PHYSICS 116A Homework 8 Solutions 1. Boas, problem 3.9–4. Given the matrix, 0 2i 1 − A = i 2 0 , − 3 0 0 find the transpose, the inverse, the complex conjugate and the transpose conjugate of A. Verify that AA−1 = A−1A = I, where I is the identity matrix. We shall evaluate A−1 by employing Eq. (6.13) in Ch. 3 of Boas. First we compute the determinant by expanding in cofactors about the third column: 0 2i 1 − i 2 det A i 2 0 =( 1) − =( 1)( 6) = 6 . ≡ − − 3 0 − − 3 0 0 Next, we evaluate the matrix of cofactors and take the transpose: 2 0 2i 1 2i 1 − − 0 0 − 0 0 2 0 0 0 2 i 0 0 1 0 1 adj A = − − − = 0 3 i . (1) − 3 0 3 0 − i 0 − 6 6i 2 i 2 0 2i 0 2i − − − 3 0 − 3 0 i 2 − According to Eq. (6.13) in Ch. 3 of Boas the inverse of A is given by Eq. (1) divided by det(A), so 1 0 0 3 −1 1 i A = 0 2 6 . (2) 1 i 1 − − 3 It is straightforward to check by matrix multiplication that 0 2i 1 0 0 1 0 0 1 0 2i 1 1 0 0 − 3 3 − i 2 0 0 1 i = 0 1 i i 2 0 = 0 1 0 . − 2 6 2 6 − 3 0 0 1 i 1 1 i 1 3 0 0 0 0 1 − − 3 − − 3 The transpose, complex conjugate and the transpose conjugate∗ can be written down by inspec- tion: 0 i 3 0 2i 1 0 i 3 − − − ∗ † AT = 2i 2 0 , A = i 2 0 , A = 2i 2 0 . − 1 0 0 3 0 0 1 0 0 − − ∗The transpose conjugate is more often referred to as the hermitian conjugate or the adjoint. 1 These have been obtained by noting that AT is obtained by A by interchanging the rows and columns, A∗ is obtained from A by complex conjugating the matrix elements, and the definition of the hermitian conjugate is A† =(A∗)T . Alternative method: One can also evaluate A−1 by employing Gauss Jordan elimination, which is described in the class handout http://young.physics.ucsc.edu/116A/gauss jordan.pdf. 0 2i 1 1 0 0 − A = i 2 0 , I = 0 1 0 − 3 0 0 0 0 1 First we interchange R R . Then we rescale the new row 1, R 1 R to obtain 1 ↔ 3 1 → 3 1 1 1 0 0 0 0 3 0 2i 1 , 1 0 0 − i 2 0 0 1 0 − Next, we perform the elementary row operations, R R + iR and R 1 iR to obtain: 3 → 3 1 2 →− 2 2 1 1 0 0 0 0 3 0 1 1 i , 1 i 0 0 2 − 2 0 2 0 0 1 1 i 3 Next, we perform the elementary row operations, R R 2R to obtain: 3 → 3 − 1 1 1 0 0 0 0 3 0 1 1 i , 1 i 0 0 2 − 2 0 0 i i 1 1 i − 3 Finally, we perform the elementary row operations, R R + 1 R , followed by R iR to 2 → 2 2 3 3 → 3 obtain: 1 1 0 0 0 0 3 −1 1 i 0 1 0 , A = 0 2 6 0 0 1 1 i 1 − − 3 The final step produces the inverse, which is indicated above. Note that we have reproduced Eq. (2). 2. Boas, problem 3.9–5. Show that the product AAT is a symmetric matrix. Using Eq. (9.10) on p. 139 of Boas, (AB)T = BTAT for any two matrices A and B. Hence, (AAT )T =(AT )T AT = AAT , (3) where we have used the fact† that (AT )T = A for any matrix A. Eq. (3) implies that AAT is a symmetric matrix, since by definition a symmetric matrix is equal to its transpose [cf. the table at the top of p. 138 of Boas]. †The transpose of a matrix interchanges the rows and columns. Thus, if one performs the transpose operation twice, the original matrix is recovered. 2 3. Boas, problem 3.9–17. (a) Show that if A and B are symmetric, then AB is not symmetric unless A and B commute. If A and B are symmetric, then A = AT and B = BT . We now examine (AB)T = BT AT = BA, after using the fact that A and B are symmetric matrices. We conclude that (AB)T = AB if and only if AB = BA. That is, AB is not symmetric unless A and B commute. (b) Show that a product of orthogonal matrices is orthogonal. Consider orthogonal matrices Q1 and Q2. By definition [cf. the table at the top of p. 138 of −1 T −1 T Boas], we have Q1 = Q1 and Q2 = Q2 . We now compute −1 −1 −1 T T T (Q1Q2) = Q2 Q1 = Q2 Q1 =(Q1Q2) , (4) after using the fact that Q1 and Q2 are orthogonal. In deriving Eq. (4), we have used the following properties of the inverse and the transpose (AB)−1 = B−1A−1 , and (AB)T = BT AT , for any pair of matrices A and B. Thus, we have shown that −1 T (Q1Q2) =(Q1Q2) , which implies that Q1Q2 is orthogonal. (c) Show that if A and B are Hermitian, then AB is not Hermitian unless A and B commute. If A and B are Hermitian, then A = A† and B = B†. We now examine (AB)† = B†A† = BA, (5) after using the fact that A and B are Hermitian matrices. In deriving Eq. (5), we have used the fact that: (AB)† = ((AB)∗)T =(A∗ B∗)T =(B∗)T (A∗)T = B†A† . (6) We conclude that (AB)† = AB if and only if AB = BA. That is, AB is not Hermitian un- less A and B commute. (d) Show that a product of unitary matrices is unitary. Consider unitary matrices U1 and U2. By definition [cf. the table at the top of p. 138 of −1 † −1 † Boas], we have U1 = U1 and U2 = U2 . We now compute −1 −1 −1 † † † (U1U2) = U2 U1 = U2 U1 =(U1U2) , after using the fact that U1 and U2 are unitary and employing the property of the Hermitian −1 † conjugation given in Eq. (6). Thus, we have shown that (U1U2) =(U1U2) , which implies that U1U2 is orthogonal. 3 4. Boas, problem 3.10-5(a). Given two vectors, A~ = (3+ i, 1 , 2 i, 5i, i + 1) and B~ = (2i, 4 3i, 1+ i, 3i, 1) , − − − find the norms of A~ and B~ and the inner product of A~ and B~ , and note that the Schwarz inequality is satisfied. Using eq. (1.07) on p. 146 of Boas, the norms of A~ and B~ are given by: 1/2 A~ = 3+ i 2 + 1 2 + 2 i 2 + 5i 2 + i + 1 2 =(10+1+5+25+2)1/2 = √43 , k k | | | | | − | |− | | | 1/2 B~ = 2i 2 + 4 3i 2 + 1+ i 2 + 3i 2 + 1 =(4+25+2+9+1)1/2 = √41 . k k | | | − | | | | | Using eq. (10.6) on p. 146 of Boas, the inner product of A~ and B~ is given by: A~ · B~ = (3 i)(2i) + (1)(4 3i)+(2+ i)(1 + i)+(5i)(3i)+(1 i)(1) − − − =(2+6i)+(4 3i)+(1+3i) 15 + (1 i) − − − = 7 + 5i. (7) − The Schwarz inequality [see eq. (10.9) on p. 146 of Boas] is: A~ · B~ A~ B~ , | | ≤ k k k k where indicates the magnitude of the corresponding complex number. Using Eq. (7), it |···| follows that A~ · B~ = 7 + 5i = √49 + 25 = √74 . | | |− | Thus, the Schwarz inequality is satisfied for this example since √74 √43 √41 = √1763 . ≤ 5. Boas, problem 3.11–9. Show that det(C−1MC) = det M. What is the product of det(C−1) and det C? Thus, show that the product of the eigenvalues of M is equal to det M. Eq. (6.6) on p. 118 of Boas states that det(AB) = det A det B for any two matrices A and B. It follows that det(C−1)det C = det(C−1C) = det I = 1 , where I is the identity matrix. Hence,‡ 1 det(C−1)= . det C Using Eq. (6.6) of Boas again along with the result just obtained, det M det C det(C−1MC) = det(C−1)det M det C = = det M. (8) det C 1 ‡By assumption, C− exists, in which case det C 6= 0 and so it is permissible to divide by det C. 4 Finally, Boas asks you to show that the product of the eigenvalues of M is equal to det M. What she is expecting you to do is to use eq. (11.11) on p. 150 of Boas, which states that λ 0 0 0 1 ··· 0 λ2 0 0 ··· −1 0 0 λ 0 C MC = D = 3 ··· , (9) . . .. 0 0 0 λ ··· n where D is a diagonal matrix whose diagonal elements are the eigenvalues of M, denoted by λ1,λ2,...,λn above. It follows that det(C−1MC) = det D = λ λ λ . (10) 1 2 ··· n The results of Eqs. (8) and (10) then imply that det M = λ λ λ . (11) 1 2 ··· n Note that this proof is not completely general, since not all matrices M are diagonalizable and, for these, the matrix of eigenvectors C does not have an inverse since two or more eigenvectors are equal, with the result that det(C) = 0.