2 Vector Products. 2.0 Complex Numbers. (a~e 1 + b~e 2 )(c~e 1 + d~e 2 )=(ac bd)~e 1 +(ad + bc)~e 2

Transcription

1 Vector Products Copyright 07, Gregory G. Smith 8 September 07 Unlike for a pair of scalars, we did not define multiplication between two vectors. For almost all positive n N, the coordinate space R n cannot be equipped with a second binary operation which behaves like multiplication a reasonable vector product would be nontrivial, associative, and distributive. Nevertheless, this chapter focuses on the special cases in which there exists such a vector product..0 Complex Numbers How is vector multiplication on R defined? The complex numbers may viewed as R with a vector product..0.0 Theorem (Complex Numbers). The coordinate space R together with vector product defined by (a~e + b~e )(c~e + d~e )=(ac bd)~e +(ad + bc)~e, ~e + ~e ~e +~e for all a, b, c, d R, forms a field of scalars. Proof. Since Section. already demonstrates that vector addition in R satisfies the four properties for a field of scalars that only involve addition, it suffices to verify the five properties that involve multiplication. Let a, b, c, d, e, f R denote arbitrary real numbers. The commutativity, associativity, and distributivity of addition and multiplication in R give Figure.0: Complex multiplication: ~e (~e +~e )= ~e + ~e (a~e + b~e )(c~e + d~e )=(ac bd)~e +(ad + bc)~e =(ca db)~e +(da + cb)~e =(c~e + d~e )(a~e + b~e ), (a~e + b~e )(c~e + d~e ) (e~e + f ~e )= (ac bd)e (ad + bc) f ~e + (ac bd) f +(ad + bc)e ~e = a(ce df) b(cf + de) ~e + a(cf + de)+b(ce df) ~e =(a~e + b~e ) (c~e + d~e )(e~e + f ~e ),

2 linear algebra (a~e + b~e ) (c~e + d~e )+(e~e + f ~e ) = a(c + e) b(d + f ) ~e + a(d + f )+b(c + e) ~e i i i i Figure.: Multiplication table for the complex units (a~e + b~e ) = (ac bd)+(ae bf) ~e + (ad + bc)+(af + be) ~e =(a~e + b~e )(c~e + d~e )+(a~e + b~e )(e~e + f ~e ), which establishes the commutativity, associativity, and distributivity for vector multiplication on R. To prove the existence of a multiplicative identity, we observe that (~e + 0~e )(a~e + b~e )=(a 0)~e +(0 + b)~e = a~e + b~e. Lastly, if a~e + b~e 6=~0, then we have a + b 6= 0 and b a a + b ~e = + b ab + ab a + b ~e + a + b a a + b ~e ~e = ~e + 0~e, b = In(z) Im z = a + b i a = Re(z) Re Figure.: Points in the complex plane correspond to complex numbers so every nonzero vector in R has a multiplicative inverse..0. Notation. The complex numbers are denoted by C. Traditionally, one identifies the vectors ~e,~e R with the complex units, i C, so a~e + b~e = a + b i and the multiplicative identity is. When a single symbol z := a + b i is used, its real part is a := Re(z) R and its imaginary part is b := Im(z) R. We identify R with the subset of complex number in which the imaginary part is zero. Alternatively, we have C = R( p )..0. Problem. Determine the square roots of 7 4 i. Solution. Suppose that the complex number a + b i, where a, b R, satisfies the equation 7 4 i = (a + b i) = (a b )+ab i. It follows that a b = 7 and ab = 4, so we have a = /b and b b = 7. Multiplying by b and gathering terms gives 0 = b 4 7b 44 =(b 6)(b + 9) =(b 4)(b + 4)(b + 9). Since b R, we deduce that b = ±4 and a = 3, which means that the square roots are 3 4 i and i..0.3 Proposition. If z C then there exists w C such that w = z. Moreover, the additive inverse w also satisfies this equation. Figure.3: Intersection of the hyperbolas x y = a and xy = b when b > 0 Proof. Suppose that z = a + b i where a, b R. Consider w = x + y i, where x, y R, and (x y )+xy i =(x + y i) = w = z = a + b i. It follows that x y = a and xy = b, so we have x = y b and b y y = a. Multiplying by y and gathering terms gives 0 = y 4 + ay b = y a+p a +b y a+p a +b q q = y y + y + a+p a +b a+ p a +b a+ p a +b,

3 vector products 3 because p a + b > p a = a > a. Since y R, we conclude that s a + p a y = ± + b, s x = b y = ± sgn(b) b and w = ± s a + p s a sgn(b) + b + s a + p a + b = ± sgn(b) a + p a + b a + p! a + b i. The signum function of a real number x is defined to be ( if x < 0, sgn(x) := if x > 0. For all x R, we have x = sgn(x) x..0.4 Problem. Solve the equation t +( + i)t +( + 5i)=0. Solution. Completing the square yields 0 = t +( + i)t +( + 5i)= t + (i )t +(i ) i +( + 5i), so we obtain t +(i ) = i + 4 5i = 4 ( 7 4 i). The previous problem shows that the square roots of 7 4 i are ±(3 4i), so t + i = (±3 4i) and t equals 3 i or + i..0.5 Theorem (Fundamental theorem of algebra). Every polynomial in one variable with coefficients in the complex numbers is has a complex root. More explicitly, for any a positive integer n N and any complex numbers a 0, a,...,a n C, there exists z C such that z n + a n z n + a n z n + + a z + a 0 = 0. Comment on the proof. Despite its name, there is no purely algebraic proof; any proof must use the "completeness" of the real numbers. Exercises.0.6 Problem. Determine which of the following statements are true. If a statement is false, then provide a counterexample. i. Each complex number corresponds to exactly one vector in R. ii. Real numbers are not complex numbers. iii. The imaginary part of a complex number is a real number. iv. The real number has a unique square root in C..0.7 Problem. Show that complex multiplication is compatible with scalar multiplication; for all a, b, c R, we have c(a + b i) =ac + bc i..0.8 Problem. Determine every complex number such that its multiplicative inverse equals its additive inverse..0.9 Problem (Impossibility of order). Show that the following two conditions cannot both be satisfied.

4 4 linear algebra For all z C, one and only one of the relations z > 0, z > 0, and z > 0 is valid. If w > 0 and z > 0, then w + z > 0 and wz > Problem. Prove that the three numbers z, z, z 3 C with z 6= z are collinear if and only if z 3 z z z R..0. Problem. Prove that the four numbers z, z, z 3, z 4 C with z 6= z 4, z 6= z 3, and not all on the same line, lie on a circle if and only if (z z )(z 3 z 4 ) (z z 4 )(z z 3 ) R..0. Problem. Find all w C such that b Im z = a + b i w +( + 5i)w +( 0 + 5i)=0. Express your solution(s) in the form w = a + b i where a, b R. Re a b z = a b i Figure.4: Complex conjugate.0.3 Problem (Properties of the complex conjugate). For any number z = x + y i C with x, y R, its complex conjugate is the number z := a b i C. Geometrically, the complex conjugate is the reflection of the corresponding vector in the real axis. For all z, w C, establish the following properties of the complex conjugate. i. z + w = z + w ii. zw = z w iii. Re(z) = (z + z) and Im(z) = i (z z) iv. z z = z v. z = z if and only if z is a real number.

5 vector products 5. Complex Geometry How can we visualize multiplication of complex numbers? For a complex number z = a + b i where a, b R, the absolute value is z := p a + b which is the magnitude of the underlying vector in R. The argument of z is the angle arg(z) that the vector in R makes with the positive real axis. Since the argument is measured in radians, arg(z) can be changed be any integer multiple of p without changing the angle. Often, one chooses the angle such that p 6 arg(z) 6 p. Since z = a + b i = z cos arg(z) + z sin arg(z) i, we see that the absolute value and argument determine a complex number...0 Problem. Find the absolute value and argument of the complex number z := + i. Solution. By definition, we have z = p ( ) +() = p. Since tan arg(z) = = and Re(z) < 0, it follows that z sin arg(z) Im z arg(z) z cos arg(z) Figure.5: Polar form of z C Re arg(z) =arctan( )+p = p 4 + p = 3p 4... Proposition (Geometric multiplication). For all z, w C, we have zw = z w and arg(zw) = arg(z) +arg(w), so multiplication by a complex number is given by a rotation in the counterclockwise direction by the argument and a rescaling of the magnitude by the absolute value. Proof. Setting q := arg(z) and f := arg(z), we have + i p Im 3p 4 Re z = z cos(q) + sin(q) i and w = w cos(f) + sin(f) i. Figure.6: Polar form of Hence, the addition formula for trigonometric functions gives zw = z cos(q) +sin(q) i w cos(f) +sin(f) i = z w cos(q) cos(f) sin(q) sin(f) + cos(q) sin(f) +sin(q) cos(f) i + i C = z w cos(q + f) +sin(q + f) i, which yields the required formula... Corollary (De Moivre s Formula). Let z = r cos(q) + sin(q) i be a complex number with z = r and arg(z) =q. For all m N, we have z m = z m and arg(z m )=m arg(z), so z m = r m cos(mq)+sin(mq) i. The lowercase "theta" q is the eighth letter in the Greek alphabet, and is often used to denote an angle. Proof. We proceed by induction on m. The base case has m = 0 and z 0 =, so z 0 = = z 0 and arg(z 0 )=0 = 0 arg(z). Suppose that m > 0. The geometry of complex multiplication and the induction hypothesis imply z m = zz m = z z m = z z m = z m and arg(z m )=arg(zz m )=arg(z)+arg(z m ) = arg(z) +(m ) arg(z) = m arg(z).

6 6 linear algebra..3 Problem. Express cos(3q) in terms of cos(q) and sin(q). Solution. De Moivre s Formula for r = and m = 3 gives cos(3q)+sin(3q) i = cos(q)+sin(q)i 3 = cos 3 (q)+3 cos (q) sin(q) i 3 cos(q) sin (q) sin 3 (q) i, p + 3 i p 3 i Im Figure.7: Third roots of unit Re so taking real parts produces cos(3q) =cos 3 (q) 3 cos(q) sin (q)...4 Problem. Find three solutions to the equation z 3 =. Solution. Since = cos(p) + sin(p) i, consider the complex number z = cos pk 3 + sin pk 3 i for 0 6 k 6. De Moivre s Formula shows that z 3 = and arg(z 3 )=pk, so the solutions corresponding p p to k = 0,, are, + 3 i, and 3 i respectively. What is the exponential of a complex number? The exponential function exp(x) =e x equals the power series x  j j! = + x + x! + x3 3! + j=0 for all x R. To extend this function to C, it would be most natural to define e y i (y i) (y i) (y i)3 to be +! +! + 3! + for y R. Of course, to make this rigorous, we would need to discuss convergence in C which we will not do here. Nevertheless, a slight rearrangement and recognizing the power series for cos(y) and sin(y) yield exp(y i) =e y i = y! + y4 4! + y y 3 3! + y5 5! i = cos(y)+sin(y) i. This discussion can be strengthened to prove the next theorem. The equation e p i + = 0 unites the five most important numbers and the seven most important symbols in mathematics...5 Theorem (Euler s formula). For a complex number z = x + y i where x, y R, we have exp(z) = e z = e x cos(y) +sin(y) i, so we obtain e z = e x = exp Re(z), arg(e z )=y = Im(z), and z = z exp arg(z) i...6 Proposition (Properties of exp(z)). For all z, w C, we have the following. i. The exponential function is multiplicative: e z+w = e z e w. ii. The exponential function is never zero: e z 6= 0. iii. We have e z = if and only if z = pm i for some m Z. Proof. i. Euler s Formula and the multiplicative property of the real exponential function yield e z+w = exp Re(z + w) = exp Re(z)+Re(w) = e z e w, arg(e z+w )=Im(z + w) =Im(z)+Im(w) =arg(e z )+arg(e w ), so the geometry of complex multiplication shows that e z+w = e z e w.

7 vector products 7 ii. The multiplicative property implies that e z e z = e z z = e 0 =, so e z 6= 0. iii. If z = x + yi with x, y R, then the equation e z =, together with Euler s Formula, implies that e x = and x = 0. Hence, we have = e y i = cos(y)+sin(y) i, so cos(y) = and sin(y) =0 which means that y = pm for some m Z. Exercises..7 Problem. Determine which of the following statements are true. If a statement is false, then provide a counterexample. i. The absolute value of a complex number is a real number. ii. The argument of a complex number is a real number. iii. The argument of a sum of complex numbers equals the sum of the arguments. iv. The exponential function is periodic...8 Problem. Compute ( + i) Problem. Given two complex numbers z := r cos(q) + sin(q) i with r, q R and w := s cos(f)+sin(f) i 6= 0 with s, f R, show that z w = r cos(q f)+sin(q f) i. s..0 Problem. Simplify ( i)0 ( p 3 + i) p 5. ( 3i) 0.. Problem. For z, w C, show that z + w + z w = ( z + w )... Problem. If z, w C satisfy z = w = and zw 6=, then demonstrate that z + w + zw R...3 Problem. Find all z C such that z = and z z + z z =...4 Problem. For 0 < q < p, calculate the polar representation of z = + cos(q)+sin(q) i...5 Problem. Prove the following related identities. i. For all z, w C, show that zw = z w by using complex conjugates. ii. For all a, b, c, d R, show that (ac bd) +(ad + bc) =(a + b )(c + d ).

8 8 linear algebra p..6 Problem. Consider the complex numbers z := 3 3i and w := + p 3 i. i. Find zw and z/w. Give your answer in the form x + y i where x, y R. ii. Put z and w into polar form re q i = r cos(q)+sin(q) i. Find zw and z/w using the polar form and verify that you get the same answer as in part (a)...7 Problem. For any complex number z := r cos(q) +sin(q) i where r, q R, show that the n-th roots are r /n cos q+pk n + sin q+pk n i for k N with k < n...8 Problem. For all z C, we define sin(z) := ez i e z i i and For z, w C, prove the following identities. i. sin (z)+cos (z) =; ii. sin(z + w) = sin(z) cos(w) +cos(z) sin(w); iii. cos(z + w) = cos(z) cos(w) sin(z) sin(w). cos(z) := ez i + e z i.

9 vector products 9. The Cross Product How can we multiply vectors in R 3? Although significantly A unit vector has magnitude equals. different than multiplication in a field of scalars, there is a useful ~v vector product on R 3. Geometrically, we associate a magnitude (or ~w nonnegative real number) and a direction (or unit vector) to any pair q ~w of vectors ~v, ~w R 3. The associated magnitude is the area of the parallelogram formed by the two vectors. If 0 6 q 6 p denotes the ~v Figure.8: Area of parallelogram angle between ~v and ~w, then the height of the parallelogram equals k~wk sin(q) and the area of the parallelogram equals k~vk k~wk sin(q). ~w To visualize the associated direction, position the vectors ~v and ~w so that their tails coincide. Orient your right hand so that its edge and all of your finger point in the same direction as ~v. With a flat hand, extend q ~v your thumb so that it perpendicular to your fingers. When curling your fingers through the angle q, from ~v to ~w, your thumb points in the direction of the unit vector ~n. By construction, the normal vector ~n is perpendicular to the plane containing ~v and ~w, and this right-hand ~n Figure.9: The right-hand rule: if ~v and ~w lie in the plane the page, then ~n points out of the page toward the reader. rule chooses one side of the plane. The following two definitions of the cross product are equivalent. (geometry) If the vectors ~v, ~w R 3 are not parallel, then we define ~v ~w := k~vk k~wk sin(q) ~n The geometric definition was made in 878 by W.K. Clifford. The name "cross product" and the notation were introduced in 88 by J.W. Gibbs. where 0 < q < p is the angle between ~v and ~w, and ~n is the unit vector determined by the right-hand rule. If the vectors ~v and ~w are parallel, then we set ~v ~w :=~0. (algebra) For the two vectors ~v := v ~e + v ~e + v 3 ~e 3 R 3 and ~w := w ~e + w ~e + w 3 ~e 3 R 3, we have ~v ~w := (v w 3 v 3 w )~e +(v 3 w v w 3 )~e +(v w v w )~e Problem. Compute ~e j ~e k for all 6 j < k 6 3. Geometric Solution. Since the standard basis ~e,~e,~e 3 consists of pairwise perpendicular unit vectors, the geometric definition of the cross product gives ~e ~e = k~e kk~e k sin p ~e 3 = ~e 3, ~e ~e ~e 3 ~e ~e v v v 3 v v w w w 3 w w Figure.0: The cross product is the sum of the product along the solid diagonals minus the sume of the produces along the dotted diagonals. This mnemonic is named after P.F. Sarrus. ~e ~e 3 = k~e kk~e 3 k sin p ( ~e )= ~e, and ~e ~e 3 = k~e kk~e 3 k sin p ~e = ~e. Algebraic Solution. The algebraic definition of the cross product yields ~e ~e = (0)(0) (0)() ~e + (0)(0) ()(0) ~e + ()() (0)(0) ~e 3 = ~e 3, ~e ~e 3 = (0)() (0)(0) ~e + (0)(0) ()() ~e + ()(0) (0)(0) ~e 3 = ~e, ~e ~e 3 = ()() (0)(0) ~e + (0)(0) (0)() ~e + (0)(0) ()(0) ~e 3 = ~e, because ~e j has in the j-th entry and zero elsewhere for 6 j 6 3.

10 30 linear algebra.. Proposition (Properties of the cross product). For all ~u,~v, ~w R 3 and all c R, we have the following: (anti-commutativity) ~v ~w = (~w ~v) (compatibility with scalar multiplication) ~v (c ~w) =c(~v ~w) =(c ~v) ~w (distributivity) ~u (~v + ~w)=~u ~v + ~u ~w ~e ~e ~e 3 ~e ~0 ~e 3 ~e ~e ~e 3 ~0 ~e ~e 3 ~e ~e ~0 Figure.: Cross product multiplication table for the standard basis vectors Geometric Proof. For ~v R 3, we have ~v ~v = ~0 because ~v is parallel to itself. The right-hand rule tells us that ~v ~w and ~w ~v point in opposite directions. Since the magnitudes of ~v ~w and ~w ~v both equal the area of the parallelogram formed by ~v and ~w, we have ~v ~w = (~w ~v) which establishes anti-commutativity property. Exploiting anti-commutativity, we may assume that c > 0 by interchanging ~v and ~w if necessary. Hence, the compatibility of magnitude with scalar multiplication gives kc~vk k~wk sin(q) =c k~vk k~wk sin(q) =k~vk kc ~wk sin(q), so the compatibility with scalar multiplication follows from the geometric definition. We postpone the proof of distributivity until we have introduced the triple product; see Problem Algebraic Proof. The algebraic definition of the cross product gives ~v ~w =(v w 3 v 3 w )~e +(v 3 w v w 3 )~e +(v w v w )~e 3 = (v 3 w v w 3 )~e +(v w 3 v 3 w )~e +(v w v w )~e 3 = (~w ~v), ~v (c ~w) =(cv w 3 cv 3 w )~e +(cv 3 w cv w 3 )~e +(cv w cv w )~e 3 =(c~v) ~w = c (v 3 w v w 3 )~e +(v w 3 v 3 w )~e +(v w v w )~e 3 = c(~v ~w), ~u (~v + ~w) = u (v 3 + w 3 ) u 3 (v + w ) ~e + u 3 (v + w ) u (v 3 + w 3 ) ~e + u (v + w ) u (v + w ) ~e 3 =(u v 3 u 3 v )~e +(u 3 v u v 3 )~e +(u v u v )~e 3 +(u w 3 u 3 w )~e +(u 3 w u w 3 )~e +(u w u w )~e 3 = ~u ~v + ~u ~w, which establishes the three properties of the cross product. Why do the two definitions of the cross product agree? For any ~u R 3, the anti-commutativity of the cross product means that ~u ~u = (~u ~u), which implies that (~u ~u) =~0 and ~u ~u = ~0. Now, for ~v := v ~e + v ~e + v 3 ~e 3 and ~w := w ~e + w ~e + w 3 ~e 3, the properties of cross product establish that ~v ~w =(v ~e + v ~e + v 3 ~e 3 ) (w ~e + w ~e + w 3 ~e 3 ) = v w (~0 )+v w (~e ~e )+v w 3 (~e ~e 3 ) v w (~e ~e )+v w (~0 )+v w 3 (~e ~e 3 ) v 3 w (~e ~e 3 ) v 3 w (~e ~e 3 )+v 3 w 3 (~0 ). Hence, it suffices to know that geometric and algebraic definitions of the cross products agree on ~e j ~e k where 6 j < k 6 3; see Problem..0.

11 vector products 3 Exercises.. Problem. Determine which of the following statements are true. If a statement is false, then provide a counterexample. i. The cross product of two vectors in R 3 is another vector in R 3. ii. The cross product is defined for any two vectors in R n. iii. The cross product of two vectors is zero if and only if one vector is parallel to the other. iv. The cross product is commutative...3 Problem. For ~u,~v, ~w R 3, simplify the following expressions. i. (~v ~w) (~v + ~w) ii. (~u + ~v + ~w) (~v + ~w)..4 Problem. Show that the cross product is not associative by exhibiting three vectors ~u,~v, ~w R 3 such that (~u ~v) ~w 6= ~u (~v ~w)...5 Problem. Show that the cross product is not cancellative by exhibiting three vectors ~u,~v, ~w R 3 such that ~u 6= ~0, ~v 6= ~w, and ~u ~v = ~u ~w...6 Problem. Find the area of the triangle with edges corresponding to the vectors ~v := ~e +~e 3~e 3, ~w := ~e + 3~e + ~e 3, and ~w ~v...7 Problem. If~r is the position vector locating a particle relative to a fixed point has mass m and velocity ~v, then the angular momentum ~` of the particle relative to the fixed is defined to be ~` := m(~r ~v). A kg object has position vector ~r =(~e + 4~e 3~e 3 ) m and velocity vector ~v =( 6~e + 3~e + 3~e 3 ) m s. Determine the angular momentum of the object about the origin...8 Problem. If a force ~F is applied to a particle and ~r is a position vector locating the particle relative to a fixed point, then the torque on the particle relative to the fixed point is defined to be ~t :=~r ~F. A particle moves in R 3 while a force acts on it. When the particle has the position vector ~r =(~e 5~e +~e 3 ) m, the force is given by ~F =(F ~e + 4~e 3~e 3 ) N, and the corresponding torque about the origin is ~t =(~e + 5~e + t 3 ~e 3 ) N m. Find the scalars F and t Problem. A particle of charge q moving with velocity ~v in the presence of an electric field ~E and a magnetic field ~B experiences the Lorentz force ~F = q(~e + ~v ~B). A particle with 3Ccharge moves in R 3. At one instant, the particle has velocity ~v =(3~e + v ~e ~e 3 ) m s, the electric field is given by ~E =( 4~e + E ~e 8~e 3 ) N C, the magnetic field is given by ~B =(5~e +~e ~e 3 ) N m s C, and the corresponding force is ~F =( 6~e ~e + F 3 ~e 3 ) N. Find the scalars v, E, and F 3.

12 3 linear algebra a c A b C Figure.: The angles a, b, g in the triangle are opposite to the sides a, b, c. b B g a..0 Problem (Law of sines). If a, b, and c are the lengths of the sides in a triangle and a, b, and g are the opposite angles, then prove that using the cross product. sin(a) a = sin(b) b = sin(g) c,

13 vector products 33.3 Quaternions* How is vector multiplication on R 4 defined? The quaternions may viewed as R 4 with a vector product. The quaternions were introduced in 843 by W.R. Hamilton..3.0 Theorem (Quaternions). The coordinate space R 4, together with the vector multiplication defined by ~v~w =(v ~e + v ~e + v 3 ~e 3 + v 4 ~e 4 )(w ~e + w ~e + w 3 ~e 3 + w 4 ~e 4 ) =(v w v w v 3 w 3 v 4 w 4 )~e +(v w + v w + v 3 w 4 v 4 w 3 )~e +(v w 3 v w 4 + v 3 w + v 4 w )~e 3 +(v w 4 + v w 3 v 3 w + v 4 w )~e 4 satisfies the defining properties a field of scalars, except that multiplication is not commutative. Proof. Since Section. already demonstrates that vector addition in R 4 satisfies the four properties for a field of scalars that only involve addition, it suffices to verify the four of the properties that involve vector multiplication. Let ~u,~v, ~w R 4 be arbitrary vectors. The associativity and distributivity of addition and multiplication in R give (~u~v)~w = (u v u v u 3 v 3 u 4 v 4 )w (u v + u v + u 3 v 4 u 4 v 3 )w (u v 3 u v 4 + u 3 v + u 4 v )w 3 (u v 4 + u v 3 u 3 v + u 4 v )w 4 ~e + (u v u v u 3 v 3 u 4 v 4 )w +(u v + u v + u 3 v 4 u 4 v 3 )w +(u v 3 u v 4 + u 3 v + u 4 v )w 4 (u v 4 + u v 3 u 3 v + u 4 v )w 3 ~e + (u v u v u 3 v 3 u 4 v 4 )w 3 (u v + u v + u 3 v 4 u 4 v 3 )w 4 +(u v 3 u v 4 + u 3 v + u 4 v )w +(u v 4 + u v 3 u 3 v + u 4 v )w ~e 3 + (u v u v u 3 v 3 u 4 v 4 )w 4 +(u v + u v + u 3 v 4 u 4 v 3 )w 3 (u v 3 u v 4 + u 3 v + u 4 v )w +(u v 4 + u v 3 u 3 v + u 4 v )w ~e 4 = u (v w v w v 3 w 3 v 4 w 4 ) u (v w + v w + v 3 w 4 v 4 w 3 ) u 3 (v w 3 v w 4 + v 3 w + v 4 w ) u 4 (v w 4 + v w 3 v 3 w + v 4 w ) ~e + u (v w + v w + v 3 w 4 v 4 w 3 )+u (v w v w v 3 w 3 v 4 w 4 ) +u 3 (v w 4 + v w 3 v 3 w + v 4 w ) u 4 (v w 3 v w 4 + v 3 w + v 4 w ) ~e + u (v w 3 v w 4 + v 3 w + v 4 w ) u (v w 4 + v w 3 v 3 w + v 4 w ) +u 3 (v w v w v 3 w 3 v 4 w 4 )+u 4 (v w + v w + v 3 w 4 v 4 w 3 ) ~e 3 + u (v w 4 + v w 3 v 3 w + v 4 w )+u (v w 3 v w 4 + v 3 w + v 4 w ) u 3 (v w + v w + v 3 w 4 v 4 w 3 )+u 4 (v w v w v 3 w 3 v 4 w 4 ) ~e 4 = ~u(~v~w), ~e ~v = ()(v ) (0)(v ) (0)v 3 (0)v 4 ~e + ()(v )+(0)(v )+(0)(v 4 ) (0)(v 3 ) ~e + ()(v 3 ) (0)(v 4 )+(0)(v )+(0)(v ) ~e 3 + ()(v 4 )+(0)(v 3 ) (0)(v )+(0)(v ) ~e 4 = v ~e + v ~e + v 3 ~e 3 + v 4 ~e 4 = ~v,

14 34 linear algebra ~u(~v + ~w) = u (v + w ) u (v + w ) u 3 (v 3 + w 3 ) u 4 (v 4 + w 4 ) ~e + u (v + w )+u (v + w )+u 3 (v 4 + w 4 ) u 4 (v 3 + w 3 ) ~e + u (v 3 + w 3 ) u (v 4 + w 4 )+u 3 (v + w )+u 4 (v + w ) ~e 3 + u (v 4 + w 4 )+u (v 3 + w 3 ) u 3 (v + w )+u 4 (v + w ) ~e 4 = (u v u v u 3 v 3 u 4 v 4 )+(u w u w u 3 w 3 u 4 w 4 ) ~e + (u v + u v + u 3 v 4 u 4 v 3 )+(u w + u w + u 3 w 4 u 4 w 3 ) ~e + (u v 3 u v 4 + u 3 v + u 4 v )+(u w 3 u w 4 + u 3 w + u 4 w ) ~e 3 + (u v 4 + u v 3 u 3 v + u 4 v )+(u w 4 + u w 3 u 3 w + u 4 w ) ~e 4 = ~u~v + ~u~w, which establishes the associativity, existence of a multiplicative identity, and distributivity for vector multiplication in R 4. Lastly, if ~v 6=~0, then we have v + v + v 3 + v 4 6= 0. Hence, commutativity of multiplication in R yields ~v(v ~e v ~e v 3 ~e 3 v 4 ~e 4 )=(v + v + v 3 + v 4 )~e +( v v + v v v 3 v 4 + v 4 v 3 )~e +( v v 3 + v v 4 + v 3 v v 4 v )~e 3 ( v v 4 v v 3 + v 3 v v 4 v )~e 4 =(v + v + v 3 + v 4 )~e, so (v + v + v 3 + v 4 ) (v ~e v ~e v 3 ~e 3 v 4 ~e 4 ) R 4 is the multiplicative inverse of the vector ~v. i j k i j k i i k j j j k i k k j i Figure.3: The multiplication table for the quaternion units.3. Notation. The quaternions are denoted by H. Traditionally, one identifies the standard basis vectors ~e,~e,~e 3,~e 4 R with the quaternion units, i, j, k H, so a~e + b~e + c~e 3 + d~e 4 = a + b i + c j + d k and the multiplicative identity is..3. Problem. Show that quaternion units satisfy the relations i = j = k = ijk=. Solution. The definition of the multiplication in H implies that i = (0)(0) ()() (0)(0) (0)(0) + (0)()+()(0)+(0)(0) (0)(0) i + (0)(0) ()(0)+(0)(0)+(0)() j + (0)(0)+()(0) (0)()+(0)(0) k =, j = (0)(0) (0)(0) ()() (0)(0) + (0)(0)+(0)(0)+()(0) (0)() i + (0)() (0)(0)+()(0)+(0)(0) j + (0)(0)+(0)() ()(0)+(0)(0) k =, k = (0)(0) (0)(0) (0)(0) ()() + (0)(0)+(0)(0)+(0)() ()(0) i + (0)(0) (0)()+(0)(0)+()(0) j + (0)()+(0)(0) (0)(0)+()(0) k =, ij= (0)(0) ()(0) (0)() (0)(0) + (0)(0)+()(0)+(0)(0) (0)() i + (0)() ()(0)+(0)(0)+(0)(0) j + (0)(0)+()() (0)(0)+(0)(0) k = k, and i j k = k =.

15 vector products 35 Exercises.3.3 Problem. Determine which of the following statements are true. If a statement is false, then provide a counterexample. i. The quaternions form a field of scalars. ii. The quaternion units satisfy j i = i j..3.4 Problem. From the basic relations i = j = k = ijk = deduce the following relations: ii=, ij= k, ik= j, ji= k, jj=, jk= i, ki= j, kj= i, kk=..3.5 Problem. For any numbers v, v, v 3, w, w, w 3 R, consider the quaternions p := v i + v j + v 3 k H and q := w i + w j + w 3 k H, and the corresponding vectors ~p := v ~e + v ~e + v 3 ~e 3 R 3 and ~q := w ~e + w ~e + w 3 ~e 3 R 3. How is the quaternion product pq related to the cross product ~p ~q?.3.6 Problem (Properties of the quaternion conjugate). For any quaternion p := v + v i + v 3 j + v 4 k H, with v, v, v 3, v 4 R, its conjugate q is p := v v i v 3 j v 4 k H and its absolute value is p := v + v + v 3 + v 4. For all p, q H, verify the following properties of the quaternion conjugate. i. pq = q p ii. p = p iii. p p = p iv. p + p R.3.7 Problem. Prove the following related identities. i. For all p, q H, show that pq = p q. ii. For all v, v, v 3, v 4, w, w, w 3, w 4 R, show that (v + v + v 3 + v 4 )(w + w + w 3 + w 4 )=(v w v w v 3 w 3 v 4 w 4 ).3.8 Problem. If h H, then show that there exists a, b R such that h = ah+ b..3.9 Problem. If h H, then evaluate the expression h i h i + i h i h i h i h h i h i..3.0 Problem. For any h := a i + b j + c k H such that a, b, c R and a + b + c =, then prove that h = 0, which show that quadratic polynomials may have infinitely many roots over H. +(v w + v w + v 3 w 4 v 4 w 3 ) +(v w 3 v w 4 + v 3 w + v 4 w ) +(v w 4 + v w 3 v 3 w + v 4 w ).