EN221 - Fall HW # 1 Solutions

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1 EN221 - Fall HW # 1 Solutions Prof. Vivek Shenoy 1. Let A be an arbitrary tensor. i). Show that II A = 1 2 {(tr A)2 tr A 2 } ii). Using the Cayley-Hamilton theorem, deduce that Soln. i). Method-1 III A = 1 6 {(tr A)3 3 tr A tr A tr A 3 } Using the definition of II A, i.e. Eqn(1.26) from Chadwick II A [a,b,c] = [a,ab,ac] + [Aa,b,Ac] + [Aa,Ab,c] (1) Let, a = e 1, b = e 2 and c = e 3 II A [e 1,e 2,e 3 ] = [e 1,Ae 2,Ae 3 ] + [Ae 1,e 2,Ae 3 ] + [Ae 1,Ae 2,e 3 ] (2) II A ɛ 123 = [e 1,Ae 2,Ae 3 ] + [Ae 1,e 2,Ae 3 ] + [Ae 1,Ae 2,e 3 ] (3) since, ɛ 123 = 1, and Ae i = A mn (e m e n )e i = A mn (e n e i )e m = A mn δ ni e m = A mi e m II A = A q2 A r3 [e 1,e q,e r ] + A p1 A r3 [e p,e 2,e r ] + A p1 A q2 [e p,e q,e 3 ] since p, q and r are dummy indices (and rearranging) II A = A p2 A q3 [e 1,e p,e q ] + A p1 A q3 [e p,e 2,e q ] + A p1 A q2 [e p,e q,e 3 ] = A p2 A q3 ɛ 1pq A p1 A q3 ɛ 2pq + A p1 A q2 ɛ 3pq (4) we can as well write it by interchanging p and q II A = A q2 A p3 ɛ 1qp A q1 A p3 ɛ 2qp + A q1 A p2 ɛ 3qp = A q2 A p3 ɛ 1pq + A q1 A p3 ɛ 2pq A q1 A p2 ɛ 3pq (5) 1

2 on summing Eqn(4) and Eqn(5) 2II A = (A p2 A q3 A q2 A p3 )ɛ 1pq +... (6) (A p1 A q3 A q1 A p3 )ɛ 2pq (A p1 A q2 A q1 A p2 )ɛ 3pq 2II A = (A p2 A q3 ɛ A p3 A q2 ɛ 132 )ɛ 1pq +... (7) Eqn(7) can be re written as +(A p1 A q3 ɛ A p3 A q1 ɛ 231 )ɛ 2pq (A p1 A q2 ɛ A p2 A q1 ɛ 321 )ɛ 3pq II A = 1 2 A pia qj ɛ rij ɛ rpq (8) If we closely examine Eqn(8), we observe that r is different from {i,j} and the sign of ɛ rij determines the sign of the each term. All the existing six terms of the Eqn(7) are obtained by the all the {i,j,r} permutations of the Eqn(8) using the fact (see P1.3b Chadwick) Eqn(8) can be written as, Method-2 ɛ rij ɛ rpq = δ ip δ jq δ iq δ jp II A = 1 2 A pia qj (δ ip δ jq δ iq δ jp ) = 1 2 (A ii A jj A ij A ji ) = 1 2 {(tr A)2 tr A 2 } (9) II A [a,b,c] = [a,ab,ac] + [Aa,b,Ac] + [Aa,Ab,c] (10) we know that A (a b) = Aa Ab and also, B (a b) c = (a b) B T c II A [a,b,c] = [(A ) T a,b,c] + [a, (A ) T b,c] + [a,b, (A ) T c] Using Cayley-Hamilton theorem = tr (A ) T [a,b,c] = tr A [a,b,c] (11) 2

3 A 3 I A A 2 + II A A III A = 0 (12) Multiply above equation with A 1 to get, A 2 I A A + II A A 1 III A = 0 (13) knowing the fact that III A = deta and A = A 1 deta, the above Eqn. can be re-written as Now, take the trace of the above Eqn. A 2 I A A + II A A = 0 (14) tr A 2 I A tr A + II A tr I tr (A ) = 0 tr A 2 (tr A) 2 + 2II A = 0 II A = 1 2 {(tr A)2 tr A 2 } (15) 3

4 ii) Cayley-Hamilton theorem states that, A 3 I A A 2 + II A A III A = 0 (16) Taking the trace of the above Eqn. we obtain, tr A 3 I A tr A 2 + II A tr A III A tr I = 0 (17) using the fact that II A = 1 2 {(tr A)2 tr A 2 } and tr I = 3 we obtain, on simplifying, tr A 3 I A tr A {(tr A)2 tr A 2 }tr A 3III A = 0 (18) 3III A = tr A 3 tr Atr A {(tr A)2 tr A 2 }tr A = 1 2 [2tr A3 3tr Atr A 2 + (tr A) 3 ] (19) or III A = 1 6 [2tr A3 3tr A tr A 2 + (tr A) 3 ] (20) 4

5 2. Using the result or otherwise, show that deta detb = det(a T B) A, B L, ɛ ijk ɛ lmn = δ il (δ jm δ kn δ jn δ km ) + δ im (δ jn δ kl δ jl δ kn ) Hence derive the formula, (a) ɛ ijp ɛ lmp = δ il δ jm δ im δ jl, (b) ɛ ipq ɛ lpq = 2 δ il Soln. +δ in (δ jl δ km δ jm δ kl ) eg. e 1 = (δ 11, δ 12, δ 13 ) = (1, 0, 0) δ i1 δ j1 δ k1 ɛ ijk = [e i, e j, e k ] = δ i2 δ j2 δ k2 δ i3 δ j3 δ k3 = e i = (δ i1, δ i2, δ i3 ) (21) δ i1 δ i2 δ i3 δ j1 δ j2 δ j3 δ k1 δ k2 δ k3 (22) ɛ ijk ɛ lmn = = = δ i1 δ j1 δ k1 δ l1 δ m1 δ n1 δ i2 δ j2 δ k2 δ l2 δ m2 δ n2 δ i3 δ j3 δ k3 δ l3 δ m3 δ n3 δ i1 δ i2 δ i3 δ l1 δ m1 δ n1 δ j1 δ j2 δ j3 δ l2 δ m2 δ n2 δ k1 δ k2 δ k3 δ l3 δ m3 δ n3 δ il δ im δ in δ jl δ jm δ jn δ kl δ km δ kn (23) (24) (25) on expanding the determinant in the Eqn(25), we obtain ɛ ijk ɛ lmn = δ il (δ jm δ kn δ jn δ km ) + δ im (δ jn δ kl δ jl δ kn ) (a) set k = p and n = p, in the Eqn(26) +δ in (δ jl δ km δ jm δ kl ) (26) ɛ ijp ɛ lmp = δ il (δ jm δ pp δ jp δ pm ) + δ im (δ jp δ pl δ jl δ pp ) +δ ip (δ jl δ pm δ jm δ pl ) (27) = 3 δ il δ jm δ il δ jm + δ im δ jl 3 δ im δ jl +δ im δ jl δ il δ jm (28) = δ il δ jm δ im δ jl (29) 5

6 (b) since p is a dummy index in Eqn(29) replace it by the q ɛ ijq ɛ lmq = δ il δ jm δ im δ jl (30) in the above equation now set j = p, and m = p, to obtain ɛ ipq ɛ lpq = δ il δ pp δ ip δ pl = 3 δ il δ il = 2 δ il (31) 6

7 3. Let A be an arbitrary tensor and A its adjugate. (i) Given that A ij are the components of A relative to an orthonormal basis e, show that the components of A are 1 2 ɛ ipqɛ jrs A pr A qs. Deduce that A T = A (ii) Show that (a) (A ) = (deta)a, (b) tr A = II A, (c) A{a (A T b)} = (A b) a, b E Soln. (i) using the defintion of an adjugate, A (a b) = (Aa) (Ab) (32) replacing a and b with the orthonormal basis e i and e j we get, A (e i e j ) = (Ae i ) (Ae j ) (33) A pq (e p e q )(e i e j ) = A mn (e m e n )e i A rs (e r e s )e j (34) Using e i e j = ɛ kij e k, and (a b)c = (c.b)a, we obtain A pq (e p e q )ɛ kij e k = A mn (e i e n )e m A rs (e j e s )e r (35) Using e i e j = δ ij, we obtain A pq (e k e q )e p ɛ kij = A mn δ in e m A rs δ js e r A pq δ kq e p ɛ kij = A mn δ in e m A rs δ js e r A pq e pɛ qij = A mi A rj e m e r (36) A pq e p ɛ qij = A mi A rj ɛ smr e s (37) taking the dot product with e k on both sides, we obtain A pqɛ qij (e p e k ) = A mi A rj ɛ smr (e s e k ) Multiplying both sides with ɛ sij we obtain A pq ɛ qijδ pk = A mi A rj ɛ smr δ sk A kqɛ qij = A mi A rj ɛ kmr (38) ɛ sij ɛ qij A kq = ɛ sij ɛ kmr A mi A rj (39) using, ɛ sij ɛ qij = 2δ sq 2 δ sq A kq = ɛ sij ɛ kmr A mi A rj (40) A ks = 1 2 ɛ sijɛ kmr A mi A rj (41) which, by replacing the dummy varibles can be written as, A ij = 1 2 ɛ ipqɛ jrs A pr A qs (42) 7

8 Now, (A T ) ij = 1 2 ɛ ipqɛ jrs A T pra T qs = 1 2 ɛ ipqɛ jrs A rp A sq (43) on replacing the dummy variables Eqn(43) can be rewritten as (A T ) ij = 1 2 ɛ irsɛ jpq A pr A qs = A ji = (A T ) ij (44) Thus, (A T ) = A T 8

9 (ii) (a) consider the case when det A 0, In that case there exists an inverse A 1 and we know that or We need to find (A ), (A T ) = (A ) T = A 1 = (det A)A 1 (deta) 1 A = (det A)A T = B det(a T ) = det(a 1 ) = 1 deta (45) now, detb = det(deta A T ) = (deta) 3 det(a T ) = (deta) 2 (46) B T = (A ) T = (A T deta) T = = 1 deta (A T ) T A deta (47) (A ) = B = (detb)b T (48) on substuting Eqns(46 and 47) in Eqn(48) we obtain (A ) = B = deta A (49) 9

10 (b) II A [a,b,c] = [a,ab,ac] + [Aa,b,Ac] + [Aa,Ab,c] (50) we know that A (a b) = Aa Ab and also, B (a b) c = (a b) B T c II A [a,b,c] = [(A ) T a,b,c] + [a, (A ) T b,c] + [a,b, (A ) T c] = tr (A ) T [a,b,c] = tr A [a,b,c] (51) 10

11 (c) Method-1 By definition or A = (A ) deta and using B (a b) = (Ba) (Bb) { } A a (A T b) (A ) = deta A (52) = (A ) { } a (A T b) deta 1 { } = A a A (A T b) deta (53) (54) Using A A T = deta { } A a (A T b) = 1 deta {A a detab} = deta deta {A a b} = {A a b} (55) Method-2 { } consider A a (A T b) Take dot product with an arbitrary non zero vector c and using Ba b = a B T b A{a (A T b)} c = {a (A T b)} A T c = [a, (A T b), (A T c)] = {a (A ) T (b c)} = {A a (b c)} = {A a b} c (56) 11