17. Vandermonde determinants

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1 17. Vadermode determiats 17.1 Vadermode determiats 17.2 Worked examples 1. Vadermode determiats A rigorous systematic evaluatio of Vadermode determiats below) of the followig idetity uses the fact that a polyomial rig over a UFD is agai a UFD. A Vadermode matrix is a square matrix of the form i the theorem. [1.0.1] Theorem: det x 1 x 2... x x 2 1 x x 3 1 x x 2 x 3... x 1 1 x x 1 = 1) 1)/2 x i x j ) i<j [1.0.2] Remark: The most uiversal versio of the assertio uses idetermiates x i, ad proves a idetity i Z[x 1,..., x ] Proof: First, the idea of the proof. Whatever the determiat may be, it is a polyomial i x 1,..., x. The most uiversal choice of iterpretatio of the coefficiets is as i Z. If two colums of a matrix are the same, the the determiat is 0. From this we would wat to coclude that for i j the determiat is divisible by [1] x i x j i the polyomial rig Z[x 1,..., x ]. If we ca coclude that, the, sice these polyomials [1] If oe treats the x i merely as complex umbers, for example, the oe caot coclude that the product of the expressios x i x j with i < j divides the determiat. Attemptig to evade this problem by declarig the x i as somehow variable complex umbers is a impulse i the right directio, but is made legitimate oly by treatig geuie idetermiates. 223

2 224 Vadermode determiats are pairwise relatively prime, we ca coclude that the determiat is divisible by x i x j ) i<j Cosideratios of degree will show that there is o room for further factors, so, up to a costat, this is the determiat. To make sese of this lie of argumet, first observe that a determiat is a polyomial fuctio of its etries. Ideed, the formula is det M = σp) M 1p1) M 2p2)... M p) p where p rus over permutatios of thigs ad σp) is the sig or parity of p, that is, σp) is +1 if p is a product of a eve umber of 2-cycles ad is 1 if p is the product of a odd umber of 2-cycles. Thus, for ay Z-algebra homomorphism f to a commutative rig R with idetity, we have f : Z[x 1,..., x ] R fdet V ) = det fv ) where by fv ) we mea applicatio of f etry-wise to the matrix V. Thus, if we ca prove a idetity i Z[x 1,..., x ], the we have a correspodig idetity i ay rig. Rather tha talkig about settig x j equal to x i, it is safest to try to see divisibility property as directly as possible. Therefore, we do ot attempt to use the property that the determiat of a matrix with two equal colums is 0. Rather, we use the property [2] that if a elemet r of a rig R divides every elemet of a colum or row) of a square matrix, the it divides the determiat. Ad we are allowed to add ay multiple of oe colum to aother without chagig the value of the determiat. Subtractig the j th colum from the i th colum of our Vadermode matrix with i < j), we have From the idetity x i x j... x j x 2 i x2 j... x 2 j... det V = det... x 3 i x3 j... x 3 j x 1 i x 1 j... x 1 j... x m y m = x y)x m 1 + x m 2 y y m 1 ) it is clear that x i x j divides all etries of the ew i th colum. Thus, x i x j divides the determiat. This holds for all i < j. Sice these polyomials are liear, they are irreducible i Z[x 1,..., x ]. Geerally, the uits i a polyomial rig R[x 1,..., x ] are the uits R i R, so the uits i Z[x 1,..., x ] are just ±1. Visibly, the various irreducible x i x j are ot associate, that is, do ot merely differ by uits. Therefore, their least commo multiple is their product. Sice Z[x 1,..., x ] is a UFD, this product divides the determiat of the Vadermode matrix. To fiish the computatio, we wat to argue that the determiat ca have o further polyomial factors tha the oes we ve already determied, so up to a costat which we ll determie) is equal to the latter [2] This follows directly from the just-quoted formula for determiats, ad also from other descriptios of determiats, but from ay viewpoit is still valid for matrices with etries i ay commutative rig with idetity.

3 Garrett: Abstract Algebra 225 product. [3] To prove this, we eed the otio of total degree: the total degree of a moomial x m x m is m m, ad the total degree of a polyomial is the maximum of the total degrees of the moomials occurrig i it. We grat for the momet the result of the propositio below, that the total degree of a product is the sum of the total degrees of the factors. The total degree of the product is 1 i<j 1 = 1 i< i = 1 1) 2 To determie the total degree of the determiat, ivoke the usual formula for the determiat of a matrix M with etries M ij, amely det M = π σπ) i M i,πi) where π is summed over permutatios of thigs, ad where σπ) is the sig of the permutatio π. I a Vadermode matrix all the top row etries have total degree 0, all the secod row etries have total degree 1, ad so o. Thus, i this permutatio-wise sum for a Vadermode determiat, each summad has total degree ) = 1 1) 2 so the total degree of the determiat is the total degree of the product Thus, det 1 i<j 1 = 1 i< x 1 x 2... x x 2 1 x x 3 1 x x 2 x 3... x 1 1 x x 1 i = 1 1) 2 = costat x i x j ) i<j Gratig this, to determie the costat it suffices to compare a sigle moomial i both expressios. For example, compare the coefficiets of x 1 1 x 2 2 x x 1 1x 0 I the product, the oly way x 1 1 appears is by choosig the x 1 s i the liear factors x 1 x j with 1 < j. After this, the oly way to get x 2 2 is by choosig all the x 2 s i the liear factors x 2 x j with 2 < j. Thus, this moomial has coefficiet +1 i the product. I the determiat, the oly way to obtai this moomial is as the product of etries from lower left to upper right. The idices of these etries are, 1), 1, 2),..., 2, 1), 1, ). Thus, the coefficiet of this moomial is 1) l where l is the umber of 2-cycles ecessary to obtai the permutatio p such that pi) = + 1 i Thus, for eve there are /2 two-cycles, ad for odd 1)/2 two-cycles. For a closed form, as these expressios will appear oly as expoets of 1, we oly care about values modulo 2. Because of the divisio by 2, we oly care about modulo 4. Thus, we have values /2 = 0 mod 2 for = 0 mod 4) 1)/2 = 0 mod 2 for = 1 mod 4) /2 = 1 mod 2 for = 3 mod 4) 1)/2 = 1 mod 2 for = 1 mod 4) [3] This is more straightforward tha settig up the right viewpoit for the first part of the argumet.

4 226 Vadermode determiats After some experimetatio, we fid a closed expressio Thus, the leadig costat is 1)/2 mod 2 1) 1)/2 i the expressio for the Vadermode determiat. /// Verify the property of total degree: [1.0.3] Lemma: Let fx 1,..., x ) ad gx 1,..., x ) be polyomials i k[x 1,..., x ] where k is a field. The the total degree of the product is the sum of the total degrees. Proof: It is clear that the total degree of the product is less tha or equal the sum of the total degrees. Let x e xe ad x f xf be two moomials of highest total degrees s = e e ad t = f f occurrig with o-zero coefficiets i f ad g, respectively. Assume without loss of geerality that the expoets e 1 ad f 1 of x 1 i the two expressios are the largest amog all moomials of total degrees s, t i f ad g, respectively. Similarly, assume without loss of geerality that the expoets e 2 ad f 2 of x 2 i the two expressios are the largest amog all moomials of total degrees s, t i f ad g, respectively, of degrees e 1 ad f 1 i x 1. Cotiuig similarly, we claim that the coefficiet of the moomial M = x e1+f1... x e+f is simply the product of the coefficiets of x e xe ad x f xf, so o-zero. Let x u xu ad x v xv be two other moomials occurrig i f ad g such that for all idices i we have u i + v i = e i + f i. By the maximality assumptio o e 1 ad f 1, we have e 1 u 1 ad f 1 v 1, so the oly way that the ecessary power of x 1 ca be achieved is that e 1 = u 1 ad f 1 = v 1. Amog expoets with these maximal expoets of x 1, e 2 ad f 2 are maximal, so e 2 u 2 ad f 2 v 2, ad agai it must be that e 2 = u 2 ad f 2 = v 2 to obtai the expoet of x 2. Iductively, u i = e i ad v i = f i for all idices. That is, the oly terms i f ad g cotributig to the coefficiet of the moomial M i f g are moomials x e xe ad x f xf. Thus, the coefficiet of M is o-zero, ad the total degree is as claimed. /// 2. Worked examples [17.1] Show that a fiite itegral domai is ecessarily a field. Let R be the itegral domai. The itegral domai property ca be immediately paraphrased as that for 0 x R the map y xy has trivial kerel as R-module map of R to itself, for example). Thus, it is ijective. Sice R is a fiite set, a ijective map of it to itself is a bijectio. Thus, there is y R such that xy = 1, provig that x is ivertible. /// [17.2] Let P x) = x 3 + ax + b k[x]. Suppose that P x) factors ito liear polyomials P x) = x α 1 )x α 2 )x α 3 ). Give a polyomial coditio o a, b for the α i to be distict. Oe might try to do this as a symmetric fuctio computatio, but it s a bit tedious.) If P x) = x 3 + ax + b has a repeated factor, the it has a commo factor with its derivative P x) = 3x 2 + a. If the characteristic of the field is 3, the the derivative is the costat a. Thus, if a 0, gcdp, P ) = a k is ever 0. If a = 0, the the derivative is 0, ad all the α i are the same. Now suppose the characteristic is ot 3. I effect applyig the Euclidea algorithm to P ad P, x 3 + ax + b ) x 3 3x 2 + a ) = ax + b x 3 a = 2 3 ax + b

5 Garrett: Abstract Algebra 227 If a = 0 the the Euclidea algorithm has already termiated, ad the coditio for distict roots or factors is b 0. Also, possibly surprisigly, at this poit we eed to cosider the possibility that the characteristic is 2. If so, the the remaider is b, so if b 0 the roots are always distict, ad if b = 0 Now suppose that a 0, ad that the characteristic is ot 2. The we ca divide by 2a. Cotiue the algorithm 3x 2 + a ) 9x ) 2 2a 3 ax + b = a + 27b2 4a 2 Sice 4a 2 0, the coditio that P have o repeated factor is 4a b 2 0 [17.3] The first three elemetary symmetric fuctios i idetermiates x 1,..., x are σ 1 = σ 1 x 1,..., x ) = x 1 + x x = i x i Express x x x 3 i terms of σ 1, σ 2, σ 3. σ 2 = σ 2 x 1,..., x ) = i<j σ 3 = σ 3 x 1,..., x ) = i<j<l x i x j x i x j x l Execute the algorithm give i the proof of the theorem. Thus, sice the degree is 3, if we ca derive the right formula for just 3 idetermiates, the same expressio i terms of elemetary symmetric polyomials will hold geerally. Thus, cosider x 3 + y 3 + z 3. To approach this we first take y = 0 ad z = 0, ad cosider x 3. This is s 1 x) 3 = x 3. Thus, we ext cosider x 3 + y 3) s 1 x, y) 3 = 3x 2 y + 3xy 2 As the algorithm assures, this is divisible by s 2 x, y) = xy. Ideed, x 3 + y 3) s 1 x, y) 3 = 3x + 3y)s 2 x, y) = 3s 1 x, y) s 2 x, y) The cosider x 3 + y 3 + z 3) s 1 x, y, z) 3 3 s 2 x, y, z) s 1 x, y, z) ) = 3xyz = 3s 3 x, y, z) Thus, agai, sice the degree is 3, this formula for 3 variables gives the geeral oe: where s i = s i x 1,..., x ). x x 3 = s 3 1 3s 1 s 2 + 3s 3 [17.4] Express i j x2 i x j as a polyomial i the elemetary symmetric fuctios of x 1,..., x. We could as i the previous problem) execute the algorithm that proves the theorem assertig that every symmetric that is, S -ivariat) polyomial i x 1,..., x is a polyomial i the elemetary symmetric fuctios. But, also, sometimes ad hoc maipulatios ca yield shortcuts, depedig o the cotext. Here, x 2 i x j = x 2 i x j ) ) x 2 i x j = x 2 i x j x 3 i i,j i=j i j i i j

6 228 Vadermode determiats A easier versio of the previous exercise gives ad the previous exercise itself gave Thus, i x 2 i = s 2 1 2s 2 x 3 i = s 3 1 3s 1 s 2 + 3s 3 i x 2 i x j = s 2 1 2s 2 ) s 1 s 3 ) 1 3s 1 s 2 + 3s 3 = s 3 1 2s 1 s 2 s s 1 s 2 3s 3 = s 1 s 2 3s 3 i j [17.5] Suppose the characteristic of the field k does ot divide. Let l > 2. Show that is irreducible i k[x 1,..., x l ]. P x 1,..., x ) = x x l First, treatig the case l = 2, we claim that x + y is ot a uit ad has o repeated factors i ky)[x]. We take the field of ratioal fuctios i y so that the resultig polyomial rig i a sigle variable is Euclidea, ad, thus, so that we uderstad the behavior of its irreducibles.) Ideed, if we start executig the Euclidea algorithm o x + y ad its derivative x 1 i x, we have x + y ) x x 1 ) = y Note that is ivertible i k by the characteristic hypothesis. Sice y is ivertible beig o-zero) i ky), this says that the gcd of the polyomial i x ad its derivative is 1, so there is o repeated factor. Ad the degree i x is positive, so x + y has some irreducible factor due to the uique factorizatio i ky)[x], or, really, due idirectly to its Noetheria-ess). Thus, our iductio o ) hypothesis is that x 2 + x x l is a o-uit i k[x 2, x 3,..., x ] ad has o repeated factors. That is, it is divisible by some irreducible p i k[x 2, x 3,..., x ]. The i k[x 2, x 3,..., x ][x 1 ] k[x 1, x 2, x 3,..., x ] Eisestei s criterio applied to x as a polyomial i x 1 with coefficiets i k[x 2, x 3,..., x ] ad usig the irreducible p yields the irreducibility. [17.6] Fid the determiat of the circulat matrix x 1 x 2... x 2 x 1 x x x 1 x 2... x 2 x 1 x 1 x x 1 x 2... x x 3 x 1 x 2 x 2 x 3... x x 1 Hit: Let ζ be a th root of 1. If x i+1 = ζ x i for all idices i <, the the j + 1) th row is ζ times the j th, ad the determiat is 0. ) Let C ij be the ij th etry of the circulat matrix C. The expressio for the determiat det C = σp) C 1,p1)... C,p) p S

7 Garrett: Abstract Algebra 229 where σp) is the sig of p shows that the determiat is a polyomial i the etries C ij with iteger coefficiets. This is the most uiversal viewpoit that could be take. However, with some hidsight, some itermediate maipulatios suggest or require elargig the costats to iclude th roots of uity ω. Sice we do ot kow that Z[ω] is a UFD ad, ideed, it is ot, i geeral), we must adapt. A reasoable adaptatio is to work over Qω). Thus, we will prove a idetity i Qω)[x 1,..., x ]. Add ω i 1 times the i th row to the first row, for i 2. The ew first row has etries, from left to right, x 1 + ωx 2 + ω 2 x ω 1 x x 2 + ωx 3 + ω 2 x ω 1 x 1 x 3 + ωx 4 + ω 2 x ω 1 x 2 x 4 + ωx 5 + ω 2 x ω 1 x 3... x 2 + ωx 3 + ω 2 x ω 1 x 1 The t th of these is ω t x 1 + ωx 2 + ω 2 x ω 1 x ) sice ω = 1. Thus, i the rig Qω)[x 1,..., x ], x 1 + ωx 2 + ω 2 x ω 1 x ) divides this ew top row. determiat. Therefore, from the explicit formula, for example, this quatity divides the Sice the characteristic is 0, the roots of x 1 = 0 are distict for example, by the usual computatio of gcd of x 1 with its derivative). Thus, there are superficially-differet liear expressios which divide det C. Sice the expressios are liear, they are irreducible elemets. If we prove that they are o-associate do ot differ merely by uits), the their product must divide det C. Ideed, viewig these liear expressios i the larger rig Qω)x 2,..., x )[x 1 ] we see that they are distict liear moic polyomials i x 1, so are o-associate. Thus, for some c Qω), det C = c 1 l x 1 + ω l x 2 + ω 2l x 3 + ω 3l x ω 1)l x ) Lookig at the coefficiet of x 1 o both sides, we see that c = 1. Oe might also observe that the product, whe expaded, will have coefficiets i Z.) Exercises 17.[2.0.1] A k-liear derivatio D o a commutative k-algebra A, where k is a field, is a k-liear map D : A A satisfyig Leibiz idetity Dab) = Da) b + a Db) Give a polyomial P x), show that there is a uique k-liear derivatio D o the polyomial rig k[x] sedig x to P x).

8 230 Vadermode determiats 17.[2.0.2] Let A be a commutative k-algebra which is a itegral domai, with field of fractios K. Let D be a k-liear derivatio o A. Show that there is a uique extesio of D to a k-liear derivatio o K, ad that this extesio ecessarily satisfies the quotiet rule. 17.[2.0.3] Let fx 1,..., x ) be a homogeeous polyomial of total degree, with coefficiets i a field k. Let / x i be partial differetiatio with respect to x i. Prove Euler s idetity, that i=1 x i f x i = f 17.[2.0.4] Let α be algebraic over a field k. Show that ay k-liear derivatio D o kα) ecessarily gives Dα = 0.

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