46 CHAPTER 1. REVIEW OF SOME LINEAR ALGEBRA
2. A and AT are one to one.
3. A is onto.
Proof: This follows immediately from the above theorem.
1.9.9 An Identity of Cauchy
Theorem 1.9.28 Both the left and the right sides in the following yield the samepolynomial in the variables ai,bi for i≤ n.
∏i, j
(ai +b j)
∣∣∣∣∣∣∣1
a1+b1· · · 1
a1+bn...
...1
an+b1· · · 1
an+bn
∣∣∣∣∣∣∣= ∏j<i
(ai−a j)(bi−b j) . (1.26)
Proof: The theorem is true if n = 2. This follows from some computations. Suppose itis true for n−1, n≥ 3.∣∣∣∣∣∣∣∣∣∣
1a1+b1
1a1+b2
· · · 1a1+bn
...... · · ·
...1
an−1+b11
an−1+b21
an−1+bn1
an+b11
an+b2· · · 1
an+bn
∣∣∣∣∣∣∣∣∣∣=
∣∣∣∣∣∣∣∣∣∣
an−a1(a1+b1)(b1+an)
an−a1(a1+b2)(b2+an)
· · · an−a1(a1+bn)(an+bn)
...... · · ·
...an−an−1
(an−1+b1)(an+b1)an−an−1
(b2+an)(b2+an−1)an−an−1
(an+bn)(bn+an−1)1
an+b11
an+b2· · · 1
an+bn
∣∣∣∣∣∣∣∣∣∣Continuing to use the multilinear properties of determinants, this equals∣∣∣∣∣∣∣∣∣∣
1(a1+b1)(b1+an)
1(a1+b2)(b2+an)
· · · 1(a1+bn)(an+bn)
...... · · ·
...1
(an−1+b1)(an+b1)1
(b2+an)(b2+an−1)1
(an+bn)(bn+an−1)1
an+b11
an+b2· · · 1
an+bn
∣∣∣∣∣∣∣∣∣∣n−1
∏k=1
(an−ak)
and this equals ∣∣∣∣∣∣∣∣∣∣
1(a1+b1)
1(a1+b2)
· · · 1(a1+bn)
...... · · ·
...1
(an−1+b1)1
(b2+an−1)1
(bn+an−1)
1 1 · · · 1
∣∣∣∣∣∣∣∣∣∣∏
n−1k=1 (an−ak)
∏nk=1 (an +bk)
Now take −1 times the last column and add to each previous column. Thus it equals∣∣∣∣∣∣∣∣∣∣
bn−b1(a1+b1)(a1+bn)
bn−b2(a1+b2)(a1+bn)
· · · 1(a1+bn)
...... · · ·
...bn−b1
(b1+an−1)(bn+an−1)bn−b2
(b2+an−1)(bn+an−1)1
(an−1+bn)
0 0 · · · 1
∣∣∣∣∣∣∣∣∣∣∏
n−1k=1 (an−ak)
∏nk=1 (an +bk)
46 CHAPTER 1. REVIEW OF SOME LINEAR ALGEBRA2. Aand A! are one to one.3. A is onto.Proof: This follows immediately from the above theorem.1.9.9 An Identity of CauchyTheorem 1.9.28 Both the left and the right sides in the following yield the samepolynomial in the variables a;,b; for i <n.1 eee 1a, +b; ai +bn[](atei| : > | =[][(a—aj) (b:—8)). (1.26)ij 1 _ 1 j<ian+b, antbnProof: The theorem is true if n = 2. This follows from some computations. Suppose itis true forn—1,n> 3.1 1 lay+b, ay+b2 ay+bn—1 _1 _ _ 1nV +1 anh an—1+Pnan +b, a@ntby an+bnan—a| an—4a cee an—4a4(a, +b1)(b1 +an) (a, +b2)(b2+4n) (a) +bn)(an+bn)= an—4n-1 an—4n-1 Gn—4n-1(@n—1+b1 )(an+b1) (b2-+an)(bo+an—1) (an-+bn)(bn+an—1)ant, an+b2 ant+bnContinuing to use the multilinear properties of determinants, this equals1 1 1(a, +b1)(b1 +4n) (aj+b2)(b2+an) (a1 +bn)(an+bn): : : nol(an — ax)(4n—1+b1 )(Qn+b1) (b2+an)(bo+an—1) (an+bn)(bn+an—1) k=lGn+b Gn+b2 “ an+bnand this equals1 1 1(a, +51) (a+b) (a, +bn): Do, > FINE] @n — ax)1 1 I tan b(an—1 +61) (b2+an_1) (bn+an—1) Mit (an + ‘)1 1 tee 1Now take —1 times the last column and add to each previous column. Thus it equalsbn—b bn—b wee 1(a) +b) )(a1 +n) (a) +b2)(a1 +n) (a) +bn)ee TET Gn =a)bn—by bn—by ! Tai (Gn + dx(b1+4n—1)(bnt+4n—1) (b2+4n—1)(bnt+4n—1) (Qn—1+bn) kl ( " )0 0 1