Problems from the Book – Problem 19.9 Let n ∈ N Let w1 , w2 , , wn be n reals Prove the inequality n n i=1 j=1 ijwi wj ≥ i+j−1 n wi i=1 Solution by Darij Grinberg Notations j the entry in the j-th column and the i i-th row of A [This is usually denoted by Aij or by Ai,j ] • For any matrix A, we denote by A • Let k be a field Let u ∈ N and v ∈ N, and let ai,j be an element of k for every (i, j) ∈ {1, 2, , u} × {1, 2, , v} Then, we denote by (ai,j )1≤j≤v 1≤i≤u the u × v matrix j A which satisfies A = ai,j for every (i, j) ∈ {1, 2, , u} × {1, 2, , v} i • Let n ∈ N Let t1 , t2 , , tn be n objects Let m ∈ {1, 2, , n} Then, we let t1 , t2 , , tm , , tn denote the (n − 1)-tuple (t1 , t2 , , tm−2 , tm−1 , tm+1 , tm+2 , , tn ) (that is, the (n − 1)-tuple (s1 , s2 , , sn−1 ) defined by si = ti , if i < m; for ti+1 , if i ≥ m all i ∈ {1, 2, , n − 1}) • Let R be a commutative ring with unity Let a1 , a2 , , am be m elements of R Then, we define an element σk (a1 , a2 , , am ) of R by σk (a1 , a2 , , am ) = a (i) S⊆{1,2, ,m}; i∈S |S|=k This element σk (a1 , a2 , , am ) is simply the k-th elementary symmetric polynomial evaluated at a1 , a2 , , am The Viete theorem states that m (−1)k σk (a1 , a2 , , am ) xm−k (x − a ) = ∈{1,2, ,m} k=0 for every x ∈ R If we choose some i ∈ {1, 2, , m} and apply this equality to the m − elements a1 , a2 , , , , am in lieu of the m elements a1 , a2 , , am , then we obtain m−1 (−1)k σk (a1 , a2 , , , , am ) xm−1−k (x − a ) = ∈{1,2, ,m}\{i} k=0 (1) Theorem (Sylvester) Let n ∈ N, and let A ∈ Rn×n be a symmetric n × n matrix Then, the matrix A is positive definite if and only if every 1≤j≤m j m ∈ {1, 2, , n} satisfies det A > i 1≤i≤m For a proof of Theorem 1, see any book on symmetric or Hermitian matrices Theorem (Cauchy determinant) Let k be a field Let m ∈ N Let a1 , a2 , , am be m elements of k Let b1 , b2 , , bm be m elements of k Assume that aj = bi for every (i, j) ∈ {1, 2, , m}2 Then, ((ai − aj ) (bj − bi )) det aj − bi 1≤j≤m = (i,j)∈{1,2, ,m}2 ; i>j (aj − bi ) 1≤i≤m (i,j)∈{1,2, ,m}2 In the following, I attempt to give the most conceptual proof of Theorem First we recall a known fact we are not going to prove: Theorem (Vandermonde determinant) Let S be a commutative ring with unity Let m ∈ N Let a1 , a2 , , am be m elements of S Then, det aj−1 i 1≤j≤m 1≤i≤m (ai − aj ) = (i,j)∈{1,2, ,m} ; i>j Besides, a trivial fact: Lemma Let S be a commutative ring with unity Let a ∈ S In the ring S [X] (the polynomial ring over S in one indeterminate X), the element X − a is not a zero divisor And a consequence of this fact: Lemma Let R be a commutative ring with unity Let m ∈ N In the ring R [X1 , X2 , , Xm ] (the polynomial ring over R in m indeterminates X1 , X2 , , Xm ), the element (Xi − Xj ) is not a zero divisor (i,j)∈{1,2, ,m}2 ; i>j Proof of Lemma We will first show that: For any (i, j) ∈ {1, 2, , m}2 satisfying i > j, the element Xi − Xj of the ring R [X1 , X2 , , Xm ] is not a zero divisor (2) Proof of (2) Let R X1 , X2 , , Xi , , Xm denote the sub-R-algebra of R [X1 , X2 , , Xm ] generated by the elements X1 , X2 , , Xi−2 , Xi−1 , Xi+1 , Xi+2 , , Xm (that is, the m elements X1 , X2 , , Xm except of Xi ) Consider the ring R X1 , X2 , , Xi , , Xm [X] (this is the polynomial ring over the ring R X1 , X2 , , Xi , , Xm in one indeterminate X) It is known that there exists an R-algebra isomorphism φ : R X1 , X2 , , Xi , , Xm R [X1 , X2 , , Xm ] such that φ (X) = Xi and φ (Xk ) = Xk for every k ∈ {1, 2, , m} \ = Xi − Xj Since X − Xj is not a zero {i} Hence, φ (X − Xj ) = φ (X) − φ (Xj ) =Xi [X] → =Xj , as j∈{1,2, ,m}\{i} divisor in R X1 , X2 , , Xi , , Xm [X] (by Lemma 4, applied to S = R X1 , X2 , , Xi , , Xm and a = Xj ), it thus follows that φ (X − Xj ) = Xi − Xj is not a zero divisor in R [X1 , X2 , , Xm ] (since φ is an R-algebra isomorphism) This proves (2) It is known that if we choose some elements of a ring such that each of these elements is not a zero divisor, then the product of these elements is not a zero divisor Hence, (2) yields that the element (Xi − Xj ) of the ring R [X1 , X2 , , Xm ] is not (i,j)∈{1,2, ,m}2 ; i>j a zero divisor This proves Lemma Now comes a rather useful fact: Theorem Let R be a commutative ring with unity Let m ∈ N Consider the ring R [X1 , X2 , , Xm ] (the polynomial ring over R in m indeterminates X1 , X2 , , Xm ) Then, det 1≤j≤m (−1)m−j σm−j X1 , X2 , , Xi , , Xm (Xi − Xj ) = 1≤i≤m (i,j)∈{1,2, ,m} ; j>i Proof of Theorem Let V = Xij−1 1≤j≤m 1≤i≤m Then, V j i = Xij−1 for every i ∈ {1, 2, , m} and j ∈ {1, 2, , m}, and det V = det Xij−1 1≤j≤m (Xi − Xj ) = 1≤i≤m (3) (i,j)∈{1,2, ,m} ; i>j (by Theorem 3, applied to S = R [X1 , X2 , , Xm ] and = Xi ) Let W = (−1)m−j σm−j X1 , X2 , , Xi , , Xm 1≤j≤m Then, 1≤i≤m j = (−1)m−j σm−j X1 , X2 , , Xi , , Xm for every i ∈ {1, 2, , m} and j ∈ i {1, 2, , m} For every i ∈ {1, 2, , m} and j ∈ {1, 2, , m}, we can apply (1) to x = Xj and ak = Xk , and obtain W m−1 (−1)k σk X1 , X2 , , Xi , , Xm Xjm−1−k (Xj − X ) = ∈{1,2, ,m}\{i} k=0 Now, for every i ∈ {1, 2, , m} and j ∈ {1, 2, , m}, we have (4) m j i WV T k i W = k=1 ci , ,Xm ) =(−1)m−k σm−k (X1 ,X2 , ,X j k VT · k j =V 5=X k−1 j m (−1)m−k σm−k X1 , X2 , , Xi , , Xm Xjk−1 = k=1 m−1 (−1)k σk X1 , X2 , , Xi , , Xm Xjm−1−k = (here, we substituted k for m − k in the sum) k=0 (Xj − X ) = (by (4)) (5) ∈{1,2, ,m}\{i} j j = (since W V T = (Xj − X ) , i i ∈{1,2, ,m}\{i} (Xj − X ) contains the factor Xj − Xj = and thus Thus, if j = i, then W V T but the product ∈{1,2, ,m}\{i} equals 0) Hence, the matrix W V T is diagonal Therefore, m det W V T = WV T i=1 i i m (Xi − X ) = i=1 ∈{1,2, ,m}\{i} since (5), applied to j = i, yields W V T (Xi − Xj ) = = (i, )∈{1,2, ,m}2 ; =i i i (Xi − X ) = ∈{1,2, ,m}\{i} (Xi − Xj ) = (i,j)∈{1,2, ,m}2 ; j=i (Xi − Xj ) · (i,j)∈{1,2, ,m}2 ; j>i (Xi − Xj ) (i,j)∈{1,2, ,m}2 ; i>j since the set (i, j) ∈ {1, 2, , m}2 | j = i is the union of the two disjoint sets (i, j) ∈ {1, 2, , m}2 | j > i and (i, j) ∈ {1, 2, , m}2 | i > j But on the other hand, det W V T = det W · det V T = det W · (Xi − Xj ) (i,j)∈{1,2, ,m} ; i>j (since det V T = det V = (Xi − Xj )) Hence, (i,j)∈{1,2, ,m}2 ; i>j (Xi − Xj ) = det W V T = det W · (Xi − Xj )· (i,j)∈{1,2, ,m} ; i>j (i,j)∈{1,2, ,m} ; j>i (Xi − Xj ) (i,j)∈{1,2, ,m} ; i>j (Xi − Xj ) of the ring R [X1 , X2 , , Xm ] is not a But since the element (i,j)∈{1,2, ,m}2 ; i>j zero divisor (according to Lemma 5), this yields (Xi − Xj ) det W = (i,j)∈{1,2, ,m} ; j>i Since W = (−1)m−j σm−j X1 , X2 , , Xi , , Xm det (−1)m−j σm−j X1 , X2 , , Xi , , Xm 1≤j≤m , this becomes 1≤i≤m 1≤j≤m (Xi − Xj ) = 1≤i≤m (i,j)∈{1,2, ,m} ; j>i Thus, Theorem is proven Next, we show: Theorem Let R be a commutative ring with unity Let m ∈ N Let a1 , a2 , , am be m elements of R Let b1 , b2 , , bm be m elements of R Then, 1≤j≤m (aj − b ) det ∈{1,2, ,m}\{i} ((ai − aj ) (bj − bi )) = (i,j)∈{1,2, ,m} ; i>j 1≤i≤m Proof of Theorem Consider the ring R [X1 , X2 , , Xm ] (the polynomial ring over R in m indeterminates X1 , X2 , , Xm ) 1≤j≤m j Let V = aj−1 Then, V for every i ∈ {1, 2, , m} and j ∈ = aj−1 i i 1≤i≤m i {1, 2, , m} Let W = (−1)m−j σm−j X1 , X2 , , Xi , , Xm 1≤j≤m Then, 1≤i≤m j = (−1)m−j σm−j X1 , X2 , , Xi , , Xm for every i ∈ {1, 2, , m} and j ∈ i {1, 2, , m} For every i ∈ {1, 2, , m} and j ∈ {1, 2, , m}, we can apply (1) to x = aj and ak = Xk , and obtain W m−1 (−1)k σk X1 , X2 , , Xi , , Xm am−1−k j (aj − X ) = ∈{1,2, ,m}\{i} k=0 Now, for every i ∈ {1, 2, , m} and j ∈ {1, 2, , m}, we have (6) j i WV T m = W k=1 k i j k · VT ci , ,Xm ) =(−1)m−k σm−k (X1 ,X2 , ,X =Ve k j 5=ak−1 j m (−1)m−k σm−k X1 , X2 , , Xi , , Xm ak−1 j = k=1 m−1 (−1)k σk X1 , X2 , , Xi , , Xm am−1−k j = (here, we substituted k for m − k in the sum) k=0 (aj − X ) = (by (6)) ∈{1,2, ,m}\{i} Hence, 1≤j≤m WV T = (aj − X ) ∈{1,2, ,m}\{i} Thus, 1≤i≤m 1≤j≤m det (aj − X ) = det W V T ∈{1,2, ,m}\{i} 1≤i≤m = det W · det V T =W Ve T (Xi − Xj ) · = (ai − aj ) (i,j)∈{1,2, ,m} ; j>i (i,j)∈{1,2, ,m} ; i>j 1≤j≤m m−j (−1) σm−j X1 , X2 , , Xi , , Xm = (Xi − Xj ) since det W = det 1≤i≤m (i,j)∈{1,2, ,m}2 ; j>i j−1 1≤j≤m T by Theorem and det V = det V = det 1≤i≤m = (ai − aj ) (i,j)∈{1,2, ,m}2 ; i>j by Theorem (Xj − Xi ) · = (ai − aj ) (j,i)∈{1,2, ,m} ; i>j (i,j)∈{1,2, ,m} ; i>j (here, we renamed i and j as j and i in the first product) (Xj − Xi ) · = (i,j)∈{1,2, ,m} ; i>j (ai − aj ) (i,j)∈{1,2, ,m} ; i>j ((Xj − Xi ) (ai − aj )) = = (i,j)∈{1,2, ,m} ; i>j ((ai − aj ) (Xj − Xi )) (i,j)∈{1,2, ,m} ; i>j Both sides of this identity are polynomials over the ring R in m indeterminates X1 , X2 , , Xm Evaluating these polynomials at X1 = b1 , X2 = b2 , , Xm = bm , we obtain 1≤j≤m (aj − b ) det ∈{1,2, ,m}\{i} ((ai − aj ) (bj − bi )) = (i,j)∈{1,2, ,m} ; i>j 1≤i≤m This proves Theorem Proof of Theorem For every i ∈ {1, 2, , m} and j ∈ {1, 2, , m} , we have (aj − b ) = aj − bi ∈{1,2, ,m}\{i} (aj − b ) ∈{1,2, ,m} Hence, the matrix 1≤j≤m 1≤j≤m aj − b i (aj − b ) is what we obtain if we take the matrix ∈{1,2, ,m}\{i} 1≤i≤m (aj − b ) for every j ∈ {1, 2, , m} Therefore, and divide its j-th column by ∈{1,2, ,m} det aj − b i 1≤j≤m (aj − b ) det 1≤j≤m ∈{1,2, ,m}\{i} (aj − b ) j∈{1,2, ,m} ∈{1,2, ,m} 1≤j≤m (aj − b ) det 1≤i≤m = 1≤i≤m ∈{1,2, ,m}\{i} 1≤i≤m = (aj − b ) ( ,j)∈{1,2, ,m} ((ai − aj ) (bj − bi )) = (i,j)∈{1,2, ,m}2 ; i>j (aj − bi ) (i,j)∈{1,2, ,m} (by Theorem 7) Thus, Theorem is proven Theorem Let n ∈ N Let a1 , a2 , , an be n pairwise distinct reals Let c be a real such that + aj + c > for every (i, j) ∈ {1, 2, , n}2 Then, 1≤j≤n the matrix ∈ Rn×n is positive definite + aj + c 1≤i≤n 1≤j≤n 1 j Proof of Theorem Let A = Then, A = i + aj + c 1≤i≤n + aj + c for every i ∈ {1, 2, , m} and j ∈ {1, 2, , m} Thus, A ∈ Rn×n is a symmetric n × n matrix Define n reals b1 , b2 , , bn by bi = −ai − c for every i ∈ {1, 2, , n} Then, aj = bi for every (i, j) ∈ {1, 2, , n}2 (since aj − bi = aj − (−ai − c) = + aj + c > 0) 1≤i≤m Now, every m ∈ {1, 2, , n} satisfies det A 1≤j≤m j i = det 1≤i≤m aj − bi 1≤j≤m 1≤i≤m 1 j = = = i + aj + c aj − (−ai − c) aj − bi ((ai − aj ) (bj − bi )) since A = (i,j)∈{1,2, ,m}2 ; i>j by Theorem 2, since aj = bi for every (i, j) ∈ {1, 2, , m}2 (aj − bi ) (i,j)∈{1,2, ,m}2 (ai − aj )2 (i,j)∈{1,2, ,m} ; i>j = (ai + aj + c) (i,j)∈{1,2, ,m}2 since (ai − aj ) (bj − bi ) = (ai − aj ) ((−aj − c) − (−ai − c)) = (ai − aj )2 and aj − bi = aj − (−ai − c) = + aj + c >0 (since (ai − aj )2 > for every (i, j) ∈ {1, 2, , m}2 satisfying i > j (because a1 , a2 , , an are pairwise distinct, so that = aj , thus − aj = 0), and + aj + c > for every (i, j) ∈ {1, 2, , m}2 ) Hence, according to Theorem 1, the symmetric matrix A is positive definite Since 1≤j≤n 1≤j≤n 1 A= , this means that the matrix is positive + aj + c 1≤i≤n + aj + c 1≤i≤n definite Thus, Theorem is proven Corollary Let n ∈ N Let a1 , a2 , , an be n pairwise distinct reals Let c be a real such that + aj + c > for every (i, j) ∈ {1, 2, , n}2 Let n n vi vj ≥ holds, v1 , v2 , , be n reals Then, the inequality i=1 j=1 + aj + c with equality if and only if v1 = v2 = = = v1 v2 Proof of Corollary Define a vector v ∈ Rn by v = Then, v T + aj + c n 1≤j≤n n v= 1≤i≤n i=1 j=1 vi vj = + aj + c n n i=1 j=1 vi vj + aj + c (7) Also, obviously, v = holds if and only if v1 = v2 = = = Now, since the matrix 8), we have v T + aj + c + aj + c (8) 1≤j≤n ∈ Rn×n is positive definite (by Theorem 1≤i≤n 1≤j≤n v ≥ 0, with equality if and only if v = According 1≤i≤n n n vi vj ≥ 0, with equality if and only if i=1 j=1 + aj + c v1 = v2 = = = Thus, Corollary is proven to (7) and (8), this means that Corollary 10 Let n ∈ N Let a1 , a2 , , an be n pairwise distinct reals Let c be a real such that + aj + c > for every (i, j) ∈ {1, 2, , n}2 n n aj wi wj ≥ Let w1 , w2 , , wn be n reals Then, the inequality i=1 j=1 + aj + c n −c wi holds, with equality if and only if (c + a1 ) w1 = (c + a2 ) w2 = i=1 = (c + an ) wn = Proof of Corollary 10 Define n reals v1 , v2 , , by vi = (c + ) wi for every i ∈ {1, 2, , n} Then, n n n aj wi wj − −c wi a + a + c i j i=1 j=1 i=1 n n = i=1 j=1 wi n i=1 j=1 n n = i=1 j=1 i=1 j=1 n aj wi wj +c + aj + c n n wi wj i=1 j=1 n wi wj = i=1 j=1 aj wi wj + cwi wj + aj + c = n = i=1 i=1 n n wi n since n aj wi wj +c + aj + c (c + ) (c + aj ) wi wj = + aj + c n n = n i=1 j=1 n i=1 j=1 aj + c wi wj + aj + c (c + ) wi (c + aj ) wj = + aj + c n n i=1 j=1 vi vj + aj + c (since (c + ) wi = vi and (c + aj ) wj = vj ) Hence, n n i=1 j=1 aj wi wj ≥ −c + aj + c n wi n n holds if and only if i=1 i=1 j=1 vi vj ≥ + aj + c (9) Also, clearly, v1 = v2 = = = holds if and only if (c + a1 ) w1 = (c + a2 ) w2 = = (c + an ) wn = (10) n n vi vj By Corollary 9, the inequality ≥ holds, with equality if and only i=1 j=1 + aj + c n n aj wi wj ≥ if v1 = v2 = = = According to (9) and (10), this means that i=1 j=1 + aj + c n −c wi , with equality if and only if (c + a1 ) w1 = (c + a2 ) w2 = = (c + an ) wn = i=1 Thus, Corollary 10 is proven The problem follows from Corollary 10 (applied to c = −1 and = i) ... , , Xi , , Xm for every i ∈ {1, 2, , m} and j ∈ i {1, 2, , m} For every i ∈ {1, 2, , m} and j ∈ {1, 2, , m}, we can apply (1) to x = Xj and ak = Xk , and obtain W m−1 (−1)k σk X1 , X2 , , Xi... , m} and j ∈ {1, 2, , m}, we can apply (1) to x = aj and ak = Xk , and obtain W m−1 (−1)k σk X1 , X2 , , Xi , , Xm am−1−k j (aj − X ) = ∈{1,2, ,m}{i} k=0 Now, for every i ∈ {1, 2, , m} and j... the inequality ≥ holds, with equality if and only i=1 j=1 + aj + c n n aj wi wj ≥ if v1 = v2 = = = According to (9) and (10), this means that i=1 j=1 + aj + c n −c wi , with equality if and