Goldstein_4.1, 4.2, 4.10, 4.14, 4.15

Goldstein_4.1, 4.2, 4.10, 4.14, 4.15 - Homework 6 4.1 4.2...

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Homework 6: # 4.1, 4.2, 4.10, 4.14, 4.15 Michael Good Oct 4, 2004 4.1 Prove that matrix multiplication is associative. Show that the product of two orthogonal matrices is also orthogonal. Answer: Matrix associativity means A ( BC ) = ( AB ) C The elements for any row i and column j , are A ( BC ) = k A ik ( m B km C mj ) ( AB ) C = m ( k A ik B km ) C mj Both the elements are the same. They only differ in the order of addition. As long as the products are defined, and there are finite dimensions, matrix multiplication is associative. Orthogonality may be defined by AA = I The Pauli spin matrices, σ x , and σ z are both orthogonal. ˜ σ x σ x = 0 1 1 0 0 1 1 0 = 1 0 0 1 = I ˜ σ z σ z = 1 0 0 - 1 1 0 0 - 1 = 1 0 0 1 = I The product of these two: σ x σ z = 0 1 1 0 1 0 0 - 1 = 0 - 1 1 0 q 1
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is also orthogonal: ˜ qq = 0 1 - 1 0 0 - 1 1 0 = 1 0 0 1 = I More generally, if AA = 1 BB = 1 then both A , and B are orthogonal. We can look at ABAB = k ( AB ) ik ( AB ) kj = k AB ki AB kj = k,s,r a ks b si a kr b rj The elements are k,s,r a ks b si a kr b rj = b si a ks a kr b rj = b si ( ˜ AA ) sr b rj This is ABAB = b si δ sr b rj = ˜ BB ij = δ ij Therefore the whole matrix is I and the product ABAB = I is orthogonal. 4.2 Prove the following properties of the transposed and adjoint matrices: AB = BA ( AB ) = B A Answer: For transposed matrices AB = AB ij = AB ji = a js b si = b si a js = B is A sj = ( BA ) ij = BA As for the complex conjugate, ( AB ) = ( AB ) * From our above answer for transposed matrices we can say AB = BA 2
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And so we have ( AB ) = ( AB ) * = ( BA ) * = B * A * = B A 4.10 If B is a square matrix and A is the exponential of B, defined by the infinite series expansion of the exponential, A e B = 1 + B + 1 2 B 2 + ... + B n n ! + ..., then prove the following properties: e B e C = e B + C , providing B and C commute. A - 1 = e - B e CBC - 1 = CAC - 1 A is orthogonal if B is antisymmetric Answer: Providing that B and C commute; BC - CB = 0 BC = CB we can get an idea of what happens: (1+ B + B 2 2 + O ( B 3 ))(1+ C + C 2 2 + O ( C 3 )) = 1+ C + C 2 2 + B + BC + B 2 2 + O (3) This is 1+( B + C )+ 1 2 ( C 2 +2 BC + B 2 )+ O (3) = 1+( B + C )+ ( B + C ) 2 2 + O (3) = e B + C Because BC = CB and where O (3) are higher order terms with products of 3 or more matrices. Looking at the k th order terms, we can provide a rigorous proof.
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