Determinants and Their Applications in Mathematical Physics
46 3. Intermediate Determinant Theory The proof is completed by applying (C) and (A 1 ) withn→n−1. Theorem 3.7 is applied in Sec ...
3.7 Bordered Determinants 47 and letBndenote the determinant of order (n+ 1) obtained by bordering Anby the column X= [ x 1 x 2 ...
48 3. Intermediate Determinant Theory Proof. Bij=(−1) i+j ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ a 11 a 12 ··· a 1 ,j− 1 a 1 ,j+1 ··· a 1 ...
3.7 Bordered Determinants 49 Proof. It follows from (3.7.2) that n ∑ j=1 n ∑ s=1 Air,js=0, 1 ≤i, r≤n. ExpandingBijby elements fr ...
50 3. Intermediate Determinant Theory =− n ∑ q=1 n ∑ s=1 vqysAqs. Similarly, Bn+1,n+2=+ n ∑ p=1 n ∑ s=1 upysAps, Bn+2,n+1=+ n ∑ ...
4 Particular Determinants 4.1 Alternants............................ 4.1.1 Introduction Any function ofnvariables which changes ...
52 4. Particular Determinants If thex’s are not distinct, the determinant has two or more identical rows. If they’s are not dist ...
4.1 Alternants 53 When any two of thexrare equal,Xnhas two identical rows and therefore vanishes. Hence, very possible differenc ...
54 4. Particular Determinants Postmultiply the VandermondianVn(x)orVn(x 1 ,x 2 ,...,xn)byAn, prove the reduction formula Vn(x 1 ...
4.1 Alternants 55 Hence, [xij]n[Fji]n=I, [Fji]n=[xij] − 1 =[X ji n ]n. The theorem follows. Theorem 4.2. X (n) nj =(−1) n−j Xn ...
56 4. Particular Determinants This completes the proof of Theorem 4.3 which can be expressed in the form Hrs Xn = ∏n i=1 (ys−xi) ...
4.1 Alternants 57 4.1.5 The Cauchy Double Alternant............ The Cauchy double alternant is the determinant An= ∣ ∣ ∣ ∣ 1 xi− ...
58 4. Particular Determinants Exercises 1.Prove the reduction formula A (n) ij =A (n−1) ij n− 1 ∏ r=1 r=i ( xn−xr xr−yn )n− 1 ∏ ...
4.1 Alternants 59 Wn=(−1) n(n+1)/ 2 XnYn n ∏ i=1 (xi+ 1)(yi−1). Removingf(x 1 ),f(x 2 ),...,f(xn), from the firstnrows inVnandWn ...
60 4. Particular Determinants where qij= i ∑ r=1 U (i) ir Pr(xj) = i ∑ r=1 U (i) ir r→i ∑ s=1 asrx s− 1 j (asr=0, s>r) = i ∑ ...
4.1 Alternants 61 b.Vn=V(x 2 ,x 3 ,...,xn) n ∏ r=2 (xr−x 1 ). c. V(xt,xt+1,...,xn)=V(xt+1,xt+2,...,xn) n ∏ r=t+1 (xr−xt). d.V (n ...
62 4. Particular Determinants referring to Lemma (a) withn→s+ 1 and Lemma (c) withm→s+1, s+1 ∏ r=1 Mr= Vn ∏s r=1 (xs+1−x ...
4.1 Alternants 63 4.1.9 Further Vandermondian Identities.......... The notation Nm={ 12 ···m}, Jm={j 1 j 2 ···jm}, Km={k 1 k 2 · ...
64 4. Particular Determinants Proof. Denote the left side of (a) bySm. Then, applying Lemma 4.6, Sm= Nm ∑ Jm ( m ∏ r=1 x r− 1 jr ...
4.3 Skew-Symmetric Determinants 65 = ∣ ∣ ∣ ∣ Apr Aps Aqr Aqs ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ Apq Aqs Apr Ars ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ Aqr Ars Apq Aps ...
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