1 The Geometry and Algebra of Vectors 11 By property (b) of Theorem 1. 1 , we may unambiguously write u + v + w without parentheses, since we may group the summands in whichever way we please. By (a), we may also rearrange the summands-for example, as w + u + v-if we choose. Likewise, sums of four or more vectors can be calculated without regard to order or grouping. In general, if v1 , v2 , , vk are vectors in !R n , we will write such sums with­. out parentheses: The next example illustrates the use of Theorem 1. 1 in performing algebraic calculations with vectors. Exa m p l e 1. 5 Let a, b, and x denote vectors in !R n. (a) Simplify 3a + (Sb - 2a) + 2(b - a). (b) If Sx - a = 2(a + 2x), solve for x in terms of a. Solution We will give both solutions in detail, with reference to all of the properties in Theorem 1. 1 that we use. It is good practice to justify all steps the first few times you do this type of calculation. Once you are comfortable with the vector properties, though, it is acceptable to leave out some of the intermediate steps to save time and space. (a) We begin by inserting parentheses. 3a + (Sb - 2a) + 2(b - a) = (3a + (Sb - 2a)) + 2(b - a) (3a + ( - 2a + Sb)) + (2b - 2a) (a), (e) ((3a + ( - 2a)) + Sb) + (2b - 2a) (b) ((3 + ( - 2))a + Sb) + (2b - 2a) (f) ( la + Sb) + (2b - 2a) ((a + Sb) + 2b) - 2a (b ), (h) (a + (Sb + 2b)) - 2a (b) (a + (S + 2)b) - 2a (f) (7b + a) - 2a (a) = 7b + (a - 2a) (b) = 7b + ( 1 - 2)a (f), (h) = 7b + ( - l )a = 7b - a You can see why we will agree to omit some of these steps! In practice, it is acceptable to simplify this sequence of steps as 3a + (Sb - 2a) + 2(b - a) = 3a + Sb - 2a + 2b - 2a (3a - 2a - 2a) + (Sb + 2b) = - a + 7b or even to do most of the calculation mentally. 12 Chapter 1 Vectors (b) In detail, we have Sx - a = 2 ( a + 2x) Sx - a = 2a + 2 ( 2x) (e) Sx - a = 2a + ( 2 · 2 ) x (g) Sx - a = 2a + 4x ( Sx - a ) - 4x = ( 2a + 4x) - 4x ( - a + Sx) - 4x = 2a + ( 4x - 4x) (a), (b) - a + ( Sx - 4x) = 2a + 0 (b), (d) - a + ( 5 - 4)x = 2a (f), (c) - a + ( l )x = 2a a + ( - a + x) = a + 2a (h) ( a + ( - a )) + x = ( 1 + 2)a (b), (f) 0+x = 3a (d) x = 3a (c) Again, we will usually omit most of these steps. Linear Combin alions and Coordin a1es A vector that is a sum of scalar multiples of other vectors is said to be a linear combi­ nation of those vectors. The formal definition follows. D e f i n i l i O D A vector v is a linear combination of vectors v1 , v2 ,... , vk if there are scalars c 1 , c2 ,... , c k such that v = c 1 v1 + c2v2 + · ·+ ckvk. The scalars · c 1 , c2 ,... , ck are called the coefficients of the linear combination. Exa m p l e 1. 6 Remark Determining whether a given vector is a linear combination of other vectors is a problem we will address in Chapter 2. In IR 2 , it is possible to depict linear combinations of two (nonparallel) vectors quite conveniently. Exa m p l e 1. 1 Let u = [ ] and v = [ ]. We can use u and v to locate a new set of axes (in the same way that e 1 = [ ] and e2 = [ ] locate the standard coordinate axes). We can use  Section 1.1 The Geometry and Algebra of Vectors 13 y I u I I I- I Figure 1. 1 9 these new axes to determine a coordinate grid that will let us easily locate linear combinations of u and v. As Figure 1. 1 9 shows, w can be located by starting at the origin and traveling - u followed by 2v. That is, w = -u + 2v We say that the coordinates of w with respect to u and v are - 1 and 2. (Note that this is just another way of thinking of the coefficients of the linear combination.) It follows that (Observe that - 1 and 3 are the coordinates of w with respect to e 1 and e2. ) Switching from the standard coordinate axes to alternative ones is a useful idea. It has applications in chemistry and geology, since molecular and crystalline structures often do not fall onto a rectangular grid. It is an idea that we will encounter repeatedly in this book. Binarv vec1ors and Modular Arilhmelic We will also encounter a type of vector that has no geometric interpretation-at least not using Euclidean geometry. Computers represent data in terms of Os and ls (which can be interpreted as off/on, closed/open, false/true, or no/yes). Binary vectors are vectors each of whose components is a 0 or a 1. As we will see in Chapter 8, such vectors arise naturally in the study of many types of codes. In this setting, the usual rules of arithmetic must be modified, since the result of each calculation involving scalars must be a 0 or a 1. The modified rules for addition and multiplication are given below. The only curiosity here is the rule that 1 + 1 = 0. This is not as strange as it appears; if we replace 0 with the word "even" and 1 with the word "odd," these tables simply 14 Chapter 1 Vectors summarize the familiar parity rules fo r the addition and multiplication of even and odd integers. For example, 1 + 1 = 0 expresses the fact that the sum of two odd inte­ gers is an even integer. With these rules, our set of scalars {O, l } is denoted by 22 and is called the set of integers modulo 2. Exa m p l e 1. 8 In 22, 1 + 1 + 0 + 1 = 1 and 1 + 1 + 1 + 1 = 0. (These calculations illustrate the parity rules again: The sum of three odds and an even is odd; the sum of four.+ odds is even.) We are using the term length dif­ ferently from the way we used it in !FR". This should not be confusing, With 22 as our set of scalars, we now extend the above rules to vectors. The set of since there is no geometric notion all n-tuples of Os and l s (with all arithmetic performed modulo 2) is denoted by 2. of length for binary vectors. The vectors in 2 are called binary vectors of length n. Exa m p l e 1. 9 The vectors in 2 are [O, OJ , [O, l ] , [ l , OJ , and [ l , l J. (How many vectors does 2 contain, in general?) Exa m p l e 1. 1 0 Let u = [l, 1, 0, 1, OJ and v = [O, 1, 1, 1, OJ be two binary vectors oflength 5. Find u + v. Solulion The calculation of u + v takes place over 22 , so we have u + v = [ 1 , i , o, i , o ] + [o, I , I , i , o ] = [ 1 + o, I + I , o + I , I + I , o + o ] = [ 1 , 0, 1 , 0, 0 ] It is possible to generalize what we have just done for binary vectors to vectors whose components are taken from a finite set {O, 1, 2,... , k} for k 2: 2. To do so, we must first extend the idea of binary arithmetic. Exa m p l e 1. 1 1 The integers modulo 3 is the set 2 3 = {O, 1, 2} with addition and multiplication given by the following tables: + 0 2 0 1 2 0 0 1 2 0 0 0 0 1 2 0 1 0 2 2 2 0 2 0 2 1 Observe that the result of each addition and multiplication belongs to the set {O, 1, 2}; we say that 2 3 is closed with respect to the operations of addition and multi­ plication. It is perhaps easiest to think of this set in terms of a 3-hour clock with 0, 1 , and 2 o n its face, as shown i n Figure 1.20. The calculation 1 + 2 = 0 translates as follows: 2 hours after 1 o'clock, it is 0 o'clock. Just as 24:00 and 1 2:00 are the same on a 1 2-hour clock, so 3 and 0 are equivalent on this 3-hour clock. Likewise, all multiples of 3-positive and negative­ are equivalent to 0 here; 1 is equivalent to any number that is 1 more than a multiple of 3 (such as - 2, 4, and 7); and 2 is equivalent to any number that is 2 more than a  Section 1.1 The Geometry and Algebra of Vectors 15 multiple of 3 (such as - 1 , 5, and 8). We can visualize the number line as wrapping around a circle, as shown in Figure 1.2 1. 0... , - 3 , 0, 3,... 2... , 1 , 2, 5 ,...... , - 2, 1 , 4,... Figure 1. 2 0 Arithmetic modulo 3 Figure 1. 2 1 Exa m p l e 1. 1 2 To what is 3548 equivalent in Z/ Solution This is the same as asking where 3548 lies on our 3-hour clock. The key is to calculate how far this number is from the nearest (smaller) multiple of 3; that is, we need to know the remainder when 3548 is divided by 3. By long division, we find that 3548 = 3 · 1 1 82 + 2, so the remainder is 2. Therefore, 3548 is equivalent to 2 in l'.. 3 4 In courses in abstract algebra and number theory, which explore this concept in greater detail, the above equivalence is often written as 3548 = 2 (mod 3) or 3548 = 2 (mod 3), where = is read "is congruent to." We will not use this notation or termi­ nology here. Exa m p l e 1. 1 3 In l'.. 3 , calculate 2 + 2 + 1 + 2. Solution 1 We use the same ideas as in Example 1. 12. The ordinary sum is 2 + 2 + 1 + 2 = 7, which is 1 more than 6, so division by 3 leaves a remainder of 1. Thus, 2 + 2 + 1 + 2 = 1 in l'.. 3. Solution 2 A better way to perform this calculation is to do it step by step entirely in l'.. 3 2+2+ 1 +2= (2 + 2) + 1 + 2 = 1 + 1 +2 = (1 + 1) + 2 = 2+2 = 1 Here we have used parentheses to group the terms we have chosen to combine. We could speed things up by simultaneously combining the first two and the last two terms: (2 + 2) + ( 1 + 2) = 1 + 0 = 1 16 Chapter 1 Vectors Repeated multiplication can be handled similarly. The idea is to use the addition and 4 multiplication tables to reduce the result of each calculation to 0, 1 , or 2. Extending these ideas to vectors is straightforward. Exa m p l e 1. 1 4 In Z , let u = [2, 2, 0, 1 , 2] and v = [ l , 2, 2, 2, l ]. Then m - 1 0 + v = [ 2, 2, 0, 1, 2 ] + [ 1, 2, 2, 2, l ] u _.,---._ m - 2 2 [ 2 + 1, 2 + 2, 0 + 2, 1 + 2, 2 + l ] 3 [ 0, 1 , 2, 0, 0 ] Vectors in Z are referred to as ternary vectors of length 5. In general, we have the set Z m = {O, 1, 2,... , m - l } of integers modulo m (cor­ responding to an m-hour clock, as shown in Figure 1.22). A vector of length n whose Figure 1.22 entries are in Z m is called an m-ary vector of length n. The set of all m-ary vectors of Arithmetic modulo m length n is denoted by z::i... I Exercises 1. 1 1. Draw the following vectors in standard position 6. A hiker walks 4 km north and then 5 km northeast. in IR 2 : Draw displacement vectors representing the hiker's (a) a = [] 3 0 (b) b = [] 2 3 trip and draw a vector that represents the hiker's net displacement from the starting point. Exercises 7- 1 0 refer to the vectors in Exercise 1. Compute (d) d = [] -2 3 the indicated vectors and also show how the results can be obtained geometrically. 2. Draw the vectors in Exercise 1 with their tails at the 7. a + b 8. b - c point (2, - 3). 9. d - c 10. a + d 3. Draw the following vectors in standard position in IR 3 : Exercises 1 1 and 12 refer to the vectors in Exercise 3. (a) a = [O, 2, OJ (b) b = [ 3 , 2, l ] Compute the indicated vectors. (c) c = [ l , - 2, l ] (d) d = [ - 1, - 1 , - 2] 1 1. 2a + 3c 12. 3b - 2c + d 4. If the vectors in Exercise 3 are translated so that their 13. Find the components of the vectors u, v, u + v, and heads are at the point (3, 2, 1), find the points that u - v, where u and v are as shown in Figure 1.23. correspond to their tails. 14. In Figure 1. 2 4 , A, B, C, D, E, and F are the vertices of a 5. For each of the following pairs of points, draw the ----> ----> regular hexagon centered at the origin. vector AB. Then compute and redraw AB as a vector Express each of the following vectors in terms of in standard position. ----> ----> a = OA and b = OB : (a) A = ( 1, - 1 ), B = (4, 2) (b) A = (O, - 2), B = ( 2, - 1 ) (a) Ai (b) BC (c) A = (2, f), B = (t, 3 ) (c) Ai5 (d) a (d) A = (t, t), B = (i, t) (e) xc (f) BC + ill + PX  Section 1.1 The Geometry and Algebra of Vectors 11 y 21. u = [- ], v= [ ] , [ ] w= 22. u = [ - ], v= [ ] , [ ] w= 23. Draw diagrams to illustrate properties (d) and (e) of Theorem 1. 1. 24. Give algebraic proofs of properties ( d) through (g) of Theorem 1. 1. In Exercises 25-28, u and v are binary vectors. Find u + v in each case. Figure 1. 2 3 25. u = [ ] ,v [ ] = 27. u = [ 1 , 0, 1 , 1 ] , v = [ 1, 1 , 1, 1 ] y 28. u = [ 1 , 1, 0, 1, 0 ] , v = [ O, 1, 1 , 1, 0 ] 29. Write out the addition and multiplication tables for Z 4. c B 30. Write out the addition and multiplication tables for Zs. In Exercises 3 1 -43, perform the indicated calculations. 31. 2 + 2 + 2 in Z3 32. 2 · 2 · 2 in Z3 ( ) 33. 2 2 + 1 + 2 in Z3 34. 3 + 1 + 2 + 3 in Z4 35. 2 · 3 · 2 in Z4 36. 3 ( 3 + 3 + 2 ) in Z4 37. 2 + 1 + 2 + 2 + 1 in Z3 , Z4 , and Zs E F 38. ( 3 + 4 ) ( 3 + 2 + 4 + 2 ) in Zs Figure 1. 2 4 39. 8 ( 6 + 4 + 3 ) in Z9 40. 2 100 in Z 1 1 41. [ 2 , 1 , 2 ] + [ 2, 0, 1 ] in Z 42. 2 [ 2, 2, 1 ] in Z 43. 2 ( [ 3, l , 1 , 2 ] + [ 3, 3, 2, l ] ) in Z! and Z In Exercises 1 5 and 1 6, simplify the given vector expression. Indicate which properties in Theorem 1. 1 you use. In Exercises 44-55, solve the given equation or indicate that there is no solution. 15. 2 (a - 3b) + 3 (2b + a) 44. x + 3 = 2 in Zs 45. x + 5 = 1 in Z6 16. - 3(a - c) + 2(a + 2b) + 3(c - b) 46. 2x = 1 in Z3 47. 2x = 1 in Z4 In Exercises 1 7 and 1 8, solve for the vector x in terms of the 48. 2x = 1 in Zs 49. 3x = 4 in Zs vectors a and b. 50. 3x = 4 in Z6 51. 6x = 5 in Zs 17. x - a = 2(x - 2a) 52. Bx = 9 in Z 1 1 53. 2x + 3 = 2 in Z5 18. x + 2a - b = 3 (x + a) - 2 (2a - b) 54. 4x + 5 = 2 in Z6 55. 6x + 3 = 1 in Zs 56. (a) For which values of a does x + a = 0 have a solu­ In Exercises 19 and 20, draw the coordinate axes relative to tion in Zs? u and v and locate w. (b) For which values of a and b does x + a = b have a [ ] [ l w 19. u = _ , v = = 2 u + 3v solution in Z6? (c) For which values of a, b, and m does x + a = b [ - l [ l have a solution in Z m ? 57. (a) For which values of a does ax = 1 have a solution 20. u = v= _ w = - u - 2v in Zs? (b) For which values of a does ax = 1 have a solution In Exercises 21 and 22, draw the standard coordinate axes in Z6? on the same diagram as the axes relative to u and v. Use (c) For which values of a and m does ax = 1 have a these to find w as a linear combination of u and v. solution in z m ? 18 Chapter 1 Vectors Length a n d A n g l e : T h e D o t P ro d u ct It is quite easy to reformulate the familiar geometric concepts of length, distance, and angle in terms of vectors. Doing so will allow us to use these important and powerful ideas in settings more general than IR 2 and IR 3 In subsequent chapters, these simple geometric tools will be used to solve a wide variety of problems arising in applications-even when there is no geometry apparent at all! The Doi Producl The vector versions of length, distance, and angle can all be described using the notion of the dot product of two vectors. Definilion If then the dot product u · v of u and v is defined by In words, u · v is the sum of the products of the corresponding components of u and v. It is important to note a couple of things about this "product" that we have just defined: First, u and v must have the same number of components. Second, the dot product u · v is a number, not another vector. (This is why u · v is sometimes called the scalar product of u and v.) The dot product of vectors in !R n is a special and im­ portant case of the more general notion of inner product, which we will explore in Chapter 7. Exa m p l e 1. 1 5 Solution u · v = l · ( - 3) + 2 · 5 + ( - 3) · 2 = 1 Notice that if we had calculated v · u in Example 1. 15, we would have computed v · u = ( - 3) · 1 + 5 · 2 + 2 · ( - 3) = 1 That u · v = v · u in general is clear, since the individual products of the components commute. This commutativity property is one of the properties of the dot product that we will use repeatedly. The main properties of the dot product are summarized in Theorem 1.2.  Section 1.2 Length and Angle: The Dot Product 19 Theorem 1. 2 Let u, v, and w be vectors in n and let c be a scalar. Then a. u·v = v·u Commutativity b. u. ( v + w) = u. v + u. w Distributivity c. ( cu). v = c ( u. v) d. u · u 2: 0 and u · u = 0 if and only if u = 0 Proof We prove (a) and (c) and leave proof of the remaining properties for the exercises. (a) Applying the definition of dot product to u · v and v · u, we obtain u · v = U 1 V 1 + U z Vz +.. + U n V n · = V 1 U 1 + Vz U z +... + Vn U n = v·u where the middle equality follows from the fact that multiplication of real numbers is commutative. (c) Using the definitions of scalar multiplication and dot product, we have (cu) · v = [ cu 1 , cu 2 ,... , cu " ] [ v 1 , v2 , , v" ].. CU 1 V 1 + CU z V2 + · · + CU n Vn · c( U 1 V 1 + U z Vz + · · + U n Vn ) · c(u · v) R e m a rlls Property (b) can be read from right to left, in which case it says that we can factor out a common vector u from a sum of dot products. This property also has a "right-handed" analogue that follows from properties (b) and (a) together: (v + w) · u = v · u + w · u. Property (c) can be extended to give u · (cv) = c(u · v) (Exercise 58). This extended version of ( c) essentially says that in taking a scalar multiple of a dot product of vectors, the scalar can first be combined with whichever vector is more convenient. For example, ( [ - 1, -3, 2 ] ) · [ 6, - 4, 0 ] = [ - 1 , -3, 2 ] · ( [6, - 4, 0 ] ) = [ - 1, -3, 2 ] · [3, - 2, 0] = 3 With this approach we avoid introducing fractions into the vectors, as the original grouping would have. The second part of ( d) uses the logical connective if and only if. Appendix A dis­ cusses this phrase in more detail, but for the moment let us just note that the wording signals a double i mplication- namely, if u = 0, then u · u = 0 and if u · u = 0, then u = 0 Theorem 1.2 shows that aspects of the algebra of vectors resemble the algebra of numbers. The next example shows that we can sometimes find vector analogues of familiar identities. 20 Chapter 1 Vectors Exa m p l e 1. 1 6 Prove that (u + v) · (u + v) = u · u + 2(u · v) + v · v for all vectors u and v in !R n. Solulion (u + v) · (u + v) = ( u + v) · u + ( u + v) · v = u·u + v·u + u·v + v·v = u·u + u·v + u·v + v·v = u · u + 2(u · v) + v · v (Identify the parts of Theorem 1.2 that were used at each step.) y Length b v = [ ] To see how the dot product plays a role in the calculation oflengths, recall how lengths are computed in the plane. The Theorem of Pythagoras is all we need. In IR 2 , the length of the vector v = [ ] is the distance from the origin to the point (a, b), which, by Pythagoras' Theorem, is given by Va 2 + b 2 , as in Figure 1.25. Observe that a 2 + b 2 = v · v. This leads to the following definition. a Figure 1. 2 5 Definition The length (or norm) of a vector v = tive scalar ll v ll defined by [ : ] vn · in !R n is the nonnega- ll v ll = VV:V = Vvi + v + · · · + v In words, the length of a vector is the square root of the sum of the squares of its components. Note that the square root of v · v is always defined, since v · v 2': 0 by Theorem l.2 (d). Note also that the definition can be rewritten to give ll v ll 2 = v · v, which will be useful in proving further properties of the dot product and lengths of vectors. Exa m p l e 1. 1 1 vT3 11 [2, 3 J 11 = v2 2 + 3 2 = Theorem 1. 3 lists some of the main properties of vector length. Theorem 1. 3 Let v be a vector in IR " and let c be a scalar. Then a. ll v ll = 0 if and only ifv = 0 b. ll cv ll = l c l ll v ll Proof Property (a) follows immediately from Theorem l.2(d). To show (b), we have ll cv ll 2 = (cv) · (cv) = c 2 (v · v) = c 2 ll v ll 2 using Theorem l.2 (c). Taking square roots ofboth sides, using the fact that W = I c l for any real number c, gives the result.  Section 1.2 Length and Angle: The Dot Product 21 A vector of length 1 is called a unit vector. In IR 2 , the set of all unit vectors can be identified with the unit circle, the circle of radius 1 centered at the origin (see Figure 1.26). Given any nonzero vector v, we can always find a unit vector in the same direction as v by dividing v by its own length (or, equivalently, multiplying by 1/ ll v ll ). We can show this algebraically by using property (b) of Theorem 1.3 above: If u = ( 1 / ll v ll )v, then ll u ll = 11 ( 1/ ll v ll )v ll = l l / ll v ll I ll v ll = ( 1/ ll v ll ) ll v ll = 1 and u is in the same direction as v, since 1 / I v I is a positive scalar. Finding a unit vec­ tor in the same direction is often referred to as normalizing a vector (see Figure 1.27). y v >:;:. / rr 1r Figure 1. 2 6 Figure 1. 2 7 Unit vectors in lffi 2 Normalizing a vector Exa m p l e 1. 1 8 In IR 2 , let e1 = [ ] and e2 = [ ]. Then e1 and e2 are unit vectors, since the sum of the squares of their components is 1 in each case. Similarly, in IR 3 , we can construct unit vectors Observe in Figure 1.28 that these vectors serve to locate the positive coordinate axes in IR 2 and IR 3. t z.--+ y x y Figure 1. 2 8 Standard unit vectors in lffi 2 and !ffi 3 22 Chapter 1 Vectors In general, in !R n , we define unit vectors e1, e2 , , en , where e; has 1 in its ith.. component and zeros elsewhere. These vectors arise repeatedly in linear algebra and are called the standard unit vectors. Exa m p l e 1. 1 9 Nmmalfae the vectm v [ -n Solulion ll v ll V2 2 + ( - 1 ) 2 + 3 2 = = \/14, so a unit vector in the same direc­ ] tion as v is given by u ( 1/ ll v ll l v ( 1/ v'14 - { :J [ 2/ \/14 - 1/ \/14 3/ \/14 Since property (b) of Theorem 1.3 describes how length behaves with respect to scalar multiplication, natural curiosity suggests that we ask whether length and vec­ tor addition are compatible. It would be nice if we had an identity such as II u + v II = ll u 1 + ll v ll , but for almost any choice of vectors u and v this turns out to be false. [See Exercise 52(a).] However, all is not lost, for it turns out that if we replace the = sign by :s , the resulting inequality is true. The proof of this famous and important result-the Triangle Inequality-relies on another important inequality-the Cauchy-Schwarz Inequality-which we will prove and discuss in more detail in Chapter 7. Theorem 1. 4 The Cauchy-Schwarz Inequality For all vectors u and v in !R n , l u · vl :s ll n ll ll v ll - See Exercises 7 1 and 72 for algebraic and geometric approaches to the proof of this u inequality. In IR 2 or IR 3 , where we can use geometry, it is clear from a diagram such as Figure 1. 2 9 Figure 1.29 that ll u + v ii :s ll n ll + ll v ll for all vectors u and v. We now show that The Triangle Inequality this is true more generally. Theorem 1. 5 The Triangle Inequality For all vectors u and v in !R n , ll u + v ii :s ll u ll + ll v ll  Section 1. 2 Length and Angle: The Dot Product 23 Proof Since both sides of the inequality are nonnegative, showing that the square of the left-hand side is less than or equal to the square of the right-hand side is equiva­ lent to proving the theorem. (Why?) We compute ll u + v ll 2 = (u + v) - (u + v) = u · u + 2(u · v) + v · v By Example 1.9 ll ll 2 l ::; u + 2 u · v + v l ll ll 2 ::; ll n ll 2 + 2 ll n ll ll v ll + ll v ll 2 By Cauchy-Schwarz = ( ll n ll + ll v ll ) 2 as required. Distance The distance between two vectors is the direct analogue of the distance between two points on the real number line or two points in the Cartesian plane. On the number line (Figure 1.30), the distance between the numbers a and b is given by l a - bl. (Tak­ ing the absolute value ensures that we do not need to know which of a or b is larger.) This distance is also equal to V ( a - b ) 2 , and its two-dimensional generalization is the familiar formula for the distance d between points (a 1 , a 2 ) and (b 1 , b 2 )-namely, d = V ( a 1 - b 1 ) 2 + ( a 2 - bi ) 2 a b 4 I I I + I I + I I -2 0 3 5 Figure 1. 3 0 d = la - bl = l-2 - 31 = In terms of vectors, if a = [ :: ] and b = as shown in Figure 1. 3 1. This is the basis for the next definition. [ :], then d is just the length of a - b, I I I : a 2 - b2 d I I I I I _ _ _ _ _ _ _ _ _ _ _ _f] a1 - b1 Figure 1. 3 1 d = V( a , - b,)2 + ( a2 - b )2 2 = I l a - b ll Definition The distance d(u, v) between vectors u and v in u;g n is defined by d ( U, v) = I u - v II 24 Chapter 1 Vectors Exa m p l e 1. 2 0 Find the di,tance between u [ 1] [ _ l nd v · Solution We rnmpute u - v [ J '" d ( u , v) = ll u - v ii = V( \/2) 2 + ( - 1 ) 2 + 1 2 = V4 = 2 Angles The dot product can also be used to calculate the angle between a pair of vectors. In IR 2 or IR 3 , the angle between the nonzero vectors u and v will refer to the angle (} 'D determined by these vectors that satisfies 0 :::::: (} :::::: 1 80° (see Figure 1.3 2). v v u u. /7 (} u v4 0 u Figure 1. 3 2 The angle between u and v v In Figure 1.33, consider the triangle with sides u, v, and u - v, where (} is the angle between u and v. Applying the law of cosines to this triangle yields ll u - v ll 2 ll u ll 2 + ll v ll 2 - 2 ll u ll ll v ll cos (} = Expanding the left-hand side and using ll v ll 2 = v · v several times, we obtain Figure 1. 3 3 ll u ll 2 - 2 ( u · v) + ll v ll 2 = ll u ll 2 + ll v ll 2 - 2 ll u ll ll v ll cos (} which, after simplification, leaves us with u · v = ll u ll ll v ll cos (}. From this we obtain the following formula for the cosine of the angle (} between nonzero vectors u and v. We state it as a definition. Definition For nonzero vectors u and v in !R n , u·v cos (} = u ll ll ll v ll Exa m p l e 1. 2 1 Compute the angle between the vectors u = [2, 1 , - 2] and v = [1, 1 , 1 ].  Section 1.2 Length and Angle: The Dot Product 25 Solution We calculate u·v = 2 · 1 + l · l + ( - 2 ) · 1 = 1 , JJ u JJ = V2 2 + 1 2 + (- 2 ) 2 = V9 = 3, and JJ v JJ = Vl 2 + 1 2 + 1 2 = v3. Therefore, cos e = 1 / 3 v3, so e = cos - ( 1 /3 v3) = 1.377 radians, or 78.9°....._+ 1 Exa m p l e 1. 2 2 Compute the angle between the diagonals on two adjacent faces of a cube. Solution The dimensions of the cube do not matter, so we will work with a cube with sides of length 1. Orient the cube relative to the coordinate axes in IR 3 , as shown in Figure 1.34, and take the two side diagonals to be the vectors [ 1 , 0, 1 ] and [O, 1 , 1 ]. Then angle e between these vectors satisfies cos e = l·O + O·l + l·l ----- \/2 v'2 2 from which it follows that the required angle is n /3 radians, or 60°. z [O, 1, l] y x Figure 1. 3 4 (Actually, we don't need to do any calculations at all to get this answer. If we draw a third side diagonal joining the vertices at ( 1 , 0, 1 ) and (O, 1 , 1 ) , we get an equilateral triangle, since all of the side diagonals are of equal length. The angle we want is one of the angles of this triangle and therefore measures 60°. Sometimes, a little insight can save a lot of calculation; in this case, it gives a nice check on our work!) R e m a rks As this discussion shows, we usually will have to settle for an approximation to the angle between two vectors. However, when the angle is one of the so-called special angles (0°, 30°, 45°, 60°, 90°, or an integer multiple of these), we should be able to recognize its cosine (Table 1. 1 ) and thus give the corresponding angle exactly. In all other cases, we will use a calculator or computer to approximate the desired angle by means of the inverse cosine function. Ta b l e 1. 1 cosines or Special Angles e 30° 45° 60° 90° cos e V4 v3 v'2 1 Vi Vo -= 1 -=O 2 2 2 v'2 2 2 2 26 Chapter 1 Vectors The derivation of the formula for the cosine of the angle between two vectors is valid only in IR 2 or IR 3 , since it depends on a geometric fact: the law of cosines. In IR n , for n > 3, the formula can be taken as a definition instead. This makes sense, since the Cauc y-Schwarz ! equality mplies that - 1 to 1, J USt as the cosme funct10n does. uI I u·v I ll ll v ll :::::: 1 , so I u u·v ll ll v ll ranges from Orthogonal Veclors The word orthogonal is derived The concept of perpendicularity is fundamental to geometry. Anyone studying from the Greek words orthos, mean­ geometry quickly realizes the importance and usefulness of right angles. We now gen­ eralize the idea of perpendicularity to vectors in !R n , where it is called orthogonality. In IR 2 or IR 3 , two nonzero vectors u and v are perpendicular if the angle (J between ing "upright;' and gonia, meaning "angle:' Hence, orthogonal literally them is a right angle-that is, if (J = 1T /2 radians, or 90°. Thus, I means "right-angled'.' The Latin u·v = cos 90° = 0, equivalent is rectangular. ll u ll v ll and it follows that u · v = 0. This motivates the following definition. DefiniliOD Two vectors u and v in !R n are orthogonal to each other if u ·v = 0. Since 0 · v = 0 for every vector v in !R n , the zero vector is orthogonal to every vector. Exa m p l e 1. 2 3 In IR 3 , u = [ l , 1 , - 2 ] and v = [3, 1 , 2] are orthogonal, since u · v 3+1-4 = 0. Using the notion of orthogonality, we get an easy proof of Pythagoras' Theorem, valid in !R n. Theorem 1. 6 Pythagoras' 1heoreITI For all vectors u and v in lR n , ll u + v ll 2 = ll u ll 2 + ll v ll 2 if and only if u and v are orthogonal. Proof From Example 1. 16, we have ll u + v ll 2 = ll u ll 2 + 2 ( u · v) + ll v ll 2 for all vectors u and v in !R n. It follows immediately that ll u + v ll 2 = ll u ll 2 + ll v ll 2 if and only if u · v = 0. See Figure 1.35. The concept of orthogonality is one of the most important and useful in linear algebra, and it often arises in surprising ways. Chapter 5 contains a detailed treatment of the topic, but we will encounter it many times before then. One problem in which it clearly plays a role is finding the distance from a point to a line, where "dropping a Figure 1. 3 5 perpendicular" is a familiar step.  Section 1.2 Length and Angle: The Dot Product 21 Proieclions We now consider the problem of finding the distance from a point to a line in the context of vectors. As you will see, this technique leads to an important concept: the projection of a vector onto another vector. As Figure 1.36 shows, the problem of finding the distance from a point B to a line € (in IR 2 or IR 3 ) reduces to the problem of finding the length of the perpendicular line segment PB or, equivalently, the length of the vector PB. If we choose a point "----> A on €, then, in the----" right-angled -----> triangle fl.APB, the other -----; two vectors are the leg AP and the hypotenuse AB. AP is called the projection of AB onto the line €. We will now look at this situation in terms of vectors. B B e A Figure 1. 3 6 The distance from a point to a line vu= l Puv.l u, u l u )u u v, _..,.,.-- u Consider two nonzero vectors and Let p be the vector obtained by dropping a perpendicular from the head of onto and let e be the angle between and as figure 1. 3 1 u. u·v shown in Figure 1.37. Then clearly p where = = v ( I / I is the unit vector in I II cos e , and = u l v1. The projection of v onto u the direction of Moreover, elementary trigonometry gives I p I we know that cos e II II Thus, after substitution, we obtain = l v l C1 uil · l vl l ) (ilf)u p = ( u·v )u = ( )u u·u This is the formula we want, and it is the basis of the following definition for vec­ tors in !R n. v u Definition u v If and are vectors in !R n u * 0, and then the projection of u·v onto is the vector proju(v) defined by proj0 ( v ) = (-- u·u) u An alternative way to derive this formula is described in Exercise 73. 28 Chapter 1 Vectors R e m a rks The term projection comes from the idea of projecting an image onto a wall (with a slide projector, for example). Imagine a beam oflight with rays parallel to each other and perpendicular to u shining down on v. The projection of v onto u is just the shadow cast, or projected, by v onto u. proju(v) u It may be helpful to think of proj0(v) as a function with variable v. Then the Figure 1. 3 8 variable v occurs only once on the right-hand side of the definition. Also, it is helpful to remember Figure 1.38, which reminds us that proj0(v) is a scalar multiple of the vector u (not v). ) Although in our derivation of the definition of proj0(v) we required v as well as u to be nonzero (why?), it is clear from the geometry that the projection of the zero vector onto u is 0. The definition is in agreement with this, since " Ou = 0. u·u 0 u= (. If the angle between u and v is obtuse, as in Figure 1.38, then proj0(v) will be in the opposite direction from u; that is, proj0(v) will be a negative scalar multiple of u. If u is a unit vector then proj0(v) = ( u · v) u. (Why?) Exa m p l e 1. 2 4 Find the projection of v onto u in each case. (a) v = [ - ] [ ] and u = (c) v = [ ] [ ] 3 and u = 1 / \/2 [ ] [ - ] Solution [ ] [ ] [] [ ] (a) We compute u · v = · = 1 and u · u = · = 5, so proj u ( v) = ( u·v u·u )" = 1 2 s 1 = 2 /5 1 /5 (b) Since e3 is a unit vector, (c) We see that ll u ll = V + + t = 1. Thus, proj0 ( v) = ( u · v) u = ( + 1 + ) [ ] = 3( 1 \/2) [ ] [ l 1 / \/2 1 / \/2 3(I : V2J  Section 1.2 Length and Angle: The Dot Product 29 J Exercises 1. 2.. In Exercises 1-6, find u · v. 26. Exercise 20 GAs 27. Exercise 2 1 GAs 28. Exercise 22 GAs 29. Exercise 23 30. Let A = ( - 3, 2), B = ( 1 , 0), and C = (4, 6). Prove that MBC is a right-angled triangle. 31. Let A = ( 1 , 1, - 1 ), B = ( - 3, 2, - 2), and C = (2, 2, - 4). Prove that ABC is a right-angled triangle. GAs 32. Find the angle between a diagonal of a cube and an ad­ 5. u = [ 1 , \/2, \/3, o ] , v = [ 4, - \/2, 0, - 5 ] jacent edge. 33. A cube has four diagonals. Show that no two of them GAS 6. U = [ 1. 1 2, - 3.25, 2.07, - 1.83 ] , are perpendicular. v = [ - 2.29, 1.72, 4.33, - 1.54 ] 34. A parallelogram has diagonals determined by the vectors [H In Exercises 7-12, find II u II for the given exercise, and give a unit vector in the direction of u. GAs 7. Exercise 1 10. Exercise 4 8. Exercise 2 1 1. Exercise 5 GAs 9. Exercise 3 12. Exercise 6 d, and d, [ - - Show that the parallelogram is a rhombus (all sides of In Exercises 13-16, find the distance d(u, v) between u and equal length) and determine the side length. v in the given exercise. 35. The rectangle ABCD has vertices at A = ( 1 , 2, 3), 13. Exercise 1 14. Exercise 2 B = (3, 6, - 2), and C = (O, 5, - 4). Determine the 15. Exercise 3 GAs 16. Exercise 4 coordinates of vertex D. 1 7. If u, v, and w are vectors in ll r, n 2: 2, and c is a 36. An airplane heading due east has a velocity of scalar, explain why the following expressions make 200 miles per hour. A wind is blowing from the north no sense: at 40 miles per hour. What is the resultant velocity of the airplane? (a) ll u · v ll (b) u · v + w ( (c) u · v · w ) (d) c. ( u + w ) 37. A boat heads north across a river at a rate of 4 miles per hour. If the current is flowing east at a rate of 3 miles per hour, find the resultant velocity of In Exercises 18-23, determine whether the angle between the boat. u and v is acute, obtuse, or a right angle. 38. Ann is driving a motorboat across a river that is 2 km wide. The boat has a speed of 20 km/h in still water, and the current in the river is flowing at 5 km/h. Ann heads out from one bank of the river for a dock directly across from her on the opposite bank. She drives the boat in a 20. u = [4, 3, - 1 ] , v = [ l , - 1 , l ] direction perpendicular to the current. GAS 2 1. U = [0.9, 2. 1 , 1.2] , V = [ - 4.5, 2.6, - 0.8] (a) How far downstream from the dock will 22. u = [ 1 , 2, 3, 4], v = [ - 3, 1 , 2, - 2 ] Ann land? (b) How long will it take Ann to cross the river? 23. u = [ 1 , 2, 3, 4], v = [5, 6, 7, 8] 39. Bert can swim at a rate of 2 miles per hour in still water. The current in a river is flowing at a rate of In Exercises 24-29, find the angle between u and v in the 1 mile per hour. If Bert wants to swim across the river given exercise. to a point directly opposite, at what angle to the bank 24. Exercise 1 8 25. Exercise 1 9 of the river must he swim?  30 Chapter 1 Vectors In Exercises 40-45, find the projection of v onto u. Draw a In Exercises 48 and 49, find all values of the scalar k for sketch in Exercises 40 and 41. which the two vectors are orthogonal. 48. u [ : Jv [ : l 49. u [ -J [ _fl 50. Describe all vectors v = [;] that are orthogonal to u = [ ]. ] ] [;] [ [ 3.01 1.34 51. Describe all vectors v = that are orthogonal [ ]. CAS 45. U = - 0.33 V = , 4.25 2.52 - 1.66 to u = Figure 1.39 suggests two ways in which vectors may be used to compute the area of a triangle. 52. Under what conditions are the following true for The area A of vectors u and v in IR 2 or IR 3 ? (a) ll u + v ii = ll u ll + ll v ll (b) ll u + v ii = ll u ll - ll v ll 53. Prove Theorem 1.2 (b). 54. Prove Theorem 1.2 (d). In Exercises 55-57, prove the stated property of distance between vectors. 55. d(u, v) = d(v, u) for all vectors u and v (a) 56. d(u, w) :s d(u, v) + d(v, w) for all vectors u, v, and w 57. d(u, v) = 0 if and only if u = v 58. Prove that u · c v = c( u · v) for all vectors u and v in !R n and all scalars c. 59. Prove that ll u - v ii 2': ll u ll - ll v ll for all vectors u and v in !R n. [Hint: Replace u by u - v in the Triangle Inequality.] u 60. Suppose we know that u · v = u · w. Does it follow that (b) v = w? If it does, give a proof that is valid in !R n ; Figure 1. 3 9 otherwise, give a counterexample (i.e., a specific set of vectors u, v, and w for which u · v = u · w but v -=F w). the triangle in part (a) is given by t ll u ll ll v - proj u ( v ) II , 61. Prove that (u + v) · (u - v) = ll u ll 2 - ll v ll 2 for all vec­ tors u and v in !R n. and part (b) suggests the trigonometric form of the area of a triangle: A = t I u I I v II sin e (We can use the 62. (a) Prove that ll u + v ll 2 + ll u - v ll 2 = 2 ll u ll 2 + 2 ll v ll 2 identity sin e = v 1 - cos 2 e to find sin e.) for all vectors u and v in !R n. (h) Draw a diagram showing u, v, u + v, and u - v in IR 2 and use (a) to deduce a result about In Exercises 46 and 47, compute the area of the triangle parallelograms. with the given vertices using both methods. 1 1 46. A = ( 1 , - 1 ), B = (2, 2), C = (4, O) 63. Prove that u · v = - ll u + v ll 2 - - ll u - v ll 2 for all 4 4 47. A = (3, - 1 , 4), B = (4, - 2 , 6), C = (5, 0, 2) vectors u and v in !R n.  Section 1.2 Length and Angle: The Dot Product 31 64. (a) Prove that ll u + v ii = ll u - v ii if and only if u and 72. Figure 1.40 shows that, in IR 2 or IR 3 , v are orthogonal. ll proju ( v ) I :::::: ll v ll · (b) Draw a diagram showing u, v, u + v, and u - v (a) Prove that this inequality is true in general. [Hint: in IR 2 and use (a) to deduce a result about Prove that proju(v) is orthogonal to v - proju(v) parallelograms. and use Pythagoras' Theorem.] 65. ( a) Prove that u + v and u - v are orthogonal in !R n if (b) Prove that the inequality ll proju ( v ) I :::::: ll v ll is and only if ll u ll = ll v ll. equivalent to the Cauchy-Schwarz Inequality. (b) Draw a diagram showing u, v, u + v, and u - v in IR 2 and use (a) to deduce a result about parallelograms. 66. If ll u ll = 2, ll v ll = v'3 , and u · v = 1, find ll u + v ii. 67. Show that there are no vectors u and v such that ll u ll = 1, ll v ll = 2, and u · v = 3. 68. ( a) Prove that if u is orthogonal to both v and w, then u is orthogonal to v + w. proju(v) u (b) Prove that if u is orthogonal to both v and w, then Figure 1. 4 0 u is orthogona