Sample Solutions from the Student Solution Manual

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1 1 Sample Solutions from the Student Solution Manual 1213 If all the entries are, then the matrix is certainly not invertile; if you multiply the matrix y anything, you get the matrix, not the identity So assume that one entry is not Let us suppose d Ifad = c, [ then the first row is a multiple of the second: we can write a = d c and = d d,so the matrix is A = d c d d ] If we had supposed that any other entry was nonzero, the proof would c d a work the same way) If A is invertile, there exists a matrix B = 1 c d such that AB = 1 But if the upper left-hand corner of AB is 1, we have d a c + c d) = 1, so the lower left-hand corner, which is a c + c d, cannot e a) Every linear transformation T : R 4 R 2 is given y a 2 4 matrix, eg, A = ) Any row matrix three wide will do, for example, [1, 1, 2]; such a matrix takes a vector in R 3 and gives a numer Remark On one exam at Cornell University, the first question was What is a linear transformation T : R n R m? Several students gave answers like A function taking some vector v R n and producing some vector w R m Besides not eing a complete sentence, this is wrong! The student who gave this answer may have een thinking of matrix multiplication But a mapping from [ R] n to R m need not e given y x x 2 matrix multiplication; for example, consider the mapping y y 2 In any case, defining a linear transformation y saying that it is given y matrix multiplication really egs the question, ecause it does not explain why matrix multiplication is linear The correct answer was given y another student in the class: A linear transformation T : R n R m is a mapping R n R m such that for all a, R n, T a + )=T a)+t ), and for all a R n and all scalar r, T r a) =rt a) Linearity, and the approximation of nonlinear mappings y linear mappings, is a key motif of this ook You must know the definition, which gives you a foolproof way to check whether or not a given mapping is linear 137 It is enough to know what T gives when evaluated on the three standard asis vectors 1, The matrix of T is , 131 Set fa) =A 1 and ga) =AA + A A Then F = f g, and we wish to compute [DF A)]H =[Df ga)]h =[DfgA))][DgA)]H =[DfAA + A A)] [DgA)]H }{{} new increment for Df 1) Rememer see Example 1717) that we cannot treat Equation 1) as matrix multiplication To treat the derivatives of f and g as matrices, we would have to identify Mat n, n) with R n2 ; the derivatives would e n 2 n 2 matrices Instead we think of the derivatives as linear transformations

2 2 The linear terms in H of ga + H) ga) =A+H)A + H) +A+H) A+H) AA A A are AH + HA + A H + H A; this is [DgA)]H, which is the new increment for Df We know from Proposition 1718 that [DfA)]H = A 1 HA 1, which we will rewrite as [DfB)]K = B 1 KB 1 2) to avoid confusion We sustitute AH + HA + A H + H A for the increment K in Equation 2) and ga) =AA + A A for B This gives [DF A)]H =[DfAA + A A)][DgA)]H = AA + A A) 1 AH + HA + A H + H A) AA + A A) 1 }{{}}{{}}{{} B 1 K B 1 There is no ovious way to simplify this expression 247 Let A e an n n matrix The product A A is then a 1 a 2 } a n {{ } A A {}}{ a 1 a 2 a n a 1 2 a 1 a 2 a 1 a n a 2 a 1 a 2 2 a 2 a n a n a 1 } a n a 2 {{ a n 2 } A A The diagonal entries are given y the length squared of the columns of A, since a i a i = a i a i = a i 2 All other entries are dot products of two different columns of A IfA A=I, so that all entries not on the diagonal are, while those on the diagonal are 1, then the columns of A are orthogonal and have length 1 Thus they form an orthonormal asis of R n, and A is said to e orthogonal Similarly, if A is orthogonal, then the length of each of its column vectors is 1, so that A A has 1 s on the diagonal, and the dot product of two non-identical columns is, giving s for all other entries of A A 2511 a) If a 2, then dimker A)) =, so in that case the image has dimension 2 If a = 2, the image and the kernel have dimension 1 ) This is more complicated By row operations, we can ring the matrix B to 1 2 a a a 2a a We now separate the case and = If, then we can do further row operations to ring the matrix to the form 1 a 2a a a a 1 a 2a) a a The entry in the 3rd row, 3rd column is a 2 2a +2a)

3 3 So if, and the point a is neither on the line a = nor on the hyperola of equation 2 2a+2a =, the matrix has rank 3, whereas if and the point a is on one of these curves, the matrix has If =, the matrix is 1 2 a 2a a, which evidently has rank 3 unless a =, in which case it a has rank 1 rank 3 rank 3 rank 1 a rank 3 rank 1 a Figure for Solution 2511 Left: On the curves, the kernel of A has dimension 1 and its image has rank 1 Elsewhere, the rank is 2 ie, the kernel has dimension ) Right: Along the curves and on the -axis, the image of B has, ie, its kernel has dimension 1 At the origin the rank is 1 and the dimension of the kernel is 2 Elsewhere, the kernel has dimension and the rank of the image is This is a difficult prolem; we have criteria guaranteeing that susets X R n are manifolds, ut none guaranteeing that they are not We will outline two possile solutions First solution A k-dimensional manifold X R n has the property that for any k 1)-dimensional manifold Y R n such that Y X R n, then for any y Y and for any r> sufficiently small, Y cuts X B r y) into exactly two pieces In our case, X will e the set of positions X 2, and Y will e the set of positions where the four vertices are aligned We must check that Y is a 3-dimensional manifold, ut it is pretty clearly parametrized y the position of x 1 and the polar angle of the straight) linkage But if we remove Y from X 2, then x 1 x 3 <l 1 +l 2, so the vertices x 2 and x 4 are not on the line joining x 1 to x 3 Locally X 2 Y has four pieces: the piece where oth x 2 and x 3 are to the left of the line segment going from x 1 to x 4, the piece where they are oth to the right, the piece where x 2 is to the left and x 4 to the right, and the piece where x 4 is to the left and x 2 to the right Second solution Manifolds are invariant under rotations, so we may assume that our linkage is on the x-axis If the set of positions is to e a manifold in R 8, then it must locally e the graph of a function expressing four of the variales in terms of the others We will see that this is impossile for all the possile cominations First, we cannot use the positions of any two vertices Evidently we cannot use any pair of adjacent ones, since the distance etween them is fixed, and so they cannot oth e chosen freely We cannot choose either of the pairs of opposite ones either: x 1 and x 3 cannot e further apart than l 1 + l 2, and x 2 and x 4 cannot e closer than l 1 l 4 How aout the position of one vertex, and either the x or y coordinate of two of the others? First, if we use the position of one, we cannot use the x coordinate of any of the others why?) But how aout x 1 and y 2,y 4? Do these specify a unique position? This is harder to see Clearly x 1,y 2 and y 4 with

4 4 oth y s small) specify a unique position of x 2 and x 4 close to the original position But this leaves two positions of x 2 close to the original The situation is similar for the other vertices Finally, one coordinate for each vertex We cannot use the x-coordinates of two vertices, ecause that would allow us to put them closer together or further apart than is allowed We cannot use all four y-coordinates, as once we have found one position, we can translate it horizontally to find infinitely many positions with the same y-coordinates Finally, we must investigate the case of one x- coordinate and three y-coordinates The three y-coordinates determine three of the points, and the fourth must lie on the intersection of two circles, which intersect in two points, oth close to the original point 523 By Definition 3116, S is not a parametrization, ut it is a parametrization y the relaxed definition of a parametrization, Definition 523 See the discussion at the eginning of the section) It is not difficult to show that S is equivalent to the spherical coordinates map using latitude and longitude Definition 416) Any point Sx) is on a sphere of radius r, since r 2 sin 2 ϕ cos 2 θ + r 2 sin 2 ϕ sin 2 θ + r 2 cos 2 ϕ = r 2 Since r, the mapping S is onto The angle ϕ tells what latitude a point is on Going from to π, it covers every possile latitude The polar angle θ tells what longitude the point is on; going from to 2π, it covers every possile longitude However, S is not a parametrization y Definition 3116, ecause it is not one to one Troule occurs in several places If r =,Sx) is the origin, regardless of the values of θ and ϕ Forθ, troule occurs when θ = and θ =2π; for fixed r and ϕ, these two values of θ give the same point For ϕ, troule occurs at and π; ifϕ=π, the point is the south pole of a sphere of radius r, regardless of θ, and if ϕ =, the point is the north pole of a sphere of radius r, regardless of θ By Definition 523, S is a parametrization Put simply, the troule occurs only on a set of 3- dimensional volume In the language of Definition 523, for ϕ for r for θ {}}{{}}{{}}{ M = R 3, U = [, ) [, 2π] [,π], X = {} [, 2π] [,π] [, ) {,2π} [,π] }{{}}{{} troule when r= troule when θ= or θ=2π [, ] [, 2π] {,π} }{{} troule when ϕ= or ϕ=π Thus troule occurs on a union of five surfaces, which you can think of as five sides of a ox whose sixth side is at infinity The ase of the ox lies in the plane where r = ; the sides of the ase have length π and 2π respectively The ase of the ox represents the set laeled troule when r = Two parallel sides of the ox stretching to infinity represent the set laeled troule when θ = orθ = 2π The other set of parallel sides represents the set laeled troule when ϕ = orϕ = π By Proposition 436 and Definition 521, these surfaces have 3-dimensional volume Next let us check condition 4 of Definition 523), that the derivative is one to one for all u in U X This will e true Theorem 485) if and only if the determinant of the derivative is not The determinant is ) = det det DS r θ ϕ sin ϕ cos θ r sin ϕ sin θ rcos ϕ cos θ sin ϕ sin θ rsin ϕ cos θ rcos ϕ sin θ = 2r 2 sin ϕ, cos ϕ r sin ϕ which is only if r =,ϕ=orϕ=π, which are not in U X Next, we will show that S :U X) R 3 is of class C 1 with locally Lipschitz derivative We know the first derivatives exist, since we just computed the derivative They are continuous, since they are all polynomials in r, sine, and cosine, which are continuous see Theorem 1528 on comining continuous

5 functions) In fact, S :U X) R 3 is C, since its derivatives of all order are polynomials in r, sine, and cosine, and are thus continuous We do not need to check anything aout Lipschitz conditions ecause Proposition 271 says that the derivative of a C 2 function is Lipschitz Finally, we need to show that SX) has 3-dimensional volume If r =, then S r θ is the origin, ϕ whatever the values of θ and ϕ If θ = we have S r = ϕ x, z)-plane If θ =2π, we get the same surface If ϕ =,weget and if ϕ = π, we get the z-axis going from to r r sin ϕ r cos ϕ r 637 a) The asis v 1, v 2 is direct for V oriented y dx dz since dx dz 1, 1 1 = det = , which is a surface in the, the z-axis going from to, ) The asis w 1, w 2 is also direct since dx dz 2 3, 1 2, 2 1 = det = Since w 1 =2 v 1 3 v 2 and w 2 = v 1 +2 v 2, the change of asis matrix is, with det c) The determinant is 1/7, the inverse of the determinant in part ) This uses Theorem 4811) We can also do it y direct computation: [ since ] v 1 =2/7) w 1 +3/7) w 2 and v 2 = 1/7) w 1 +2/7) w 2, 2/7 1/7 the change of asis matrix is, with determinant 1/7 But this is unnecessary work 3/7 2/ a) The form ω does not orient M Here are two solutions Computational solution: Parametrize the circle centered at cos θ y θ A vector in the tangent space to this sin θ sin θ circle can e written, and cos θ sin θ xdy ydx) = 3 + cos θ) cos θ sin θ sin θ) =3cosθ+1 cos θ So ω vanishes on the tangent space to the circle at θ = arccos 1/3) More elegant solution: [ As ] seen in the figure elow, at two points on C 2, the tangent spaces to C 2 consist of radial vectors x x At those points, ω does not orient C y 2, since xdy ydx =xy xy = y

6 6 1 3 C 1 C 2 Figure[ for ] Solution 6317, part a) At two points on C 2, the tangent spaces to C 2 consist of radial x vectors The form ω = xdy ydx vanishes on those tangent spaces, so ω does not orient C 2 y 6519 a) No, it does not preserve orientation, since ) u det Dγ cos v cos w u cos w sin v R sin w u cos v sin w v = det cos v sin w u sin w sin v Rcos w + u cos v cos w = Ru u 2 cos v w sin v ucos v This quantity is always negative, ecause y definition u and R are positive and Ru > u 2 cos v since u u cos v u and u r<r, so that even when v is etween π/2 and 3π/2 so that cos v is negative) still Ru u 2 cos v< ) The integral ecomes = 2π 2π r 2π 2π r =2π f γ 2π r = 2π 2π u v w fγu)) det Dγ u v w det[dγ] ) du dv dw {}}{{}}{ R + u cos v) 2 ur + u cos v) du dv dw R 3 u 3R 2 u 2 cos v 3Ru 3 cos 2 v u 4 cos 3 v) du dv R 3 r 2 + R 2 r 3 cos v + 3Rr4 cos 2 v + r5 cos 3 ) v dv ) = π 2 2R 3 r 2 + 3Rr4 2 Since the parametrization reverses orientation, we should multiply that integral y a) If you approach this is a straightforward manner, integrating everything in sight, this is a long computation To make it earale one needs to 1) not compute terms that don t need to e computed, since they will disappear in the limit, and 2) take advantage of cancellations early in the game To interpret the answer, it also helps to know what one expects to find more on that later) Computing the exterior derivative from the definition means computing the integral on the right-hand side of dz 2 dx dy) P x v 1, v 2, v 3 ) ) 1 = lim h h 3 z 2 dx dy P xh v 1,h v 2,h v 3) Thus we must integrate the 2-form field z 2 dx dy over the oundary of the 3-parallelogram spanned y h v 1,h v 2,h v 3, ie, over the six faces of the 3-parallelogram shown elow

7 7 hv 3 hv 3 hv 2 x hv x 1 + hv 1 Figure for solution 675 The 3-parallelogram spanned y h v 1,h v 2,h v 3 We parametrize those six faces as follows, where s, t h, and x = x y ; we use Proposition z 6615 to determine which faces are taken with a plus sign and which are taken with a minus sign: 1 γ st = x + s v 1 + t v 2, minus This is the ase of the ox shown aove) 2 γ st = x + h v 3 + s v 1 + t v 2, plus This is the top of the ox) 3 γ st = x + s v 1 + t v 3, plus This is the front of the ox) 4 γ st = x + h v 2 + s v 1 + t v 3, minus This is the ack of the ox) 5 γ st = x + s v 2 + t v 3, minus This is the left side of the ox) 6 γ st = x + h v 1 + s v 2 + t v 3, plus This is the right side of the ox) Note that the two first faces are spanned y v 1 and v 2 or translates thereof), the ase taken with and the top taken with + The next two faces, with opposite signs, are spanned y v 1 and v 3, and the two sides, with opposite signs, are spanned y v 2 and v 3 To integrate our form field over these parametrized domains we use Definition 621 The computations are tedious, ut we do not have to integrate everything in sight For one thing, common terms that cancel can e ignored; for another, anything that amounts to a term in h 4 can e ignored, since h 4 /h 3 will vanish in the limit as h We will compute in detail the integrals over the first two faces The integral over the first face the ase) is the following, where v 1,3 denotes the third entry of v 1 and v 2,3 denotes the third entry of v 2 : z 2 dx dy eval at γ st ) D s γ, D t γ {}}{{}}{ z + sv 1,3 + tv 2,3 ) 2 dx dy v 1, v 2 ) ds dt = }{{} z 2 + s 2 v1,3 2 + t 2 v 2 ) 2,3 +2stv 1,3 v 2,3 +2szv 1,3 +2tzv 2,3 v1,1 v 2,2 v 1,2 v 2,1 ) ds dt }{{}}{{} term in h 2 terms in h 4 terms in h 3 Note that the integral z2 ds dt will give a term in h 2, and the next three terms will give a term in h 4 ; for example, h s2 v1,3 2 ds gives an h 3 and h s2 v1,3 2 dt gives an h, making h 4 in all These higher

8 8 degree terms can e disregarded; we will denote them elow y Oh 4 ) This gives the following integral over the first face: z 2 +2szv 1,3 +2tzv 2,3 + Oh 4 ) ) v 1,1 v 2,2 v 1,2 v 2,1 ) ds dt = h 2 z 2 + h 3 zv 1,3 + h 3 zv 2,3 + Oh 4 ) ) v 1,1 v 2,2 v 1,2 v 2,1 ) Before computing the integral over the second face the top), notice that it is exactly like the first face except that it also has the term h v 3 This face comes with a plus sign, while the first face comes with a minus sign, so when we integrate over the second face, the identical terms will cancel each other Thus we didn t actually have to compute the integral over the first face at all! In computing the contriution of the first two faces to the integral, we need only concern ourselves with those terms in z+hv 3,3 +sv 1,3 +tv 2,3 ) 2 that contain hv 3,3 : ie, h 2 v3,1, 2 2hsv 3,3 v 1,3,2htv 3,3 v 2,3, and 2zhv 3,3 Integrating the first three would give terms in h 4, so the entire contriution to the integral of the first two faces is 2zhv3,3 + Oh 4 ) ) v 1,1 v 2,2 v 1,2 v 2,1 )ds dt =2zh 3 v 3,3 + Oh 4 ))v 1,1 v 2,2 v 1,2 v 2,1 ) Similarly, the entire contriution of the second pair of faces is 2zhv2,3 + Oh 4 ) ) v 1,1 v 3,2 v 3,1 v 1,2 )ds dt =2zh 3 v 2,3 + Oh 4 ))v 1,1 v 3,2 v 3,1 v 1,2 ) Note that the partial derivatives are different, so we have v 1,1 v 3,2 v 3,1 v 1,2 ), not v 1,1 v 2,2 v 1,2 v 2,1 ) as efore) And the contriution of the last pair of faces is 2zhv1,3 + Oh 4 ) ) v 2,1 v 3,2 v 2,2 v 3,1 )ds dt =2zh 3 v 1,3 + Oh 4 ))v 2,1 v 3,2 v 2,2 v 3,1 ) Dividing y h 3 and taking the limit as h gives dz 2 dx dy) P x v 1, v 2, v 3 ) ) =2zv 3,3 )v 1,1 v 2,2 v 1,2 v 2,1 )+2zv 2,3 )v 1,1 v 3,2 v 3,1 v 1,2 ) +2zv 1,3 )v 2,1 v 3,2 v 2,2 v 3,1 ) Now it helps to know what one is looking for Since dz 2 dx dy) is a 3-form on R 3, it is a multiple of the determinant If you compute det[ v 1, v 2, v 3 ]= v 1,1 v 2,1 v 3,1 v 1,2 v 2,2 v 3,2, you will see that v 1,3 v 2,3 v 3,3 dz 2 dx dy) P x v 1, v 2, v 3 ) ) =2zdet[ v 1, v 2, v 3 ] Note that in this solution we have reversed our usual rule for suscripts, where the row comes first and column second) ) Using Theorem 673, we have part e) part d) dz 2 {}}{ dx dy) = dz 2 {}}{ dx dy = D 1 z 2 dx + D 2 z 2 dy + D 3 z 2 dz) dx dy Prop 6121 {}}{ =2zdz dx dy = 2zdx dy dz