EN221 - Fall HW # 2 Solutions

Transcription

1 EN1 - Fall8 - HW # Solutions Prof. Vivek Shenoy 1) Let A and B be arbitrary tensors, A and B the adjugates of A and B, and α and β arbitrary scalars. Show that det(αa + βb) = deta + β tr (B T A ) + αβ tr (A T B ) + β 3 detb det(αa + βb) = [(αa + βb)e 1, (αa + βb)e, (αa + βb)e 3 ] (1) det(αa + βb) = (αa + βb)e 1 [(αa + βb)e (αa + βb)e 3 ] = (αa + βb)e 1 [ Ae Ae 3 + αβbe Ae 3 αβae Be 3 + β Be Be 3 ] = Ae 1 Ae Ae β[be 1 Ae Ae 3 + Ae 1 Ae Be 3 + Ae 1 Be Ae 3 ] αβ [Ae 1 Be Be 3 + Be 1 Be Ae 3 + Ae 1 Be Be 3 ] β 3 Be 1 Be Be 3 Note that, Ae 1 Ae Ae 3 = deta and, Be 1 Be Be 3 = detb Now gather the terms that go with β Be 1 Ae Ae 3 = Be 1 A (e e 3 ) = [A T Be 1,e,e 3 ] And the sum of the above terms are, Ae 1 Ae Be 3 = [e 1,e,A T Be 3 ] Ae 1 Be Ae 3 = [e 1,A T Be,e 3 ] [A T Be 1,e,e 3 ] + [e 1,e,A T Be 3 ] + [e 1,A T Be,e 3 ] = tr (A T B) = tr (A T B) T = tr (B T A ) It can similarly be easily seen that the terms that go with αβ is tr (A T B ) hence, det(αa + βb) = deta + β tr (B T A ) + αβ tr (A T B ) + β 3 detb () 1

2 ) Show that cos θ sin θ sin θ cos θ 1 (3) is an improper orthogonal matrix that represents a change of basis equivalent to a reflection in the plane through e 3 inclined at a positive angle e 1. The determinant of Q is -1 hence it is improper orthogonal matrix. Consider an arbitrary vector a, a = e 1 + e + e 3 a = The components of the vector are then split into two components. One component is the plane formed by e 1 and e 3. Here e 1 is the vector in the plane of e 1 e and at an angle θ w.r.t. e 1 e 1 = cosθ e 1 + sin θ e The vector perpendicular to the plane e 1 e 3 is e = e 3 e 1 = e 3 (cos θ e 1 + sin θ e ) The component of a out of the plane e 1 e 3 is = cosθ e sin θ e 1 (4) a = a e = ( e 1 + e + e 3 ) (cos θe sin θe 1 ) and the component in the plane is = sin θ + cos θ (5) a = a a e = ( e 1 + e + e 3 ) [ sinθ + cos θ][cosθe sin θe 1 ] = [ sin θ + sin θ cos θ]e [ cos θ + sin θ cos θ]e e 3 = [ cos θ + sin θ cosθ]e 1 + [ sin θ + sin θ cosθ]e + e 3 (6)

3 The transformation Q converts a to a r a r = Qa = = cos θ sin θ sin θ cos θ 1 cosθ + sin θ sinθ cosθ The component of a r out of plane e 1 e 3 is a r = a r e = ( cosθ + sinθ, sinθ cosθ, ) ( sin θ, cosθ,) = sin θ cos θ (7) Similarly, the component of a r in the plane e 1 e 3 is a r = a r (a r e )e = ( cos θ + cos θ sin θ)e 1 + ( cosθ sin θ + sin θ)e + e 3 (8) Eqns. 5,6,7,8 tell that a r = a, In plane components are equal for a and a a r = a, Out of plane components are equal and opposite But, that precisely is the definition of reflection in 3-D. 3

4 3) Consider the following symmetric matrix, S: (9) (a) Show that S is positive definite. (b) Calculate A, the square root S, Verify that A = S. The eigenvalues of S are obtained from 5 λ 4 4 λ 4 4 λ = (1) (5 λ)[(4 λ)(5 λ)] + 4[ 4(4 λ)] = (4 λ)[(5 λ) 16] = (4 λ)[(5 λ 4)(5 λ + 4)] = (4 λ)(1 λ)(9 λ) = (11) the three roots are, λ 1 = 1, λ = 4, λ 3 = 9 Since all the eigenvalues are positive the matrix S is +ve definite To find the eigenvectors on solving we obtain = and = so the unit eigenvector is, p 1 = so the unit eigenvector is, p = = = 4 1 = 9 (1) (13) (14) (15) (16) 4

5 so the unit eigenvector is, p 3 = 1 1 (17) The square root matrix S 1/ is written as A = S 1/ = 3 i=1 λ 1/ i p i p i = 1 p 1 p p p + 9 p 3 p = = (18) 1 Checking A A = = = S (19) 5

6 4) if det T deduce using det (T λi) = that and hence show that det (T 1 λ 1 I) = λ 3 I 1 (T 1 )λ + I (T 1 )λ 1 I 3 (T 1 ) = I 1 (T 1 ) = I (T)/I 3 (T) I (T 1 ) = I 1 (T)/I 3 (T) I 3 (T 1 ) = 1/I 3 (T) Since dett, T 1 exists, and using TT 1 = I = det(t λi) = det(t λtt 1 ) = det( λ[tt 1 Tλ 1 ]) = det( λt[t 1 Iλ 1 ]) = λ 3 det(t) det(t 1 Iλ 1 ) = λ 3 I 3 (T) det(t 1 Iλ 1 ) () since dett, so λ hence, Eqn.() can be written as det (T 1 λ 1 I) = using Eqn.() and expanding the determinants on both sides det(t λi) = λ 3 I 3 (T) det(t 1 Iλ 1 ) λ 3 I 1 (T)λ + I (T)λ I 3 (T) = λ 3 I 3 (T)[λ 3 I 1 (T 1 )λ + I (T 1 )λ 1 I 3 (T 1 )] = 1 I 3 (T) λ3 I 1(T) I 3 (T) λ + I (T) I 3 (T) λ 1 = 1 + I 1(T 1 )λ I (T 1 )λ + I 3 (T 1 )λ 3 = on comparing the coefficients of λ, λ and λ 3 respectively, we get (1) I 1 (T 1 ) = I (T)/I 3 (T) () I (T 1 ) = I 1 (T)/I 3 (T) (3) I 3 (T 1 ) = 1/I 3 (T) (4) 6