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Some identities involving tensors. 5 (i) If C is a second-order tensor and a is an arbitrary vector, prove a ·C = C· a if and only if C = CT , i.e., C is symmetric. Solution: If a ·C = C· a then aiCij -Cjiai = 0 or ai(Cij -Cji) = 0. Since a is arbitrary, it follows that Cij = Cji for all i and j, or C = CT . Also, to complete the proof, we verify that if C = CT (i.e. C is symmetric) so that Cij = Cji and a · C = aiCijej = aiCjiej = C · a. (ii) If C = -CT (C is anti-symmetric), then show a · C · a = 0. Solution: a·C·a = Cijaiaj = 0 as an immediate consequence of the result given in the remark to question 1 (identifying A with C and S with aa). Alternatively, working it out, a·C·a = Cijaiaj = 1 2 (Cijaiaj+Cijaiaj), then switch subscripts i and j in the second term to get = 1 2 (Cijaiaj +Cjiajai) = 1 2 (Cijaiaj -Cijajai) = 0. Another elegant method handed in by Z. Guo: if f = a · C · a, then f = (a · C · a)T = a · CT · a = -f, hence f = 0. (iii) If B is an anti-symmetric second-order tensor, show that B · B is a symmetric second-order tensor. Solution: Since B is antisymmetric, Bij = -Bji, so B · B = BikBkjeiej = (-Bki)(-Bjk)eiej = BjkBkieiej , which demonstrates the symmetry of B · B. (iv) You have seen in linear algebra that (A·B)T = BT ·AT . Verify this expression using index notation. Evaluate (A · B · C)T . Solution: (A · B)T = (AijBjkeiek)T = AijBjkekei = (BT )kj(AT )jiekei = BT · AT . (13) For the second part, since A·B·C = AijBjkCkleiel, we can write (A·B·C)T = AijBjkCklelei = AljBjkCkieiel = (CT )ik(BT )kj(AT )jleiel = CT · BT · AT . Alternatively, note that (A·B·C)T = ((A·B)·C)T = CT ·(A·B·)T = CT ·BT ·CT .
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