Algebra, /e Front Matter Preface Companies, 004 PREFACE Linear algebra has in recent years become an essential part of the mathemtical background required by mathematicians and mathematics teachers, engineers, computer scientists, physicists, economists, and statisticians, among others. This requirement re ects the importance and wide applications of the subject matter. This book is designed for use as a textbook for a formal course in linear algebra or as a supplement to all current standard texts. It aims to present an introduction to linear algebra which will be found helpful to all readers regardless of their elds of speci cation. More material has been included than can be covered in most rst courses. This has been done to make the book more exible, to provide a useful book of reference, and to stimulate further interest in the subject. Each chapter begins with clear statements of pertinent de nitions, principles and theorems together with illustrative and other descriptive material. This is followed by graded sets of solved and supplementary problems. The solved problems serve to illustrate and amplify the theory, and to provide the repetition of basic principles so vital to effective learning. Numerous proofs, especially those of all essential theorems, are included among the solved problems. The supplementary problems serve as a complete review of the material of each chapter. The rst three chapters treat vectors in Euclidean space, matrix algebra, and systems of linear equations. These chapters provide the motivation and basic computational tools for the abstract investigation of vector spaces and linear mappings which follow. After chapters on inner product spaces and orthogonality and on determinants, there is a detailed discussion of eigenvalues and eigenvectors giving conditions for representing a linear operator by a diagonal matrix. This naturally leads to the study of various canonical forms, speci cally, the triangular, Jordan, and rational canonical forms. Later chapters cover linear functions and the dual space V*, and bilinear, quadratic and Hermitian forms. The last chapter treats linear operators on inner product spaces. For completeness, there is an appendix on polynomials over a eld. The main changes in the third edition have been for pedagogical reasons rather than in content. Speci cally, the abstract notion of a linear map and its matrix representation appears before and motivates the study of eigenvalues and eigenvectors and the diagonalization of matrices (under similarity). There are also many additional solved and supplementary problems. Finally, we wish to thank the staff of the McGraw-Hill Schaum's Outline Series, especially Barbara Gilson, for their unfailing cooperation. SEYMOUR LIPSCHUTZ MARC LARS LIPSON
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 CHAPTER 1 Vectors in R n and C n, Spatial Vectors 1.1 INTRODUCTION There are two ways to motivate the notion of a vector: one is by means of lists of numbers and subscripts, and the other is by means of certain objects in physics. We discuss these two ways below. Here we assume the reader is familiar with the elementary properties of the eld of real numbers, denoted by R. On the other hand, we will review properties of the eld of complex numbers, denoted by C. In the context of vectors, the elements of our number elds are called scalars. Although we will restrict ourselves in this chapter to vectors whose elements come from R and then from C, many of our operations also apply to vectors whose entries come from some arbitrary eld K. Lists of Numbers Suppose the weights (in pounds) of eight students are listed as follows: 156; 15; 145; 14; 178; 145; 16; 19 One can denote all the values in the list using only one symbol, say w, but with different subscripts; that is w 1 ; w ; w ; w 4 ; w 5 ; w 6 ; w 7 ; w 8 Observe that each subscript denotes the position of the value in the list. For example, w 1 ˆ 156; the first number; w ˆ 15; the second number;... Such a list of values, w ˆ w 1 ; w ; w ;...; w 8 is called a linear array or vector. 1
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 VECTORS IN R n AND C n, SPATIAL VECTORS [CHAP. 1 Vectors in Physics Many physical quantities, such as temperature and speed, possess only ``magnitude''. These quantities can be represented by real numbers and are called scalars. On the other hand, there are also quantities, such as force and velocity, that possess both ``magnitude'' and ``direction''. These quantities, which can be represented by arrows having appropriate lengths and directions and emanating from some given reference point O, are called vectors. Now we assume the reader is familiar with the space R where all the points in space are represented by ordered triples of real numbers. Suppose the origin of the axes in R is chosen as the reference point O for the vectors discussed above. Then every vector is uniquely determined by the coordinates of its endpoint, and vice versa. There are two important operations, vector addition and scalar multiplication, that are associated with vectors in physics. The de nition of these operations and the relationship between these operations and the endpoints of the vectors are as follows. (i) Vector Addition: The resultant u v of two vectors u and v is obtained by the so-called parallelogram law; that is, u v is the diagonal of the parallelogram formed by u and v. Furthermore, if a; b; c and a 0 ; b 0 ; c 0 are the endpoints of the vectors u and v, then a a 0 ; b b 0 ; c c 0 is the endpoint of the vector u v. These properties are pictured in Fig. 1-1(a). x z 0 v ( a, b, c ) u + v u ( a) Vector Addition ( a+ a, b+ b, c+ c ) (, abc, ) y x z 0 u ku (, abc, ) ( b) Scalar Multiplication ( ka, kb, kc) y Fig. 1-1 (ii) Scalar Multiplication: The product ku of a vector u by a real number k is obtained by multiplying the magnitude of u by k and retaining the same direction if k > 0 or the opposite direction if k < 0. Also, if a; b; c is the endpoint of the vector u, then ka; kb; kc is the endpoint of the vector ku. These properties are pictured in Fig. 1-1(b). Mathematically, we identify the vector u with its a; b; c and write u ˆ a; b; c. Moreover, we call the ordered triple a; b; c of real numbers a point or vector depending upon its interpretation. We generalize this notion and call an n-tuple a 1 ; a ;...; a n of real numbers a vector. However, special notation may be used for the vectors in R called spatial vectors (Section 1.6). 1. VECTORS IN R n The set of all n-tuples of real numbers, denoted by R n, is called n-space. Aparticular n-tuple in R n,say u ˆ a 1 ; a ;...; a n is called a point or vector. The numbers a i are called the coordinates, components, entries,orelements of u. Moreover, when discussing the space R n, we use the term scalar for the elements of R.
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 CHAP. 1] VECTORS IN R n AND C n, SPATIAL VECTORS Two vectors, u and v, are equal, written u ˆ v, if they have the same number of components and if the corresponding components are equal. Although the vectors 1; ; and ; ; 1 contain the same three numbers, these vectors are not equal since corresponding entries are not equal. The vector 0; 0;...; 0 whose entries are all 0 is called the zero vector, and is usually denoted by 0. Example 1.1 (a) The following are vectors: ; 5 ; 7; 9 ; 0; 0; 0 ; ; 4; 5 The rst two vectors belong to R whereas the last two belong to R. The third is the zero vector in R. (b) Find x; y; z such that x y; x y; z 1 ˆ 4; ;. By de nition of equality of vectors, corresponding entries must be equal. Thus, x y ˆ 4; x y ˆ ; z 1 ˆ Solving the above system of equations yields x ˆ, y ˆ 1, z ˆ 4. Column Vectors Sometimes a vector in n-space R n is written vertically, rather than horizontally. Such a vector is called a column vector, and, in this context, the above horizontally written vectors are called row vectors. For example, the following are column vectors with ; ;, and components, respectively: 1 ; ; 4 1 4 5 5; 6 6 4 1:5 7 5 15 We also note that any operation de ned for row vectors is de ned analogously for column vectors. 1. VECTOR ADDITION AND SCALAR MULTIPLICATION Consider two vectors u and v in R n,say u ˆ a 1 ; a ;...; a n and v ˆ b 1 ; b ;...; b n Their sum, written u v, is the vector obtained by adding corresponding components from u and v. That is, u v ˆ a 1 b 1 ; a b ;...; a n b n The scalar product or, simply, product, of the vector u by a real number k, written ku, is the vector obtained by multiplying each component of u by k. That is, ku ˆ k a 1 ; a ;...; a n ˆ ka 1 ; ka ;...; ka n Observe that u v and ku are also vectors in R n. The sum of vectors with different numbers of components is not de ned. Negatives and subtraction are de ned in R n as follows: u ˆ 1 u and u v ˆ u v The vector u is called the negative of u, and u v is called the difference of u and v.
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 4 VECTORS IN R n AND C n, SPATIAL VECTORS [CHAP. 1 Now suppose we are given vectors u 1 ; u ;...; u m in R n and scalars k 1 ; k ;...; k m in R. We can multiply the vectors by the corresponding scalars and then add the resultant scalar products to form the vector v ˆ k 1 u 1 k u k u k m u m Such a vector v is called a linear combination of the vectors u 1 ; u ;...; u m. Example 1. (a) Let u ˆ ; 4; 5 and v ˆ 1; 6; 9. Then u v ˆ 1; 4 5 ; 5 9 ˆ ; 1; 4 7u ˆ 7 ; 7 4 ; 7 5 ˆ 14; 8; 5 v ˆ 1 1; 6; 9 ˆ 1; 6; 9 u 5v ˆ 6; 1; 15 5; 0; 45 ˆ 1; 4; 60 (b) The zero vector 0 ˆ 0; 0;...; 0 in R n is similar to the scalar 0 in that, for any vector u ˆ a 1 ; a ;...; a n. u 0 ˆ a 1 0; a 0;...; a n 0 ˆ a 1 ; a ;...; a n ˆu 4 9 5 (c) Let u ˆ 4 5 and v ˆ 4 1 5. Then u v ˆ 4 6 5 4 5 ˆ 4 9 5. 4 8 6 Basic properties of vectors under the operations of vector addition and scalar multiplication are described in the following theorem. Theorem 1.1: For any vectors u; v; w in R n and any scalars k; k 0 in R, (i) u v w ˆ u v w, (v) k u v ˆvu kv, (ii) u 0 ˆ u; (vi) k k 0 u ˆ ku k 0 u, (iii) u u ˆ0; (vii) kk 0 u ˆ k k 0 u, (iv) u v ˆ v u, (viii) 1u ˆ u. We postpone the proof of Theorem 1.1 until Chapter, where it appears in the context of matrices (Problem.). Suppose u and v are vectors in R n for which u ˆ kv for some nonzero scalar k in R. Then u is called a multiple of v. Also, u is said to be the same or opposite direction as v according as k > 0ork < 0. 1.4 DOT (INNER) PRODUCT Consider arbitrary vectors u and v in R n ;say, u ˆ a 1 ; a ;...; a n and v ˆ b 1 ; b ;...; b n The dot product or inner product or scalar product of u and v is denoted and de ned by u v ˆ a 1 b 1 a b a n b n That is, u v is obtained by multiplying corresponding components and adding the resulting products. The vectors u and v are said to be orthogonal (or perpendicular) if their dot product is zero, that is, if u v ˆ 0.
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 CHAP. 1] VECTORS IN R n AND C n, SPATIAL VECTORS 5 Example 1. (a) Let u ˆ 1; ;, v ˆ 4; 5; 1, w ˆ ; 7; 4. Then: u v ˆ 1 4 5 1 ˆ4 10 ˆ 9 u w ˆ 14 1 ˆ 0; v w ˆ 8 5 4 ˆ 9 Thus u and w are orthogonal. (b) Let u ˆ 4 5 and v ˆ 4 1 5. Then u v ˆ 6 8 ˆ 11. 4 (c) Suppose u ˆ 1; ; ; 4 and v ˆ 6; k; 8;. Find k so that u and v are orthogonal. First obtain u v ˆ 6 k 4 8 ˆ 10 k. Then set u v ˆ 0 and solve for k: 10 k ˆ 0 or k ˆ 10 or k ˆ 5 Basic properties of the dot product in R n (proved in Problem 1.1) follow. Theorem 1.: For any vectors u; v; w in R n and any scalar k in R: (i) u v w ˆ u w v w; (iii) u v ˆ v u, (ii) ku v ˆ k u v, (iv) u u 0; and u u ˆ 0 iff u ˆ 0. Note that (ii) says that we can ``take k out'' from the rst position in an inner product. By (iii) and (ii), u kv ˆ kv u ˆ k v u ˆk u v That is, we can also ``take k out'' from the second position in an inner product. The space R n with the above operations of vector addition, scalar multiplication, and dot product is usually called Euclidean n-space. Norm (Length) of a Vector The norm or length of a vector u in R n, denoted by kuk, is de ned to be the nonnegative square root of u u. In particular, if u ˆ a 1 ; a ;...; a n, then p kuk ˆ q u u ˆ a 1 a a n That is, kuk is the square root of the sum of the squares of the components of u. Thus kuk 0, and kuk ˆ0 if and only if u ˆ 0. Avector u is called a unit vector if kuk ˆ1 or, equivalently, if u u ˆ 1. For any nonzero vector v in R n, the vector ^v ˆ 1 kvk v ˆ v kvk is the unique unit vector in the same direction as v. The process of nding ^v from v is called normalizing v. Example 1.4 (a) Suppose u ˆ 1; ; 4; 5;. To nd kuk, we can rst nd kuk ˆ u u by squaring each component of u and adding, as follows: p Then kuk ˆ 55. kuk ˆ 1 4 5 ˆ 1 4 16 5 9 ˆ 55
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 6 VECTORS IN R n AND C n, SPATIAL VECTORS [CHAP. 1 (b) Let v ˆ 1; ; 4; and w ˆ 1 ; 1 6 ; 5 6 ; 1 6. Then p kvk ˆ 1 9 16 4 ˆ r p 9 0 and kwk ˆ 6 1 6 5 6 1 r 6 p ˆ ˆ 1 ˆ 1 6 6 Thus w is a unit vector but v is not a unit vector. However, we can normalize v as follows: ^v ˆ v kvk ˆ 1 p ; 4 p ; p ; p 0 0 0 0 This is the unique unit vector in the same direction as v. The following formula (proved in Problem 1.14) is known as the Schwarz inequality or Cauchy± Schwarz inequality. It is used in many branches of mathematics. Theorem 1. (Schwarz): For any vectors u; v in R n, ju vj kukkvk. Using the above inequality, we also prove (Problem 1.15) the following result known as the ``triangle inequality'' or Minkowski's inequality. Theorem 1.4 (Minkowski): For any vectors u; v in R n, ku vk kuk kvk. Distance, Angles, Projections The distance between vectors u ˆ a 1 ; a ;...; a n and v ˆ b 1 ; b ;...; b n in R n is denoted and de ned by q d u; v ˆku vk ˆ a 1 b 1 a b a 1 b 1 One can show that this de nition agrees with the usual notion of distance in the Euclidean plane R or space R. The angle y between nonzero vectors u; v in R n is de ned by cos y ˆ u v kukkvk This de nition is well de ned, since, by the Schwarz inequality (Theorem 1.), 1 u v kukkvk 1 Note that if u v ˆ 0, then y ˆ 90 (or y ˆ p=). This then agrees with our previous de nition of orthogonality. The projection of a vector u onto a nonzero vector v is the vector denoted and de ned by proj u; v ˆu v kvk v We show below that this agrees with the usual notion of vector projection in physics. Example 1.5 (a) Suppose u ˆ 1; ; and v ˆ ; 4; 5. Then q d u; v ˆ 1 4 5 p ˆ p 1 6 4 ˆ 41 To nd cos y, where y is the angle between u and v, we rst nd u v ˆ 8 15 ˆ 9; kuk ˆ 1 4 9 ˆ 14; kvk ˆ 4 16 5 ˆ 45
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 CHAP. 1] VECTORS IN R n AND C n, SPATIAL VECTORS 7 Then Also, cos y ˆ u v kukkvk ˆ 9 p p 14 45 proj u; v ˆu v kvk v ˆ 9 ; 4; 5 ˆ1 ; 4; 5 ˆ 45 5 5 ; 4 5 ; 1 (b) Consider the vectors u and v in Fig. 1-(a) (with respective endpoints A and B). The (perpendicular) projection of u onto v is the vector u* with magnitude ku*k ˆkukcos y ˆkuk u v kukvk ˆ u v kvk To obtain u*, we multiply its magnitude by the unit vector in the direction of v, obtaining u* ˆku*k v kvk ˆ u v v kvk kvk ˆ u v kvk v This is the same as the above de nition of proj u; v. A z P( b1 a1, b a b a) u B( b1, b, b) 0 θ u* C v B 0 u A( a1, a, a) y Projection u* of u onto v () a x u=b A () b Fig. 1-1.5 LOCATED VECTORS, HYPERPLANES, LINES, CURVES IN R n This section distinguishes between an n-tuple P a i P a 1 ; a ;...; a n viewed as a point in R n and an n-tuple u ˆ c 1 ; c ;...; c n Š viewed as a vector (arrow) from the origin O to the point C c 1 ; c ;...; c n. Located Vectors Any pair of points A a i and B b i in R n de nes the located vector or directed line segment from A to ƒ! ƒ! B, written AB. We identify AB with the vector u ˆ B A ˆ b 1 a 1 ; b a ;...; b n a n Š
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 8 VECTORS IN R n AND C n, SPATIAL VECTORS [CHAP. 1 ƒ! since AB and u have the same magnitude and direction. This is pictured in Fig. 1-(b) for the points A a 1 ; a ; a and B b 1 ; b ; b in R and the vector u ˆ B A which has the endpoint P b 1 a 1, b a, b a. Hyperplanes A hyperplane H in R n is the set of points x 1 ; x ;...; x n which satisfy a linear equation a 1 x 1 a x... a n x n ˆ b where the vector u ˆ a 1 ; a ;...; a n Š of coef cients is not zero.thus a hyperplane H in R is a line and a hyperplane H in R is a plane. We show below, as pictured in Fig. 1-(a) for R, that u is orthogonal to any directed line segment ƒ! PQ, where P p i and Q q i are points in H: [For this reason, we say that u is normal to H and that H is normal to u:] Since P p i and Q q i belong to H; they satisfy the above hyperplane equation, that is, a 1 p 1 a p... a n p n ˆ b and a 1 q 1 a q... a n q n ˆ b Let v ˆ ƒ! PQ ˆ Q P ˆ q 1 p 1 ; q p ;...; q n p n Š Then u v ˆ a 1 q 1 p 1 a q p... a n q n p n ˆ a 1 q 1 a q... a n q n a 1 p 1 a p... a n p n ˆb b ˆ 0 Thus v ˆ ƒ! PQ is orthogonal to u; as claimed. Fig. 1- Lines in R n The line L in R n passing through the point P b 1 ; b ;...; b n and in the direction of a nonzero vector u ˆ a 1 ; a ;...; a n Š consists of the points X x 1 ; x ;...; x n that satisfy 8 x 1 ˆ a 1 t b >< 1 x X ˆ P tu or ˆ a t b or L t ˆ a :::::::::::::::::::: i t b i >: x n ˆ a n t b n where the parameter t takes on all real values. Such a line L in R is pictured in Fig. 1-(b).
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 CHAP. 1] VECTORS IN R n AND C n, SPATIAL VECTORS 9 Example 1.6 (a) Let H be the plane in R corresponding to the linear equation x 5y 7z ˆ 4. Observe that P 1; 1; 1 and Q 5; 4; are solutions of the equation. Thus P and Q and the directed line segment v ˆ ƒ! PQ ˆ Q P ˆ 5 1; 4 1; 1Š ˆ 4; ; 1Š lie on the plane H. The vector u ˆ ; 5; 7Š is normal to H, and, as expected, u v ˆ ; 5; 7Š 4; ; 1Š ˆ8 15 7 ˆ 0 That is, u is orthogonal to v. (b) Find an equation of the hyperplane H in R 4 that passes through the point P 1; ; 4; and is normal to the vector u ˆ 4; ; 5; 6Š. The coef cients of the unknowns of an equation of H are the components of the normal vector u; hence the equation of H must be of the form Substituting P into this equation, we obtain 4x 1 x 5x 6x 4 ˆ k 4 1 5 4 6 ˆk or 4 6 0 1 ˆ k or k ˆ 10 Thus 4x 1 x 5x 6x 4 ˆ 10 is the equation of H. (c) Find the parametric representation of the line L in R 4 passing through the point P 1; ; ; 4 and in the direction of u ˆ 5; 6; 7; 8Š. Also, nd the point Q on L when t ˆ 1. Substitution in the above equation for L yields the following parametric representation: or, equivalently, x 1 ˆ 5t 1; x ˆ 6t ; x ˆ 7t ; x 4 ˆ 8t 4 L t ˆ 5t 1; 6t ; 7t ; 8t 4 Note that t ˆ 0 yields the point P on L. Substitution of t ˆ 1 yields the point Q 6; 8; 4; 4 on L. Curves in R n Let D be an interval ( nite or in nite) on the real line R. Acontinuous function F: D! R n is a curve in R n. Thus, to each point t D there is assigned the following point in R n : F t ˆ F 1 t ; F t ;...; F n t Š Moreover, the derivative (if it exists) of F t yields the vector V t ˆdF t ˆ df 1 t ; df t ;...; df n t dt dt dt dt which is tangent to the curve. Normalizing V t yields T t ˆ V t kv t k Thus T t is the unit tangent vector to the curve. (Unit vectors with geometrical signi cance are often presented in bold type.)
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 10 VECTORS IN R n AND C n, SPATIAL VECTORS [CHAP. 1 Example 1.7. yields Consider the curve F t ˆ sin t; cos t; tš in R. Taking the derivative of F t [or each component of F t ] V t ˆ cos t; sin t; 1Š which is a vector tangent to the curve. We normalize V t. First we obtain kv t k ˆ cos t sin t 1 ˆ 1 1 ˆ Then the unit tangent vection T t to the curve follows: T t ˆ V t kv t k ˆ cos p t ; p sin t ; 1 p 1.6 VECTORS IN R (SPATIAL VECTORS), ijk NOTATION Vectors in R, called spatial vectors, appear in many applications, especially in physics. In fact, a special notation is frequently used for such vectors as follows: i ˆ 1; 0; 0Š denotes the unit vector in the x direction: j ˆ 0; 1; 0Š denotes the unit vector in the y direction: k ˆ 0; 0; 1Š denotes the unit vector in the z direction: Then any vector u ˆ a; b; cš in R can be expressed uniquely in the form u ˆ a; b; cš ˆai bj cj Since the vectors i; j; k are unit vectors and are mutually orthogonal, we obtain the following dot products: i i ˆ 1; j j ˆ 1; k k ˆ 1 and i j ˆ 0; i k ˆ 0; j k ˆ 0 Furthermore, the vector operations discussed above may be expressed in the ijk notation as follows. Suppose Then u ˆ a 1 i a j a k and v ˆ b 1 i b j b k u v ˆ a 1 b 1 i a b j a b k and cu ˆ ca 1 i ca j ca k where c is a scalar. Also, u v ˆ a 1 b 1 a b a b and p kuk ˆ u u ˆ a 1 a a Example 1.8 Suppose u ˆ i 5j k and v ˆ 4i 8j 7k. (a) To nd u v, add corresponding components, obtaining u v ˆ 7i j 5k (b) To nd u v, rst multiply by the scalars and then add: u v ˆ 9i 1j 6k 8i 16j 14k ˆi 9j 0k (c) To nd u v, multiply corresponding components and then add: u v ˆ 1 40 14 ˆ 4 (d) To nd kuk, take the square root of the sum of the squares of the components: p kuk ˆ p 9 5 4 ˆ 8
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 CHAP. 1] VECTORS IN R n AND C n, SPATIAL VECTORS 11 Cross Product There is a special operation for vectors u and v in R that is not de ned in R n for n 6ˆ. This operation is called the cross product and is denoted by u v. One way to easily remember the formula for u v is to use the determinant (of order two) and its negative, which are denoted and de ned as follows: a b a b c d ˆ ad bc and c d ˆ bc ad Here a and d are called the diagonal elements and b and c are the nondiagonal elements. Thus the determinant is the product ad of the diagonal elements minus the product bc of the nondiagonal elements, but vice versa for the negative of the determinant. Now suppose u ˆ a 1 i a j a k and v ˆ b 1 i b j b k. Then u v ˆ a b a b i a b 1 a 1 b j a 1 b a b 1 k ˆ a1 a a b 1 b b i a 1 a a b 1 b b j a 1 a a b 1 b b i That is, the three components of u v are obtained from the array a 1 a a b 1 b b (which contain the components of u above the component of v) as follows: (1) Cover the rst column and take the determinant. () Cover the second column and take the negative of the determinant. () Cover the third column and take the determinant. Note that u v is a vector; hence u v is also called the vector product or outer product of u and v. Example 1.8. Find u v where: (a) u ˆ 4i j 6k, v ˆ i 5j k, (b) u ˆ ; 1; 5Š, v ˆ ; 7; 6Š. 4 6 (a) Use to get u v ˆ 9 0 i 1 1 j 0 6 k ˆ 9i 4j 14k 5 1 5 (b) Use to get u v ˆ 6 5; 15 1; 14 Š ˆ 41; ; 17Š 7 6 Remark: The cross products of the vectors i; j; k are as follows: i j ˆ k; j k ˆ i; k i ˆ j j i ˆ k; k j ˆ i; i k ˆ j Thus, if we view the triple i; j; k as a cyclic permutation, where i follows k and hence k precedes i, then the product of two of them in the given direction is the third one, but the product of two of them in the opposite direction is the negative of the third one. Two important properties of the cross product are contained in the following theorem. Theorem 1.5: Let u; v; w be vectors in R. (a) The vector u v is orthogonal to both u and v. (b) The absolute value of the ``triple product'' u v w represents the volume of the parallelopiped formed by the vectors u; v, w. [See Fig. 1-4(a).]
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 1 VECTORS IN R n AND C n, SPATIAL VECTORS [CHAP. 1 Fig. 1-4 We note that the vectors u; v, u v form a right-handed system, and that the following formula gives the magnitude of u v: where y is the angle between u and v. ku vk ˆkukkvk sin y 1.7 COMPLEX NUMBERS The set of complex numbers is denoted by C. Formally, a complex number is an ordered pair a; b of real numbers where equality, addition and multiplication are de ned as follows: a; b ˆ c; d if and only if a ˆ c and b ˆ d a; b c; d ˆ a c; b d a; b c; d ˆ ac bd; ad bc We identify the real number a with the complex number a; 0 ; that is, a $ a; 0 This is possible since the operations of addition and multiplication of real numbers are preserved under the correspondence; that is, a; 0 b; 0 ˆ a b; 0 and a; 0 b; 0 ˆ ab; 0 Thus we view R as a subset of C, and replace a; 0 by a whenever convenient and possible. We note that the set C of complex numbers with the above operations of addition and multiplication is a eld of numbers, like the set R of real numbers and the set Q of rational numbers. The complex number 0; 1 is denoted by i. It has the important property that p i ˆ ii ˆ 0; 1 0; 1 ˆ 1; 0 ˆ 1 or i ˆ 1 Accordingly, any complex number z ˆ a; b can be written in the form z ˆ a; b ˆ a; 0 0; b ˆ a; 0 b; 0 0; 1 ˆa bi
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 CHAP. 1] VECTORS IN R n AND C n, SPATIAL VECTORS 1 The above notation z ˆ a bi, where a Re z and b Im z are called, respectively, the real and imaginary parts of z, is more convenient than a; b. In fact, the sum and product of complex number z ˆ a bi and w ˆ c di can be derived by simply using the commutative and distributive laws and i ˆ 1: z w ˆ a bi c di ˆa c bi di ˆ a b c d i zw ˆ a bi c di ˆac bci adi bdi ˆ ac bd bc ad i We also de ne the negative of z and subtraction in C by Warning: Section 1.6. z ˆ 1z and w z ˆ w z p The letter i representing 1 has no relationship whatsoever to the vector i ˆ 1; 0; 0Š in Complex Conjugate, Absolute Value Consider a complex number z ˆ a bi. The conjugate of z is denoted and de ned by z ˆ a bi ˆ a bi Then zz ˆ a bi a bi ˆa b i ˆ a b. Note that z is real if and only if z ˆ z. The absolute value of z, denoted by jzj, is de ned to be the nonnegative square root of zz. Namely, p jzj ˆ p zz ˆ a b Note that jzj is equal to the norm of the vector a; b in R. Suppose z 6ˆ 0. Then the inverse z 1 of z and division in C of w by z are given, respectively, by z 1 ˆ z zz ˆ a a b b a b i and w z wz zz ˆ wz 1 Example 1.9. Suppose z ˆ i and w ˆ 5 i. Then z w ˆ i 5 i ˆ 5 i i ˆ 7 i zw ˆ i 5 i ˆ10 15i 4i 6i ˆ 16 11i z ˆ i ˆ i and w ˆ 5 i ˆ 5 i w 5 i 5 i i 4 19i ˆ ˆ ˆ ˆ 4 19 z i i i 1 1 1 i p jzj ˆ p 4 9 ˆ p 1 and jwj ˆ p 5 4 ˆ 9 Complex Plane Recall that the real numbers R can be represented by points on a line. Analogously, the complex numbers C can be represented by points in the plane. Speci cally, we let the point a; b in the plane represent the complex number a bi as shown in Fig. 1-4(b). In such a case, jzj is the distance from the origin O to the point z. The plane with this representation is called the complex plane, just like the line representing R is called the real line.
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 14 VECTORS IN R n AND C n, SPATIAL VECTORS [CHAP. 1 1.8 VECTORS IN C n The set of all n-tuples of complex numbers, denoted by C n, is called complex n-space. Just as in the real case, the elements of C n are called points or vectors, the elements of C are called scalars, and vector addition in C n and scalar multiplication on C n are given by z 1 ; z ;...; z n Š w 1 ; w ;...; w n Šˆ z 1 w 1 ; z w ;...; z n w n Š z z 1 ; z ;...; z n Šˆ zz 1 ; zz ;...; zz n Š where the z i, w i, and z belong to C. Example 1.10. Consider vectors u ˆ i; 4 i; Š and v ˆ i; 5i; 4 6iŠ in C. Then u v ˆ i; 4 i; Š i; 5i; 4 6iŠ ˆ 5 i; 4 4i; 7 6iŠ 5 i u ˆ 5 i i ; 5 i 4 i ; 5 i Š ˆ 16 11i; 18 1i; 15 6iŠ Dot (Inner) Product in C n Consider vectors u ˆ z 1 ; z ;...; z n Š and v ˆ w 1 ; w ;...; w n Š in C n. The dot or inner product of u and v is denoted and de ned by u v ˆ z 1 w 1 z w z n w n This de nition reduces to the real case since w i ˆ w i when w i is real. The norm of u is de ned by p kuk ˆ p u u ˆ q z 1 z 1 z z z n z n ˆ jz 1 j jz j jv n j We emphasize that u u and so kuk are real and positive when u 6ˆ 0 and 0 when u ˆ 0. Example 1.10. Consider vectors u ˆ i; 4 i; 5iŠ and v ˆ 4i; 5i; 4 iš in C. Then u v ˆ i 4i 4 i 5i 5i 4 i ˆ i 4i 4 i 5i 5i 4 i ˆ 6 1i 5 0i 6i ˆ 9 19i u u ˆj ij j4 ij j 5ij ˆ 4 9 16 1 9 5 ˆ 64 p kuk ˆ 64 ˆ 8 The space C n with the above operations of vector addition, scalar multiplication, and dot product, is called complex Euclidean n-space. Theorem 1. for R n also holds for C n if we replace u v ˆ v u by u v ˆ u v On the other hand, the Schwarz inequality (Theorem 1.) and Minkowski's inequality (Theorem 1.4) are true for C n with no changes. Solved Problems VECTORS IN R n 1.1. Determine which of the following vectors are equal: u 1 ˆ 1; ; ; u ˆ ; ; 1 ; u ˆ 1; ; ; u 4 ˆ ; ; 1 Vectors are equal only when corresponding entries are equal; hence only u ˆ u 4.
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 CHAP. 1] VECTORS IN R n AND C n, SPATIAL VECTORS 15 1.. Let u ˆ ; 7; 1, v ˆ ; 0; 4, w ˆ 0; 5; 8. Find: (a) u 4v, (b) u v 5w. First perform the scalar multiplication and then the vector addition. (a) u 4v ˆ ; 7; 1 4 ; 0; 4 ˆ 6; 1; 1; 0; 16 ˆ 18; 1; 1 (b) u v 5w ˆ 4; 14; 9; 0; 1 0; 5; 40 ˆ 5; 9; 54 5 1 1.. Let u ˆ 4 5; v ˆ 4 5 5; w ˆ 4 1 5. Find: 4 (a) 5u v, (b) u 4v w. First perform the scalar multiplication and then the vector additioin: 5 1 5 7 (a) 5u v ˆ 54 5 4 5 5 ˆ 4 15 5 4 10 5 ˆ 4 5 5 4 0 4 4 10 4 9 (b) u 4v w ˆ 4 6 5 4 0 5 4 5 ˆ 4 17 5 8 8 6 1.4. Find x and y, where: (a) x; ˆ ; x y, (b) 4; y ˆx ;. (a) Since the vectors are equal, set the corresponding entries equal to each other, yielding (b) x ˆ ; ˆ x y Solve the linear equations, obtaining x ˆ ; y ˆ 1: First multiply by the scalar x to obtain 4; y ˆ x; x. Then set corresponding entries equal to each other to obtain 4 ˆ x; y ˆ x Solve the equations to yield x ˆ, y ˆ 6. 1.5. Write the vector v ˆ 1; ; 5 as a linear combination of the vectors u 1 ˆ 1; 1; 1, u ˆ 1; ;, u ˆ ; 1; 1. We want to express v in the form v ˆ xu 1 yu zu with x; y; z as yet unknown. First we have 1 1 1 x y z 4 5 ˆ x4 1 5 y4 5 z4 1 5 ˆ 4 x y z 5 5 1 1 x y z (It is more convenient to write vectors as columns than as rows when forming linear combinations.) Set corresponding entries equal to each other to obtain x y z ˆ 1 x y z ˆ x y z ˆ 5 or x y z ˆ 1 y z ˆ y z ˆ 4 or x y z ˆ 1 y z ˆ 5z ˆ 10 This unique solution of the triangular system is x ˆ 6, y ˆ, z ˆ. Thus v ˆ 6u 1 u u.
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 16 VECTORS IN R n AND C n, SPATIAL VECTORS [CHAP. 1 1.6. Write v ˆ ; 5; as a linear combination of u 1 ˆ 1; ;, u ˆ ; 4; 1, u ˆ 1; 5; 7. Find the equivalent system of linear equations and then solve. First, 1 1 x y z 4 5 5 ˆ x4 5 y4 4 5 z4 5 5 ˆ 4 x 4y 5z 5 1 7 x y 7z Set the corresponding entries equal to each other to obtain x y z ˆ x 4y 5z ˆ 5 x y 7z ˆ or x y z ˆ y z ˆ 1 5y 5z ˆ 1 or x y z ˆ y z ˆ 1 0 ˆ The third equation, 0x 0y 0z ˆ, indicates that the system has no solution. Thus v cannot be written as a linear combination of the vectors u 1, u, u. DOT (INNER) PRODUCT, ORTHOGONALITY, NORM IN R n 1.7. Find u v where: (a) u ˆ ; 5; 6 and v ˆ 8; ; (b) u ˆ 4; ; ; 5; 1 and v ˆ ; 6; 1; 4; 8. Multiply the corresponding components and add: (a) u v ˆ 8 5 6 ˆ16 10 18 ˆ 1 (b) u v ˆ 8 1 0 8 ˆ 5 1.8. Let u ˆ 5; 4; 1, v ˆ ; 4; 1, w ˆ 1; ;. Which pair of vectors, if any, are perpendicular (orthogonal)? Find the dot product of each pair of vectors: u v ˆ 15 16 1 ˆ 0; v w ˆ 8 ˆ 14; u w ˆ 5 8 ˆ 0 Thus u and v are orthogonal, u and w are orthogonal, but v and w are not. 1.9. Find k so that u and v are orthogonal, where: (a) u ˆ 1; k; and v ˆ ; 5; 4 (b) u ˆ ; k; 4; 1; 5 and v ˆ 6; 1; ; 7; k. Compute u v, set u v equal to 0, and then solve for k: (a) u v ˆ 1 k 5 4 ˆ 5k 10. Then 5k 10 ˆ 0, or k ˆ. (b) u v ˆ 1 k 1 7 10k ˆ 7k 7. Then 7k 7 ˆ 0, or k ˆ 1. 1.10. Find kuk, where: (a) u ˆ ; 1; 4, (b) u ˆ ; ; 8; 7. p First nd kuk ˆ u u by squaring the entries and adding. Then kuk ˆ kuk. (a) kuk ˆ 1 4 p ˆ 9 144 16 ˆ 169. Then kuk ˆ 169 ˆ 1 p (b) kuk ˆ 4 9 64 49 ˆ 16. Then kuk ˆ 16.
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 CHAP. 1] VECTORS IN R n AND C n, SPATIAL VECTORS 17 1.11. Recall that normalizing a nonzero vector v means nding the unique unit vector ^v in the same direction as v, where ^v ˆ 1 kvk v Normalize: (a) u ˆ ; 4, (b) v ˆ 4; ; ; 8, (c) w ˆ 1,, 1 4 ). p (a) First nd kuk ˆ p 9 16 ˆ 5 ˆ 5. Then divide each entry of u by 5, obtaining ^u ˆ 5, 4 p 5 ). (b) Here kvk ˆ p 16 4 9 64 ˆ 9. Then (c) 4 ^v ˆ p ; p ; p ; p 9 9 9 Note that w and any positive multiple of w will have the same normalized form. Hence rst multiply w by 1 to ``clear fractions'', that is, rst nd w 0 ˆ 1w ˆ 6; 8;. Then p kw 0 kˆ 6 64 9 ˆ p 109 and ^w ˆ bw 0 6 8 ˆ p ; p ; p 109 109 109 8 9 1.1. Let u ˆ 1; ; 4 and v ˆ ; 4; 7. Find: (a) cos y, where y is the angle between u and v; (b) proj u; v, the projection of u onto v; (c) d u; v, the distance between u and v. First nd u v ˆ 1 8 ˆ 19, kuk ˆ 1 9 16 ˆ 6, kvk ˆ 9 16 49 ˆ 74. Then (a) cos y ˆ u v kukkvk ˆ 19 p p, 6 74 (b) proj u; v ˆu v kvk v ˆ 19 57 ; 4; 7 ˆ 74 74 ; 76 74 ; 1 ˆ 57 74 74 ; 8 7 ; 1 74 p (c) d u; v ˆku vkˆ k ; 7 k ˆ p 4 49 9 ˆ 6 : 1.1. Prove Theorem 1.: For any u; v; w in R n and k in R: (i) u v w ˆ u w v w, (ii) ku v ˆ k u v, (iii) u v ˆ v u, (iv) u u 0, and u u ˆ 0 iff u ˆ 0. Let u ˆ u 1 ; u ;...; u n, v ˆ v 1 ; v ;...; v n, w ˆ w 1 ; w ;...; w n. (i) Since u v ˆ u 1 v 1 ; u v ;...; u n v n, u v w ˆ u 1 v 1 w 1 u v w u n v n w n (ii) Since ku ˆ ku 1 ; ku ;...; ku n, ˆ u 1 w 1 v 1 w 1 u w u n w n v n w n ˆ u 1 w 1 u w u n w n v 1 w 1 v w v n w n ˆ u w v w ku v ˆ ku 1 v 1 ku v ku n v n ˆ k u 1 v 1 u v u n v n ˆk u v (iii) u v ˆ u 1 v 1 u v u n v n ˆ v 1 u 1 v u v n u n ˆ v u (iv) Since u i is nonnegative for each i, and since the sum of nonnegative real numbers is nonnegative, u u ˆ u 1 u u n 0 Furthermore, u u ˆ 0 iff u i ˆ 0 for each i, that is, iff u ˆ 0.
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 18 VECTORS IN R n AND C n, SPATIAL VECTORS [CHAP. 1 1.14. Prove Theorem 1. (Schwarz): ju vj kukkvk. For any real number t, and using Theorem 1., we have 0 tu v tu v ˆt u u t u v v v ˆkuk t u v t kvk Let a ˆkuk, b ˆ u v, c ˆkvk. Then, for every value of t, at bt c 0. This means that the quadratic polynomial cannot have two real roots. This implies that the discriminant D ˆ b 4ac 0or, equivalently, b 4ac. Thus Dividing by 4 gives us our result. 4 u v 4kuk kvk 1.15. Prove Theorem 1.4 (Minkowski): ku vk kuk kvk. By the Schwarz inequality and other properties of the dot product, ku vk ˆ u v u v ˆ uu uv vv kuk kukkvk kvk ˆ kuk kvk Taking the square root of both sides yields the desired inequality. POINTS, LINES, HYPERPLANES IN R n Here we distinguish between an n-tuple P a 1 ; a ;...; a n viewed as a point in R n and an n-tuple u ˆ c 1 ; c ;...; c n Š viewed as a vector (arrow) from the origin O to the point C c 1 ; c ;...; c n. 1.16. Find the vector u identi ed with the directed line segment PQ for the points: (a) P 1; ; 4 and Q 6; 1; 5 in R, (b) P ; ; 6; 5 and Q 7; 1; 4; 8 in R 4. (a) u ˆ PQ ˆ Q P ˆ 6 1; 1 ; 5 4Š ˆ 5; ; 9Š (b) u ˆ PQ ˆ Q P ˆ 7 ; 1 ; 4 6; 8 5Š ˆ 5; ; 10; 1Š 1.17. Find an equation of the hyperplane H in R 4 that passes through P ; 4; 1; and is normal to u ˆ ; 5; 6; Š. The coef cients of the unknowns of an equation of H are the components of the normal vector u. Thus an equation of H is of the form x 1 5x 6x x 4 ˆ k. Substitute P into this equation to obtain k ˆ 6. Thus an equation of H is x 1 5x 6x x 4 ˆ 6. 1.18. Find an equation of the plane H in R that contains P 1; ; 4 and is parallel to the plane H 0 determined by the equation x 6y 5z ˆ. The planes H and H 0 are parallel if and only if their normal directions are parallel or antiparallel (opposite direction). Hence an equation of H is of the form x 6y 5z ˆ k. Substitute P into this equation to obtain k ˆ 1. Then an equation of H is x 6y 5z ˆ 1. 1.19. Find a parametric representation of the line L in R 4 passing through P 4; ; ; 1 in the direction of u ˆ ; 5; 7; 8Š. Here L consists of the points X x i that satisfy X ˆ P tu or x i ˆ a i t b i or L t ˆ a i t b i where the parameter t takes on all real values. Thus we obtain x 1 ˆ 4 t; x ˆ t; x ˆ 7t; x 4 ˆ 1 8t or L t ˆ 4 t; t; 7t; 1 8t
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 CHAP. 1] VECTORS IN R n AND C n, SPATIAL VECTORS 19 1.0. Let C be the curve F t ˆ t ; t ; t ; t 5 in R 4, where 0 t 4. (a) Find the point P on C corresponding to t ˆ. (b) Find the initial point Q and terminal point Q 0 of C. (c) Find the unit tangent vector T to the curve C when t ˆ. (a) Substitute t ˆ into F t to get P ˆ f ˆ 4; 4; 8; 9. (b) The parameter t ranges from t ˆ 0 to t ˆ 4. Hence Q ˆ f 0 ˆ 0; ; 0; 5 and Q 0 ˆ F 4 ˆ 16; 10; 64; 1. (c) Take the derivative of F t, that is, of each component of F t, to obtain a vector V that is tangent to the curve: V t ˆdF t ˆ t; ; t ; tš dt Now nd V when t ˆ ; that is, substitute t ˆ in the equation for V t to obtain V ˆ V ˆ 4; ; 1; 4Š. Then normalize V to obtain the desired unit tangent vector T. Wehave p kv kˆ 16 9 144 16 ˆ p 4 1 4 185 and T ˆ p ; p ; p ; p 185 185 185 185 SPATIAL VECTORS (VECTORS IN R ), ijk NOTATION, CROSS PRODUCT 1.1. Let u ˆ i j 4k, v ˆ i j k, w ˆ i 5j k. Find: (a) u v, (b) u v 4w, (c) u v and u w, (d) kuk and kvk. Treat the coef cients of i, j, k just like the components of a vector in R. (a) Add corresponding coef cients to get u v ˆ 5i j k. (b) First perform the scalar multiplication and then the vector addition: u v 4w ˆ 4i 6j 8k 9i j 6k 4i 0j 1k ˆ i 17j 6k (c) (d) Multiply corresponding coef cients and then add: u v ˆ 6 8 ˆ 5 and u w ˆ 15 1 ˆ 1 The norm is the square root of the sum of the squares of the coef cients: p kuk ˆ p 4 9 16 ˆ p 9 and kvk ˆ p 9 1 4 ˆ 14 1.. Find the (parametric) equation of the line L: (a) through the points P 1; ; and Q ; 5; 6 ; (b) containing the point P 1; ; 4 and perpendicular to the plane H given by the equation x 5y 7z ˆ 15: (a) First nd v ˆ ƒ! PQ ˆ Q P ˆ 1; ; 8Š ˆi j 8k. Then L t ˆ t 1; t ; 8t ˆ t 1 i t j 8t k (b) Since L is perpendicular to H, the line L is in the same direction as the normal vector N ˆ i 5j 7k to H. Thus L t ˆ t 1; 5t ; 7t 4 ˆ t 1 i 5t j 7t 4 k
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 0 VECTORS IN R n AND C n, SPATIAL VECTORS [CHAP. 1 1.. Let S be the surface xy yz ˆ 16 in R. (a) Find the normal vector N x; y; z to the surface S. (b) Find the tangent plane H to S at the point P 1; ;. (a) (b) The formula for the normal vector to a surface F x; y; z ˆ0is N x; y; z ˆF x i F y j F z k where F x, F y, F z are the partial derivatives. Using F x; y; z ˆxy yz 16, we obtain F x ˆ y ; F y ˆ xy z; F z ˆ y Thus N x; y; z ˆy i xy z j yk. The normal to the surface S at the point P is N P ˆN 1; ; ˆ4i 10j 4k Hence N ˆ i 5j k is also normal to S at P. Thus an equation of H has the form x 5y z ˆ c. Substitute P in this equation to obtain c ˆ 18. Thus the tangent plane H to S at P is x 5y z ˆ 18. 1.4. Evaluate the following determinants and negative of determinants of order two: 4 (a) (i) 5 9, (ii) 1 4, (iii) 4 5 (b) (i) 6 7 4, (ii) 5 4, (iii) 1 8 Use a b a c d ˆ ad bc and b c d ˆ bc ad. Thus: (a) (i) 7 0 ˆ 7, (ii) 6 4 ˆ 10, (iii) 8 15 ˆ 7: (b) (i) 4 6 ˆ 18, (ii) 15 14 ˆ 9, (iii) 8 1 ˆ 4: 1.5. Let u ˆ i j 4k, v ˆ i j k, w ˆ i 5j k. Find: (a) u v, (b) u w 4 (a) Use to get u v ˆ 6 4 i 1 4 j 9 k ˆ i 16j 11k 1 4 (b) Use to get u w ˆ 9 0 i 4 6 j 10 k ˆ 9i j 1k 1 5 1.6. Find u v, where: (a) u ˆ 1; ;, v ˆ 4; 5; 6 ; (b) u ˆ 4; 7;, v ˆ 6; 5;. (a) 1 Use to get u v ˆ 1 15; 1 6; 5 8Š ˆ ; 6; Š 4 5 6 (b) 4 7 Use to get u v ˆ 14 15; 18 8; 0 4Š ˆ 9; 6; Š 6 5 1.7. Find a unit vector u orthogonal to v ˆ 1; ; 4Š and w ˆ ; 6; 5Š. First nd v w, which is orthogonal to v and w. 1 4 The array gives v w ˆ 15 4; 8 5; 6 61Š ˆ 9; 1; 1Š 6 5 p Normalize v w to get u ˆ 9= p p 94,1= 94, 1= 94 Š
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 CHAP. 1] VECTORS IN R n AND C n, SPATIAL VECTORS 1 1.8. Let u ˆ a 1 ; a ; a and v ˆ b 1 ; b ; b so u v ˆ a b a b ; a b 1 a 1 b ; a 1 b a b 1. Prove: (a) (b) (a) (b) u v is orthogonal to u and v [Theorem 1.5(i)] ku vk ˆ uu v v uv (Lagrange's identity). We have u uv ˆa 1 a b a b a a b 1 a 1 b a a 1 b a b 1 ˆ a 1 a b a 1 a b a a b 1 a 1 a b a 1 a b a a b 1 ˆ 0 Thus u v is orthogonal to u. Similarly, u v is orthogonal to v. We have ku vk ˆ a b a b a b 1 a 1 b a 1 b a b 1 u u v v uv ˆ a 1 a a b 1 b b a 1 b 1 a b a b Expansion of the right-hand sides of (1) and () establishes the identity. 1 COMPLEX NUMBERS, VECTORS IN C n 1.9. Suppose z ˆ 5 i and w ˆ 4i. Find: (a) z w, (b) z w, (c) zw. Use the ordinary rules of algebra together with i ˆ 1 to obtain a result in the standard form a bi. (a) z w ˆ 5 i 4i ˆ7 i (b) z w ˆ 5 i 4i ˆ5 i 4i ˆ 7i (c) zw ˆ 5 i 4i ˆ10 14i 1i ˆ 10 14i 1 ˆ 14i 1.0. Simplify: (a) 5 i 7i, (b) 4 i,(c) 1 i. (a) 5 i 7i ˆ10 6i 5i 1i ˆ 1 9i (b) 4 i ˆ 16 4i 9i ˆ 7 4i (c) 1 i ˆ 1 6i 1i 8i ˆ 1 6i 1 8i ˆ 11 i 1.1. Simplify: (a) i 0 ; i ; i 4,(b) i 5 ; i 6 ; i 7 ; i 8,(c) i 9 ; i 174, i 5, i 17. (a) i 0 ˆ 1, i ˆ i i ˆ 1 i ˆ i; i 4 ˆ i i ˆ 1 1 ˆ1. (b) i 5 ˆ i 4 i ˆ 1 i ˆi, i 6 ˆ i 4 i ˆ 1 i ˆi ˆ 1, i 7 ˆ i ˆ i, i 8 ˆ i 4 ˆ 1. (c) Using i 4 ˆ 1 and i n ˆ i 4q r ˆ i 4 q i r ˆ 1 q i r ˆ i r, divide the exponent n by 4 to obtain the remainder r: i 9 ˆ i 4 9 ˆ i 4 9 i ˆ 1 9 i ˆ i ˆ i; i 174 ˆ i ˆ 1; i 5 ˆ i 0 ˆ 1; i 17 ˆ i 1 ˆ i 1.. Find the complex conjugate of each of the following: (a) 6 4i, 7 5i, 4 i, i, (b) 6,, 4i, 9i. (a) 6 4i ˆ 6 4i, 7 5i ˆ 7 5i, 4 i ˆ 4 i, i ˆ i. (b) 6 ˆ 6, ˆ, 4i ˆ 4i, 9i ˆ 9i. (Note that the conjugate of a real number is the original number, but the conjugate of a pure imaginary number is the negative of the original number.) 1.. Find zz and jzj when z ˆ 4i. p For z ˆ a bi, use zz ˆ a b and z ˆ p zz ˆ a b. p zz ˆ 9 16 ˆ 5; jzj ˆ 5 ˆ 5
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 VECTORS IN R n AND C n, SPATIAL VECTORS [CHAP. 1 1.4. Simpify 7i 5 i : To simplify a fraction z=w of complex numbers, multiply both numerator and denominator by w, the conjugate of the denominator: 7i 5 i ˆ 7i 5 i 5 i 5 i ˆ 11 41i 4 ˆ 11 41 4 4 i 1.5. Prove: For any complex numbers z, w C, (i) z w ˆ z w, (ii) zw ˆ z w, (iii) z ˆ z. Suppose z ˆ a bi and w ˆ c di where a; b; c; d R. (i) z w ˆ a bi c di ˆ a c b d i ˆ a c b d i ˆ a c bi di ˆ a bi c di ˆ z w (ii) zw ˆ a bi c di ˆ ac bd ad bc i ˆ ac bd ad bc i ˆ a bi c di ˆ z w (iii) z ˆ a bi ˆ a bi ˆ a b i ˆ a bi ˆ z 1.6. Prove: For any complex numbers z; w C, jzwjˆjzjjwj. By (ii) of Problem 1.5, jzwj ˆ zw zw ˆ zw z w ˆ zz w w ˆ jzj jwj The square root of both sides gives us the desired result. 1.7. Prove: For any complex numbers z; w C, jz wj jzj jwj. Suppose z ˆ a bi and w ˆ c di where a; b; c; d R. Consider the vectors u ˆ a; b and v ˆ c; d in R. Note that p jzj ˆ p a b ˆkuk; jwj ˆ c d ˆkvk and q jz wj ˆj a c b d ij ˆ a c b d ˆk a c; b d k ˆ ku vk By Minkowski's inequality (Problem 1.15), ku vk kuk kvk, and so jz wj ˆku vk kuk kvk ˆjzj jwj 1.8. Find the dot products u v and v u where: (a) u ˆ 1 i; i, v ˆ 4 i; 5 6i ; (b) u ˆ i; 4i; 1 6i, v ˆ 5 i; i; 7 i. Recall that conjugates of the second vector appear in the dot product z 1 ;...; z n w 1 ;...; w n ˆz 1 w 1 z n w n (a) u v ˆ 1 i 4 i i 5 6i ˆ 1 i 4 i i 5 6i ˆ 10i 9 i ˆ 9 1i v u ˆ 4 i 1 i 5 6i i ˆ 4 i 1 i 5 6i i ˆ 10i 9 i ˆ 9 1i (b) u v ˆ i 5 i 4i i 1 6i 7 i ˆ i 5 i 4i i 1 6i 7 i ˆ 0 5i v u ˆ 5 i i i 4i 7 i 1 6i ˆ 5 i i i 4i 7 i 1 6i ˆ 0 5i In both cases, v u ˆ u v. This holds true in general, as seen in Problem 1.40.
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 CHAP. 1] VECTORS IN R n AND C n, SPATIAL VECTORS 1.9. Let u ˆ 7 i; 5i and v ˆ 1 i; 6i. Find: (a) u v; (b) iu; (c) i v; (d) u v; (e) kuk and kvk. (a) u v ˆ 7 i 1 i; 5i 6i ˆ 8 i; 1 i (b) iu ˆ 14i 4i ; 4i 10i ˆ 4 14i; 10 4i (c) i v ˆ i i i ; 9 18i i 6i ˆ 4 i; 15 15i (d) u v ˆ 7 i 1 i 5i 6i ˆ 7 i 1 i 5i 6i ˆ 5 9i 6 i ˆ 1 1i q (e) kuk ˆ 7 p 5 ˆ q 8 and kvk ˆ 1 1 6 p ˆ 47 1.40. Prove: For any vectors u; v C n and any scalar z C, (i) u v ˆ v u, (ii) zu v ˆ z u v, (iii) u zv ˆz u v. Suppose u ˆ z 1 ; z ;...; z n and v ˆ w 1 ; w ;...; w n. (i) Using the properties of the conjugate, v u ˆ w 1 z 1 w z w n z n ˆ w 1 z 1 w z w n z n ˆ w 1 z 1 w z w n z n ˆ z 1 w 1 z w z n w n ˆ u v (ii) Since zu ˆ zz 1 ; zz ;...; zz n, zu v ˆ zz 1 w 1 zz w zz n w n ˆ z z 1 w 1 z w z n w n ˆz uv (Compare with Theorem 1. on vectors in R n.) (iii) Using (i) and (ii), u zv ˆ zv u ˆ z v u ˆz v u ˆz u v VECTORS IN R n Supplementary Problems 1.41. Let u ˆ 1; ; 4, v ˆ ; 5; 1, w ˆ ; 1;. Find: (a) u v; (b) 5u v 4w; (c) u v, u w, v w; (d) kuk, kvk; (e) cos y, where y is the angle between u and v; (f ) d u; v ; (g) proj u; v. 1 1.4. Repeat Problem 1.41 for vectors u ˆ 4 5, v ˆ 4 1 5, w ˆ 4 4 5 5. 1.4. Let u ˆ ; 5; 4; 6; and v ˆ 5; ; 1; 7; 4. Find: (a) 4u v; (b) 5u v; (c) u v; (d) kuk and kvk; (e) proj u; v ; ( f ) d u; v. 1.44. Normalize each vector: (a) u ˆ 5; 7 ; (b) v ˆ 1; ; ; 4 ; (c) w ˆ 1 ; 1 ;. 4 1.45. Let u ˆ 1; ;, v ˆ ; 1; 4, and k ˆ. (a) Find kuk, kvk, ku vk, kkuk (b) Verify that kkukˆjkjkuk and ku vkkuk kvk. 1.46. Find x and y where: (a) x; y 1 ˆ y ; 6 ; (b) x ; y ˆy 1;. 6
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 4 VECTORS IN R n AND C n, SPATIAL VECTORS [CHAP. 1 1.47. Find x; y; z where x; y 1; y z ˆ x y; 4; z. 1.48. Write v ˆ ; 5 as a linear combination of u 1 and u where: (a) u 1 ˆ 1; and u ˆ ; 5 ; (b) u 1 ˆ ; 4 and u ˆ ;. 1.49. Write v ˆ 4 9 1 5 as a linear combination of u 1 ˆ 4 5, u ˆ 4 16 1.50. Find k so that u and v are orthogonal, where: (a) u ˆ ; k;, v ˆ 6; 4; ; (b) u ˆ 5; k; 4;, v ˆ 1; ; ; k ; (c) u ˆ 1; 7; k ;, v ˆ ; k; ; k. 5, u ˆ 4 5 1 4 5. LOCATED VECTORS, HYPERPLANES, LINES IN R n 1.51. Find the vector v identi ed with the directed line segment PQ! for the points: (a) P ; ; 7 and Q 1; 6; 5 in R ; (b) P 1; 8; 4; 6 and Q ; 5; ; 4 in R 4. 1.5. Find an equation of the hyperplane H in R 4 that: (a) contains P 1; ; ; and is normal to u ˆ ; ; 5; 6Š; (b) contains P ; 1; ; 5 and is parallel to x 1 x 5x 7x 4 ˆ 4. 1.5. Find a parametric representation of the line in R 4 that: (a) passes through the points P 1; ; 1; and Q ; 5; 7; 9 ; (b) passes through P 1; 1; ; and is perpendicular to the hyperplane x 1 4x 6x 8x 4 ˆ 5. SPATIAL VECTORS (VECTORS IN R ), ijk NOTATION 1.54. Given u ˆ i 4j k, v ˆ i 5j k, w ˆ 4i 7j k. Find: (a) u v; (b) u 4v w; (c) u v, u w, v w; (d) kuk, kvk, kwk. 1.55. Find the equation of the plane H: (a) with normal N ˆ i 4j 5k and containing the point P 1; ; ; (b) parallel to 4x y z ˆ 11 and containing the point Q ; 1;. 1.56. Find the (parametric) equation of the line L: (a) through the point P ; 5; and in the direction of v ˆ 4i 5j 7k; (b) perpendicular to the plane x y 7z ˆ 4 and containing P 1; 5; 7. 1.57. Consider the following curve C in R where 0 t 5: F t ˆt i t j t k (a) Find the point P on C corresponding to t ˆ. (b) Find the initial point Q and the terminal point Q 0. (c) Find the unit tangent vector T to the curve C when t ˆ.
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 CHAP. 1] VECTORS IN R n AND C n, SPATIAL VECTORS 5 1.58. Consider a moving body B whose position at time t is given by R t ˆt i t j tk. [Then V t ˆdR t =dt and A t ˆdV t =dt denote, respectively, the velocity and acceleration of B.] When t ˆ 1, nd: (a) position; (b) velocity v; (c) speed s; (d) acceleration a of B. 1.59. Find a normal vector N and the tangent plane H to each surface at the given point: (a) surface x y yz ˆ 0 and point P 1; ; ; (b) surface x y 5z ˆ 160 and point P ; ; 1 : CROSS PRODUCT 1.60. Evaluate the following determinants and negative of determinants of order two: 5 (a) 6 ; 6 1 4 ; 4 7 (b) 6 4 1 7 5 ; 8 4 ; 6 1.61. Given u ˆ i 4j k, v ˆ i 5j k, w ˆ 4i 7j k, Find: (a) u v, (b) u w, (c) v w. 1.6. Given u ˆ ; 1; Š, v ˆ 4; ; Š, w ˆ 1; 1; 5Š, nd: (a) u v, (b) u w, (c) v w. 1.6. Find the volume V of the parallelopiped formed by the vectors u; v; w in: (a) Problem 1.60, (b) Problem 1.61. 1.64. Find a unit vector u orthogonal to: (a) v ˆ 1; ; Š and w ˆ 1; 1; Š; (b) v ˆ i j k and w ˆ 4i j k. 1.65. Prove the following properties of the cross product: (a) u v ˆ vu (d) u v w ˆ uv uw (b) u u ˆ 0 for any vector u (e) v w u ˆ vu wu (c) ku v ˆ k u v ˆu kv ( f ) u v w ˆ uw v vw u COMPLEX NUMBERS 1.66. Simplify: (a) 4 7i 9 i ; (b) 5i ; (c) 1 4 7i ; (d) 9 i 5i ; (e) 1 i. 1.67. Simplify: (a) 1 i ; (b) i 7 i ; (c) i15 ; i 5 ; i 4 ; (d) 1. i
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 6 VECTORS IN R n AND C n, SPATIAL VECTORS [CHAP. 1 1.68. Let z ˆ 5i and w ˆ 7 i. Find: (a) v w; (b) zw; (c) z=w; (d) z; w; (e) jzj, jwj. 1.69. Show that: (a) Re z ˆ 1 1 z z (b) Im z ˆ z z) (c) zw ˆ 0 implies z ˆ 0orwˆ0. VECTORS IN C n 1.70. Let u ˆ 1 7i; 6i and v ˆ 5 i; 4i. Find: (a) u v (b) i u (c) iu 4 7i v (d) u v (e) kuk and kvk. 1.71. Prove: For any vectors u; v; w in C n : (a) u v w ˆ u w v w, (b) w u v ˆw u w v. 1.7. Prove that the norm in C n satis es the following laws: N 1 Š For any vector u, kuk 0; and kuk ˆ0 if and only if u ˆ 0. N Š For any vector u and complex number z, kzuk ˆjzjkuk. N Š For any vectors u and v, ku vk kuk kvk. Answers to Supplementary Problems p p p 1.41. (a) ; 16; 4 ; (b) (6,1,5); (c) ; 1; 8; (d) 1, 5, 14 ; p (e) = p p 1 5 ; (f ) 6 ; (g) 5 ; 5; 1 ˆ 9 15 5, 5, 5 ) 1.4. (Column vectors) (a) 1; 7; ; (b) 1; 6; 9 ; (c) 15; 7; 4; p p p p p (d) 6, 0 ; (e) 15= 6 0 ; (f ) 86 ; (g) 15 0 v ˆ 1; 1 ; 5 p p 1.4. (a) 1; 14; 1; 45; 0 ; (b) 0; 9; ; 16; ; (c) 6; (d) 90 ; 95, (e) 6 95 v; (f ) p 167 p 1.44. (a) 5= p 76 ; 9= 76 ; (b) 1 5 ; 5 ; 5 ; p p p 4 5 ; (c) 6= 1 ; 4 1 ; 9 1 p 1.45. (a) ; 1; 10 ; 9 1.46. (a) x ˆ ; y ˆ 5; (b) x ˆ 0; y ˆ 0, and x ˆ 1, y ˆ 1.47. x ˆ ; y ˆ ; z ˆ 1.48. (a) v ˆ 5u 1 u ; (b) v ˆ 16u 1 u 1.49. v ˆ u 1 u u 1.50. (a) 6; (b) ; (c) 1.51. (a) v ˆ 1; 9; Š; (b) [; ; 6; 10]
Algebra, /e 1. Vectors in R^n and C^n, Spatial Vectors Companies, 004 CHAP. 1] VECTORS IN R n AND C n, SPATIAL VECTORS 7 1.5. (a) x 1 x 5x 6x 4 ˆ 5; (b) x 1 x 5x 7x 4 ˆ 16 1.5. (a) t 1; 7t ; 6t 1; 11t Š; (b) t 1; 4t 1; 6t ; 8t Š 1.54. (a) j 1k; (b) i 6j 10k; (c) 0; 1; 7; (d) 1.55. (a) x 4y 5z ˆ 0; (b) 4x y z ˆ 1 1.56. (a) 4t ; 5t 5; 7t Š; (b) t 1; t 5; 7t 7Š p p p 9 ; 8 ; 69 1.57. (a) P ˆ F ˆ8i 4j k; (b) Q ˆ F 0 ˆ k, Q 0 ˆ F 5 ˆ15i 5j 7k; p (c) T ˆ 6i j k = 41 1.58. (a) i j k; (b) i j k; (c) p 17 ; (d) i 6j 1.59. (a) N ˆ 6i 7j 9k, 6x 7y 9z ˆ 45; (b) N ˆ 6i 1j 10k, x 6y 5z ˆ 16 1.60. (a) ; 6; 6; (b) ; 10; 4 1.61. (a) i 1j k; (b) i j 7k; (c) 1i 16j 6k 1.6. (a) 5; 8; 6Š; (b) ; 7; 1Š; (c) 7; 18; 5Š 1.6. (a) 14; (b) 17 p 1.64. (a) 7; 1; = p 59 ; (b) 5i 11j k = 150 1 1.66. (a) 50 55i; (b) 16 0i; (c) 65 4 7i ; (d) 1 1 i ; (e) i 1.67. (a) 1 i; (b) 1 58 5 7i ; (c) 1; i; 1; (d) 1 50 4 i 1 1.68. (a) 9 i; (b) 9 9i; (c) 61 1 41i ; (d) 5i, 7 i; (e) p p 9, 58 1.69. (c) Hint: If zw ˆ 0, then jzwj ˆ jzjjwj ˆ j0j ˆ 0 1.70. (a) 6 5i, 5 10i ; (b) 4 i, 1 16i ; (c) 8 41i, 4 i ; p p (d) 1 i; (e) 90, 54