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Mathematics IA Algebra and Geometry (Part I) Michaelmas Term 2002 lecturer: Professor Peter Haynes (phhdamtp.cam.ac.uk) November 15, 2004 1 Complex Numbers 1.1 Introduction Real numbers, (denoted by R), consist of: integers (denoted by Z)···−3, −2, −1, 0, 1, 2, ... rationals (denoted by Q) p/q where p, q are integers √ 2 irrationals 2, π, e, π etc It is often useful to visualise real numbers as lying on a line Complex numbers (denoted by C): 2 If a, b∈R, then z =a+ib∈C (’∈’means belongs to), where i is such that i =−1. If z =a+ib, then write a =Re(z) (real part of z) b =Im(z) (imaginary part of z) Extending the number system from real (R) to complex (C) allows certain important generalisations. For example, in complex numbers the quadratic equation 2 αx +βx+γ = 0 : α, β, γ ∈R ,α6= 0 always has two roots p p 2 2 β + β −4αγ β− β −4αγ x =− x =− 1 2 2α 2α where 2 x , x ∈R if β ≥ 4αγ 1 2 2 x , x ∈C if β 4αγ, when 1 2 1 Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.P.H.Haynes Mathematics IA Algebra and Geometry Michaelmas Term 2002 2 p p 2 2 4αγ−β 4αγ−β β β x =− +i , x =− −i 1 2 2α 2α 2α 2α Note: C contains all real numbers, i.e. if a∈R then a+i.0∈C. A complex number 0+i.b is said to be ’pure imaginary’ Algebraic manipulation for complex numbers: simply follow the rules for reals, 2 adding the rule i =−1. Hence: addition/subtraction : (a+ib)±(c+id) = (a±c)+i(b±d) multiplication : (a+ib)(c+id) = ac+ibc+ida+(ib)(id) = (ac−bd)+i(bc+ad) a ib −1 inverse : (a+ib) = − 2 2 2 2 a +b a +b −1 Check from the above that z.z = 1+i.0 All these operations on elements of C result in new elements of C (This is described as ‘closure’: C is ‘closed under addition’ etc.) Wemayextendtheideaoffunctionstocomplexnumbers. Thecomplex-valuedfunctionf takes any complex number as ‘input’ and defines a new complex number f(z) as ‘output’. New definitions ∗ Complex conjugate of z =a+ib is defined as a−ib, written as z (sometimes z ). −1 The complex conjugate has the properties z ±z = z ±z , z z = z z , (z ) = 1 2 1 2 1 2 1 2 −1 (z) . 1/2 2 2 Modulus of z =a+ib defined as (a +b ) and written asz. 2 −1 2 Note thatz =zz and z =z/(z ). Theorem 1.1: The representation of a complex numberz in terms of real and imaginary parts is unique. Proof: Assume∃ a,b,c,d real such that z =a+ib =c+id. 2 2 Then a−c =i(d−b), so (a−c) =−(d−b) , so a =c and b =d. It follows that if z =z : z ,z ∈C, then Re(z ) =Re(z ) and Im(z ) =Im(z ). 1 2 1 2 1 2 1 2 Definition: Given a complex-valued function f, the complex conjugate function f is defined by f (z) =f (z) Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.P.H.Haynes Mathematics IA Algebra and Geometry Michaelmas Term 2002 3 2 2 For example, if f (z) = pz +qz +r with p,q,r∈C then f (z)≡ f (z) = pz + qz + r. 2 Hence f (z) =pz +qz+r. This example generalises to any function defined by addition, subtraction, multiplication and inverse. 1.2 The Argand diagram Consider the set of points in 2D referred to Cartesian axes. We can represent each z =x+iy∈C by the point (x,y). → Label the 2D vector OP by the complex number z. This defines the Argand diagram (or the ‘complex plane’). Invented by Caspar Wessel (1797) and re-invented by Jean Robert Argand (1806) Call the x-axis, the ‘real axis’ and the y-axis, the ‘imaginary axis’. → Modulus: the modulus of z corresponds to the magnitude of the vector OP, z = 1/2 2 2 (x +y ) . → → ′ ′ Complex conjugate: if OP represents z, then OP represents z, where P is the point (x,−y) (i.e. P reflected in the x-axis). Addition: if z = x + iy associated with P , z = x + iy associated with P , then 1 1 1 1 2 2 2 2 z +z = (x +x )+i(y +y ). 1 2 1 2 1 2 z +z = z is associated with the point P , obtained by completing the parallelogram 1 2 3 3 → → → P OP P ’ i.e.’ as vector additionOP =OP +OP (sometimes called the ‘triangle law’). 1 2 3 3 1 2 Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.P.H.Haynes Mathematics IA Algebra and Geometry Michaelmas Term 2002 4 Theorem 1.2: If z ,z ∈C then 1 2 (i)z +z ≤z +z 1 2 1 2 (ii)z −z ≥(z −z ) 1 2 1 2 (i) is the triangle inequality. By the cosine rule 2 2 2 z +z =z +z −2z z cosψ 1 2 1 2 1 2 2 2 2 ≤z +z +2z z = (z +z ) 1 2 1 2 1 2 . ′ ′ ′ ′ (ii) follows from (i), putting z + z = z , z = z , so z = z − z . Hence, by (i), 1 2 2 1 1 2 1 2 ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ z ≤ z −z +z and z −z ≥ z −z . Now interchanging z and z , we have 1 1 2 2 1 2 1 2 1 2 ′ ′ ′ ′ ′ ′ z −z ≤z −z =z −z , hence result. 2 1 2 1 1 2 Polar (modulus/argument) representation Use plane polar co-ordinates to represent position in Argand diagram. x = rcosθ and y =rsinθ, hence z =x+iy =rcosθ+isinθ =r(cosθ+isinθ) 1/2 2 2 Note that z = (x +y ) = r, so r is the modulus of z (‘ mod (z)’ for short). θ is called the ‘argument’ of z (‘arg(z)’ for short). The expression for z in terms of r and θ is called the ‘modulus/argument form’. The pair (r,θ) specifies z uniquely, but z does not specify (r,θ) uniquely, since adding 2nπ toθ (n integer) does not changez. For eachz there is a unique value of the argument θ such that−π θ≤π, sometimes called the principal value of the argument. Geometric interpretation of multiplication Consider z ,z written in modulus argument form 1 2 z =r (cosθ +isinθ ) 1 1 1 1 z =r (cosθ +isinθ ) 2 2 2 2 z z =r r (cosθ .cosθ −sinθ .sinθ 1 2 1 2 1 2 1 2 +isinθ .cosθ +sinθ .cosθ ) 1 2 2 1 =r r cos(θ +θ )+isin(θ +θ ) 1 2 1 2 1 2 Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.P.H.Haynes Mathematics IA Algebra and Geometry Michaelmas Term 2002 5 Multiplication of z by z , rotates z by θ and scales z byz 2 1 2 1 2 1 z z =z z 1 2 1 2 arg(z z ) = arg(z )+arg(z )(+2kπ,with k an arbitrary integer.) 1 2 1 2 1.3 De Moivre’s Theorem: complex exponentials n Theorem 1.3 (De Moivre’s Theorem): (cosθ+isinθ) = cosnθ+isinnθ where θ∈R and n∈Z For n 0 prove by induction p Assume true for n =p : (cosθ+isinθ) = cospθ+isinpθ p+1 p Then (cosθ+isinθ) = (cosθ+isinθ)(cosθ+isinθ) = (cosθ+isinθ)(cospθ+isinpθ) = cosθ.cospθ−sinθ.sinpθ+isinθ.cospθ+cosθ.sinpθ = cos(p+1)θ+isin(p+1)θ, hence true for n =p+1 Trivially true for n = 0, hence true∀n by induction Now consider n 0, say n =−p −1 −p p (cosθ+isinθ) =(cosθ+isinθ) −1 =cospθ+isinpθ = 1/(cospθ+isinpθ) = cospθ−isinpθ = cosnθ+isinnθ Hence true∀n∈Z x Exponential function: exp x =e ∞ P 2 n define by power series exp x = 1+x+x /2··· = x /n n=0 (This series converges for all x∈R — see Analysis course.) It follows from the series that (exp x)(exp y) = exp(x+y) for x,y∈R exercise 1 x This, plus exp1 = 1+1+ ..., may be used to justify the equivalence exp x =e 2 Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.P.H.Haynes Mathematics IA Algebra and Geometry Michaelmas Term 2002 6 ∞ P n Complex exponential defined by exp z = z /n , z ∈ C, series converges for all n=0 finitez z For short, write exp z =e as above. Theorem 1.4 iw exp(iw) =e = cos w+isin w , w∈C First consider w real, ∞ P n 2 3 exp(iw) = (iw) /n = 1+iw−w /2−iw /3... n=0 2 4 3 5 = (1−w /2+w /4...)+i(w−w /3+w /5...) ∞ ∞ P P n n 2n 2n+1 = (−1) w /(2n)+i (−1) w /(2n+1) = cos w+isin w n=0 n=0 Now define the complex functions ∞ ∞ P P n n 2n 2n+1 cos w = (−1) w /(2n) and sin w = (−1) w /(2n+1) for w∈C. n=0 n=0 iw Then exp(iw) =e = cos w+isin w , w∈C. −iw Similarly, exp(−iw) =e = cos w−isin w. 1 iw −iw 1 iw −iw It follows that cos w = (e +e ) and sin w = (e −e ). 2 2i Relation to modulus/argument form iθ Put w =θ , θ∈R , then e = cos θ+isin θ. iθ Hence, z =r(cos θ+isin θ) =re , with (again) r =z, θ = argz. Note that de Moivre’s theorem n cos nθ+isin nθ = (cos θ+isin θ) ¡ ¢ n inθ iθ may be argued to follow from e = e . Multiplication of two complex numbers: ¡ ¢¡ ¢ iθ iθ i(θ +θ ) 1 2 1 2 z z = r e r e =r r e 1 2 1 2 1 2 iθ Modulus/argument expression for 1: consider solutions of e = 1, hence cos θ + i sin θ = 1, cos θ = 1 , sin θ = 0, hence θ = 2kπ , with k∈Z, i.e. 2kπi e = 1. n Roots of Unity: a root of unity is a solution of z = 1, with z ∈ C and n a positive integer. n Theorem 1.5 There are n solutions of z = 1 (i.e. n ’nth roots of unity’) One solution is z = 1. ¡ ¢ n iθ iθ n niθ iθ Seekmoregeneralsolutionsoftheformre , re =r e = 1,hencer = 1, e = 1, hence nθ = 2kπ , k∈Z with 0≤θ 2π. Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.P.H.Haynes Mathematics IA Algebra and Geometry Michaelmas Term 2002 7 θ = 2kπ/n gives n distinct roots for k = 0,1,...,n−1, with 0≤θ 2π. 2πi/n n 2 n−1 Write ω =e , then the roots of z = 1 are 1, ω,ω ,... ,ω . n−1 n−1 P P n k n−1 k n Note ω = 1, also ω = 1+ω+···+ω = 0, because ω = (ω −1)/(ω−1) = k=0 k=0 0/(ω−1) = 0. 5 Example: z = 1. iθ 5iθ 2πki 2πi/5 Put z =e , hence e =e , hence θ = 2πk/5 , k = 0,1,2,3,4 and ω =e . 2 3 4 2 3 4 Roots are 1,ω,ω ,ω ,ω , with 1+ω+ω +ω +ω = 0 (each root corresponds to a side of a pentagon). 1.4 Logarithms and complex powers u If v∈R , v 0 , the complex equation e =v has a unique real solution, u = logv. w Definition: logz for z∈C is the solution w of e =z. u+iv iθ Set w =u+iv , u,v∈R, then e =z =re u hence e =z =r v = arg z =θ+2kπ , any k∈Z Thus, w = log z = log z+i arg z, with arg z, and hence log z a multivalued function. Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.P.H.Haynes Mathematics IA Algebra and Geometry Michaelmas Term 2002 8 Definition The principal value of logz is such that −π argz =Im(logz)≤π. Example: if z = −x , x ∈ R , x 0 then log z = log −x +i arg(−x) = log x +iπ+2kiπ k∈Z. The principal value of log(−x) is logx+iπ. Powers a Recall the definition of x , for x,a∈R , x 0 a alog x x =e = exp(alog x) w w wlog z Definition: For z = 6 0 , z,w∈C, define z by z =e . w 2πikw Note that since log z is multivalued so is z (arbitary multiple of e , k∈Z) Example: i ilog i i(logi+iarg i) i(log1+2kiπ+iπ/2) −π/2 −2kπ (i) =e =e =e =e ×e k∈Z. 1.5 Lines and circles in the complex plane Line: For fixedz andc∈C, z =z +λc, λ∈R represents points on straight line through 0 0 z and parallel to c. 0 ¯ Note that λ = (z−z )/c∈R, hence λ =λ, so 0 z−z z¯−z¯ 0 0 = c c¯ Hence zc¯−z¯c =z c¯−z¯ c 0 0 is an alternative representation of the line. Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.P.H.Haynes Mathematics IA Algebra and Geometry Michaelmas Term 2002 9 Circle: circle radius r, centre a (r∈R, a∈C) is given by S =z∈C :z−a=r, the set of complex numbers z such thatz−a =r. 2 2 2 2 If a =p+iq, z =x+iy thenz−a = (x−p) +(y−q) =r , i.e. the expression for a circle with centre (p,q), radius r in Cartesian coordinates. 2 An alternative description of the circle comes fromz−a = (z¯−a¯)(z−a), so 2 2 zz¯−a¯z−az¯+a −r = 0. 1.6 M¨obius transformations Consider a ’map’ of C→C (‘C into C’) az+b ′ z7→z =f (z) = cz+d where a,b,c,d∈C (all constant) and (i) c,d not both zero, (ii) a,c not both zero and (iii) ad = 6 bc. (i)ensuresf (z)finiteforsomez. (ii)and(iii)ensuredifferentz mapintodifferentpoints. Combine all these conditions into ad−bc6= 0. f(z) maps every point of the complex plane, except z =−d/c, into another. ′ ′ Inverse: z = (−dz +b)/(cz −a), which represents another M¨obius transformation. ′ For every z except a/c there is a corresponding z, thus f maps C\−d/c to C\a/c. Composition: consider a second M¨obius transformation ′ αz +β ′ ′′ ′ z 7→z =g(z ) = α,β,γδ∈C,αδ−βγ = 6 0. ′ γz +δ ′′ Then the combined map z7→z is also a M¨obius transformation. ′′ ′ z =g(z ) =g(f (z)) ′ αz +β α(az+b)+β(cz+d) = = ′ γz +δ γ(az+b)+γ(cz+d) (αa+βc)z+αb+βd = . (γa+δc)z+γb+δd The set of all M¨obius maps is therefore closed under composition. Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.P.H.Haynes Mathematics IA Algebra and Geometry Michaelmas Term 2002 10 Examples: ′ (i) (a = 1, c = 0, d = 1), z = z +b is translation. Lines map to parallel lines. Circles map to identical circles. ′ (ii)(b = 0, c = 0, d = 1), z =az, scales z bya and rotates by arga about O. ′ ′ ′ Line z =z +λp (λ∈R) becomes z =az +λap =z +λc — another line. 0 0 0 ′ ′ ′ ′ ′ Circlez−q =r becomesz/a−q =r, hencez −aq =ar, equivalentlyz −q =r — another circle. 1 ′ (iii) (a = 0, b = 1, c = 1, d = 0), z = , described as ‘inversion’ with respect to O. z Line z =z +λp or zp¯−z¯p =z p¯−z¯ p, becomes 0 0 0 p¯ p − =z p¯−z¯ p 0 0 ′ ′ z z¯ hence ′ ′ ′ ′ ¯ ¯ zp¯−zp = (z p¯−z¯ p)zz 0 0 ′ ′ ¯ zp¯ zp ′ ′ ¯ zz − − = 0 z p¯−z¯p z¯p−z p¯ 0 0 0 0 ¯ ¯ ¯ ¯ 2 2 ¯ ¯ ¯ ¯ p¯ p ′ ¯ ¯ ¯ ¯ z − = ¯ ¯ ¯ ¯ z p¯−z¯p z¯p−z p¯ 0 0 0 0 This is a circle through origin, except when z¯ p−z p¯ = 0 (which is the condition that 0 0 ′ ′ ¯ straight line passes through origin — exercise for reader). Then zp¯− zp = 0, i.e. a straight line through the origin. 1 ′ ′ ′ ′ ¯ Circle z−q= r becomes −q= r, i.e. 1−qz = rz, hence (1−qz )(1−q¯z ) = ′ z 2 ′ ′ 2 2 ′ ′ ′ ′ ¯ ¯ ¯ r zz , hence zz q −r −qz −q¯z +1 = 0, hence ¯ ¯ 2 2 ¯ ¯ q¯ q 1 ′ ¯ ¯ z − = − ¯ ¯ 2 2 2 2 2 2 2 q −r q −r (q −r ) 2 r = . 2 2 2 (q −r ) Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.P.H.Haynes Mathematics IA Algebra and Geometry Michaelmas Term 2002 11 2 2 2 2 2 2 This is a circle centre q¯/(q −r ), radius r/(q −r ), unless q = r (implying the ′ ′ ¯ original circle passed through the origin), when qz +q¯z = 1, i.e. a straight line. Summary: under inversion in the origin circles/straight lines → circles, except cir- cles/straight lines through origin→ straight lines (to be explained later in course). A general M¨obius map can be generated by composition of translation, scaling and rotation, and inversion in origin. Consider the sequence: scaling and rotation z7→z =cz (c = 6 0) 1 translation z 7→z =z +d 1 2 1 inversion in origin z 7→z = 1/z 2 3 2 ½ ¾ bc−ad scaling and rotation z 7→z = z (bc6=ad) 3 4 3 c translation z 7→z =z +a/c 4 5 4 Then z = (az+b)/(cz+d). (Verify for yourself.) 5 This implies that a general M¨obius map sends circles/straight lines to circles/straight lines (again see later in course for further discussion). Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.P.H.Haynes Mathematics IA Algebra and Geometry Michaelmas Term 2002 12 This is a blank page Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.P.H.Haynes Mathematics IA Algebra and Geometry Michaelmas Term 2002 13 2 Vectors 2.3 Vector Product The printed notes are not complete for this subsection – refer to notes taken in lectures for completeness. Geometrical argument for a×(b+c) =a×b+a×c. −1 ′′ Consider(a a)×b =b . Thisvectoristheprojectionofbontotheplaneperpendicular to a, rotated by π/2 clockwise about a. Consider this as two steps, first projection of b ′ ′ ′′ to give b, then rotation of b to give b . a b θ b’ is the projection of b onto the plane perpendicular to a b’ b’ = b sin θ a b’’ is the result of rotating the vector b’ through an angle π 2 b’’ clockwise about a (i.e looking in π 2 the direction of a ) b’ ′ Now note that if x is the projection of the vector x onto the plane perpendicular to a, ′ ′ ′ then b +c = (b+c). (See diagram below.) c a b b+c (b+c)’ c’ b’ (b+c)’ is the projection of b+c on to the plane perpendicular to a ′ ′ ′ Rotating b, c and (b+c) by π/2 gives the required result. Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.P.H.Haynes Mathematics IA Algebra and Geometry Michaelmas Term 2002 14 2.7 Polar Coordinates 2 Plane polars (r,θ) in R : x =rcosθ, y =rsinθ, with 0≤r ∞, 0≤θ 2π. y j i e θ e r P r θ x O e is the unit vector perpendicular to curves of constantr, in the direction ofr increasing. r e is the unit vector perpendicular to curves of constantθ, in the direction ofθ increasing. θ e =icosθ+jsinθ. r e =−isinθ+jcosθ. θ e .e = 0. r θ → x =OP=xi+yj =rcosθi+rsinθj =re . r Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.P.H.Haynes Mathematics IA Algebra and Geometry Michaelmas Term 2002 15 3 Cylindrical polars (ρ,φ,z) in R : x = ρcosφ, y = ρsinφ, z = z with 0 ≤ ρ ∞, 0≤φ 2π,−∞z ∞. z e z e φ P e ρ r z k O y j i φ e φ ρ N e ρ x e istheunitvectorperpendiculartosurfacesofconstantρ,inthedirectionofρincreasing. ρ e is the unit vector perpendicular to surfaces of constant φ, in the direction of φ increas- φ ing. e =k z e , e , e =k are a right-handed triad of mutually orthogonal unit vectors: ρ φ z e .e = 0, e .e = 0, e .e = 0. ρ φ z ρ φ z e ×e =e , e ×e =e , e ×e =e . ρ φ z z ρ φ φ z ρ e .(e ×e ) = 1. ρ φ z → → x =ON +NP=ρe +ze . ρ z Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.P.H.Haynes Mathematics IA Algebra and Geometry Michaelmas Term 2002 16 3 Spherical polars (r,θ,φ) in R : x = rsinθcosφ, y = rsinθsinφ, z = rcosθ with 0≤r ∞, 0≤θ≤π, 0≤φ 2π. z e r e φ P r e θ θ z k O y j i φ e φ ρ N x e istheunitvectorperpendiculartosurfacesofconstantr,inthedirectionofrincreasing. r e istheunitvectorperpendiculartosurfacesofconstantθ,inthedirectionofθincreasing. θ e is the unit vector perpendicular to surfaces of constant φ, in the direction of φ increas- φ ing. e , e , e are a right-handed triad of mutually orthogonal unit vectors: r θ φ e .e = 0, e .e = 0, e .e = 0. r θ θ φ φ r e ×e =e , e ×e =e , e ×e =e . r θ φ θ φ r φ r θ e .(e ×e ) = 1. r θ φ → x =OP=re . r Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.P.H.Haynes Mathematics IA Algebra and Geometry Michaelmas Term 2002 17 4 Linear Maps and Matrices 4.7 Change of basis 3 ConsiderchangefromstandardbasisofR tonewbasisη ,η ,η , linearlyindependent, 1 2 3 but not necessarily orthonormal (or even orthogonal). 3 Let x be any vector in R , then 3 3 X X x = xe = ξ η , i i k k i=1 k=1 whereξ are the components of x with respect to the new basis. k Consider x.e : j 3 X x.e =x = ξ η .e =P ξ j j k j jk k k k=1 where P is jth component of η (with respect to the standard basis). jk k We writex =Pξ (wherex andξ are to be interpreted as column vectors whose elements are the x and ξ ) where the matrix P is i i P = (η η η ) matrix with columns components of new basis vectors η 1 2 3 k Matrices are therefore a convenient way of expressing the changes in components due to a change of basis. P 3 Since the η are a basis, there exist E ∈R such that e = E η . Hence ki i ki k k k=1 3 3 3 3 3 X X XX X x = x ( E η ) = ( x E )η = ξ η . i ki i ki k k k k i=1 k=1 k=1 i=1 k=1 By uniqueness of components with respect to a given basis E x =ξ . ki i k 3 3 Thus we have Pξ = x and Ex = ξ, for all x ∈ R , so Pξ = x = PEx for all x ∈ R , −1 hence PE =I. Similarly EP =I, and hence E =P , so P is invertible. 3 3 ′ Now consider a linear map M : R → R under which x 7→ x = M(x) and (in terms ′ ′ of column vectors) x = Mx where x and x are components with respect to the i i standard basise. M is the matrix ofM with respect to the standard basis. i ′ ′ ′ From above x = Pξ and x = Pξ where ξ and ξ are components with respect to j j the new basisη . j ′ ′ −1 Thus Pξ =MPξ, hence ξ = (P MP)ξ. −1 P MP is the matrix of M with respect to the new basis η , where P = (η η η ), j 1 2 3 i.e. the columns of P are the components of new basis vectors with respect to old basis (and in this case the old basis is the standard basis). n m A similar approach may be used to deduce the matrix of the map N : R → R (where n m m = 6 n) with respect to new bases of both R and R . Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.P.H.Haynes Mathematics IA Algebra and Geometry Michaelmas Term 2002 18 n m Suppose E is standard basis of R and F is standard basis of R , and N is matrix i i ′ ′ of N with respect to these two bases, so X 7→ X = NX (where X and X are to be interpreted as column vectors of components). n m Now consider new basesη of R andφ of R , with P = (η ...η ) n×n matrix i i 1 n and Q = (φ ...φ ) m×m matrix. 1 m ′ ′ ′ Then X =Pξ, X =Qξ, where ξ and ξ are column vectors of components with respect to basesη andφ respectively. i i ′ ′ −1 −1 Hence Qξ =NPξ, implyingξ =Q NPξ. So Q NP is matrix of transformation with n m respect to new bases (of R and R ). Example: Consider simple shear in x direction within (x ,x ) plane, with magnitude 1 1 2 γ. Matrix with respect to standard basise ,e ,e is: 1 2 3   1 γ 0   0 1 0 =M 0 0 1 Now consider matrix of this transformation with respect to basisη ,η ,η , where 1 2 3 η = cosψe +sinψe 1 2 1 η =−sinψe +cosψe 1 2 2 η =e 3 3 Then     cosψ −sinψ 0 cosψ sinψ 0 −1     P = sinψ cosψ 0 which is orthogonal, so P = −sinψ cosψ 0 0 0 1 0 0 1 Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.P.H.Haynes Mathematics IA Algebra and Geometry Michaelmas Term 2002 19 −1 Matrix with respect to new basis is P MP =    cosψ sinψ 0 cosψ+γsinψ −sinψ+γcosψ 0    = −sinψ cosψ 0 sinψ cosψ 0 0 0 1 0 0 1   2 1+γsinψcosψ γcos ψ 0 2   = −γsin ψ 1−γsinψcosψ 0 0 0 1 Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.P.H.Haynes Mathematics IA Algebra and Geometry Michaelmas Term 2002 20 This is a blank page Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.

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