Determinants and Their Applications in Mathematical Physics
66 4. Particular Determinants =|cij|n, where cij= n ∑ r=1 (−1) r+1 an+1−i,ran+1−j,n+1−r (putr=n+1−s) =(−1) n+1 n ∑ s=1 (−1) s+1 ...
4.3 Skew-Symmetric Determinants 67 [ A (2n−1) ij ] 2 =A (2n−1) ii A (2n−1) jj. (4.3.9) It follows from the section on bordered d ...
68 4. Particular Determinants Proof. LetAn=|aij|nand letEn+1 andFn+1denote determinants obtained by borderingAnin different ways ...
4.3 Skew-Symmetric Determinants 69 and let Bn+1 denote the skew-symmetric determinant obtained by borderingAnby the row [ − 1 − ...
70 4. Particular Determinants In detail, Bn= ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ − 1 • 11 ··· 11 − 1 − 1 • 1 ··· 11 − 1 − 1 − 1 • ··· 11 ..... ...
4.3 Skew-Symmetric Determinants 71 b. 2 n ∑ k=i Ejk=(−1) j+1 δi,odd,i≤j =(−1) j+1 δi,even, i>j c. i− 1 ∑ k=1 Ejk=(−1) j+1 δi, ...
72 4. Particular Determinants Lemma 4.16. En=δn,even. Proof. Perform the column operation C ′ n=Cn+C^1 , expand the result by el ...
4.3 Skew-Symmetric Determinants 73 =0 λii=(−1) i+1 [ δi,odd−δi,even ] =1. This completes the proofs of the preparatory lemmas. T ...
74 4. Particular Determinants The coefficient ofar, 2 n,1≤r≤(2n−1), in Pfnis found by putting (is,js)=(r, 2 n) for any value ofs ...
4.3 Skew-Symmetric Determinants 75 In the particular case in whichaij=1,i<j, denote Pfnby pf n and denote Pf (n) r by pf (n) ...
76 4. Particular Determinants which is consistent with (4.3.22). Hence, A (2n−1) ij =(−1) i+j Pf (n) i Pf (n) j . (4.3.24) Retur ...
4.3 Skew-Symmetric Determinants 77 − a 46 ∣ ∣ a 12 a 13 a 15 a 23 a 25 a 35 ∣ ∣ ∣ ∣ ∣ ∣ + a 56 ∣ ∣ a 12 a 13 a 14 a 23 a 24 a 34 ...
78 4. Particular Determinants Applying the Laplace expansion formula (Section 3.3) in reverse, A 2 3 = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ a 1 ...
4.4 Circulants 79 that is, the Pfaffian is equal to the product of its elements. 4.4 Circulants............................ 4.4. ...
80 4. Particular Determinants ωris also a function ofn, but thenis suppressed to simplify the notation. Thennumbers 1 ,ωr,ω 2 r, ...
4.4 Circulants 81 Hence, A=zr ∣ ∣ WrC 2 C 3 ···Cn ∣ ∣ . (4.4.10) It follows that eachzr,0≤r≤n−1, is a factor ofAn. Hence, An=K n ...
82 4. Particular Determinants Hence, WW=[αrs]n, where αrs= n ∑ t=1 ω (r−1)(t−1)−(t−1)(s−1) = n ∑ t=1 ω (t−1)(r−s) , αrr=n. (4.4. ...
4.4 Circulants 83 = exp [ n ∑ r=1 n− 1 ∑ t=1 ω (r−1)t xt ] = exp [ n− 1 ∑ t=1 xt n ∑ r=1 ω (r−1)t ] = exp(0). The lemma follows. ...
84 4. Particular Determinants the simple hyperbolic functions; A(H 1 ,H 2 )= ∣ ∣ ∣ ∣ H 1 H 2 H 2 H 1 ∣ ∣ ∣ ∣ =1. (4.4.19) Whenn= ...
4.5 Centrosymmetric Determinants 85 Exercises 1.Prove that whenn= 3 and (x 1 ,x 2 )→(x, y), ∂ ∂x [H 1 ,H 2 ,H 3 ]=[H 2 ,H 3 ,H 1 ...
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