Math 4707 Sprig 2018 Darij Griberg): homewor set 4 page 1 Math 4707 Sprig 2018 Darij Griberg): homewor set 4 due date: Wedesday 11 April 2018 at the begiig of class, or before that by email or moodle Please solve at most 3 of the 6 exercises! Cotets 0.1. Permutatios σ with σ 2) = σ 1) + 1.................. 1 0.2. A itroductio to roo theory...................... 1 0.3. The sum of the widths of all iversios of σ............. 6 0.4. A hollowed-out determiat....................... 6 0.5. Arrowhead matrices............................ 7 Recall the followig: We have N = {0, 1, 2,...}. If N, the [] deotes the -elemet set {1, 2,..., }. For each N, we let S deote the set of all permutatios of []. 0.1. Permutatios σ with σ 2) = σ 1) + 1 Exercise 1. Let 2 be a iteger. Prove that there are precisely 1)! permutatios σ S satisfyig σ 2) = σ 1) + 1. 0.2. A itroductio to roo theory Defiitio 0.1. For ay x R ad N, we let x deote the -th lower factorial ) of x ; this is the real umber x x 1) x + 1). Thus, x = x!.) For example, x 0 = 1, x 1 = x, x 2 = x x 1), etc. 1. Exercise 2. Let N. Prove the followig: a) If a 0, a 1,..., a ) is a + 1)-tuple of ratioal umbers such that each x {0, 1,..., } satisfies a x = 0, =0 1 For all LATEX users: x is x^{\uderlie{}}. Feel free to create a macro for this, e.g., by puttig the followig lie ito the header of your TeX file alog with the other VARIOUS USEFUL COMMANDS ): \ewcommad{\lf}[2]{{#1}^{\uderlie{#2}}} The, you ca use \lf{x}{} to obtai x.
Math 4707 Sprig 2018 Darij Griberg): homewor set 4 page 2 the a 0, a 1,..., a ) = 0, 0,..., 0). b) If a 0, a 1,..., a ) ad b 0, b 1,..., b ) are two + 1)-tuples of ratioal umbers such that each x {0, 1,..., } satisfies a x = b x, =0 =0 the a 0, a 1,..., a ) = b 0, b 1,..., b ). [Hit: I terms of liear algebra, part a) is sayig that the + 1 vectors 0, 1,..., ) T Q +1 for {0, 1,..., } are liearly idepedet. You may fid this useful or ot; the exercise has a fully elemetary solutio.] I the followig, we will cosider each pair i, j) Z 2 of two itegers as a square o a ifiite) chessboard; we say that it lies i row i ad i colum j. A roo placemet shall mea a subset X of Z 2 such that ay two distict elemets of X lie i differet rows ad i differet colums 2. The idea behid this ame is that if we place roos ito the squares i, j) X, the o two roos will attac each other.) For example, {1, 3), 2, 2), 3, 7)} is a roo placemet, whereas {1, 3), 2, 2), 7, 3)} is ot sice the distict squares 1, 3) ad 7, 3) lie i the same colum). If X is a roo placemet, the the elemets of X are called the roos of X. Now, fix N. If u = u 1, u 2,..., u ) N is a -tuple of oegative itegers, the we defie the u-board D u) to be the set {i, j) i [] ad j [u i ]}. We visually represet this set as a chessboard cosistig of left-aliged rows 3, where the i-th row cosists of u i boxes which occupy colums 1, 2,..., u i ). For example, if = 4 ad u = 2, 1, 4, 1), the D u) is the set {1, 1), 1, 2), 2, 1), 3, 1), 3, 2), 3, 3), 3, 4), 4, 1)}, ad is visually represeted as the chessboard. The set or chessboard ) D u) is also ow as the Youg diagram or Ferrers diagram) of u; we have briefly see its use i the proof of Propositio 4.13 o March 21st. 2 I other words, a roo placemet meas a subset X of Z 2 such that ay two distict elemets i, j) ad i, j ) of X satisfy i = i ad j = j. 3 or ras, to use the termiology of chess but we label them 1, 2,..., from top to bottom, ot from bottom to top as o a actual chessboard)
Math 4707 Sprig 2018 Darij Griberg): homewor set 4 page 3 If u N, the a roo placemet i D u) meas a subset X of D u) that is a roo placemet. For example, if = 4 ad u = 2, 1, 4, 1), the {1, 2), 3, 3), 4, 1)} is a roo placemet i D u). We represet this roo placemet by puttig roos i.e., R symbols) ito the squares that belog to it; so we get R R R. Note that the empty set is always a roo placemet i D u); so is ay 1- elemet subset of D u). If u N ad N, the we let R u) deote the umber of all roo placemets i D u) of size. I terms of the picture, size meas that it cotais exactly roos.) For example, if = 3 ad u = 2, 1, 3), the R 0 u) = 1; R 1 u) = 6; R 2 u) = 7; R 3 u) = 1; R u) = 0 for all 4. It is easy to see that if u = u, u,..., u for some u N, the }{{} times R u) = ) u )! for each N. 1) Ideed, i order to place o-attacig roos i D u), we first choose the rows they will occupy, the the colums they will occupy, ad fially a appropriate permutatio of [] that will determie which colum has a roo i which row.) For other -tuples u, fidig R u) is harder. For ) example, if you try the same reasoig for u = 2 1, 2 3,..., 5, 3, 1), the you still have choices for the rows occupied by roos; but the umber of optios i the followig steps will deped o the specific rows you have chose the lower the rows, the fewer optios). This way, you get the formula R 2 1, 2 3,..., 5, 3, 1) = s 0) s 1 1) s 2 2) s 1 1)), 1 s 1 <s 2 < <s which is a far cry from the simplicity of 1). But there is a simple formula, which we ll see i Corollary 0.3! We ca restrict ourselves to -tuples u = u 1, u 2,..., u ) satisfyig u 1 u 2 u, because switchig some of the etries of u does ot chage the values of R u). I geeral, R u) = 0 for all >, because each of the rows 1, 2,..., cotais at most oe roo. For =, we have a eat formula:
Math 4707 Sprig 2018 Darij Griberg): homewor set 4 page 4 Propositio 0.2. Let u = u 1, u 2,..., u ) N be such that u 1 u 2 u. The, 1 R u) = u i i). i=0 Proof of Propositio 0.2. Each square i D u) belogs to oe of the rows 1, 2,...,. I a roo placemet, each row cotais at most oe roo. Thus, ay roo placemet i D u) cotais at most roos, ad the oly way it ca cotai roos is if it cotais oe roo i each of the rows 1, 2,...,. Hece, a roo placemet i D u) of size is the same as a roo placemet i D u) that cotais a roo i each of the rows 1, 2,...,. We ca costruct such a roo placemet by the followig algorithm: First, we place the roo i row. This roo must lie i oe of the u colums 1, 2,..., u ; so we have u optios. The, we place the roo i row 1. This roo must lie i oe of the u 1 colums 1, 2,..., u 1, but ot i the colum that cotais the previous roo; so we have u 1 1 optios. The, we place the roo i row 2. This roo must lie i oe of the u 2 colums 1, 2,..., u 2, but ot i either of the two colums that cotai the previous roos; so we have u 2 2 optios. Ad so o. Thus, the total umber of choices ad therefore the total umber of roo placemets i D u) of size ) is ) u u 1 1) u 2 2) u 1) 1) = u 0) u 1 1) u 2 2) u 1 1)) = 1 u i i). i=0 Sice the total umber of roo placemets i D u) of size has bee deoted by R u), we thus coclude that R u) = 1 u i i). i=0 This proves Propositio 0.2. [Are you woderig where we have used the coditio u 1 u 2 u?
Math 4707 Sprig 2018 Darij Griberg): homewor set 4 page 5 Aswer: 1, 2,..., u 2 sice u 2 u), ad similarly for the secod forbidde colum. colum is amog the colums 1, 2,..., u ad therefore also amog the colums u 2 2 optios for our roo). This fact is true because the first forbidde colums 1, 2,..., u 2 if this was ot the case, the we would have more tha colums i.e., the two colums that cotai the previous roos) are amog the u 2 cotai the previous roos. This tacitly relied o the fact that the two forbidde placed i ay of the u 2 colums 1, 2,..., u 2 except for the two colums that i row 2, we argued that we had u 2 2 optios, because the roo could be Aswer: Cosider the algorithm above. Whe we placed the roo ] Exercise 3. Let u = u 1, u 2,..., u ) N be such that u 1 u 2 u. a) For each x N, defie a -tuple u + x N by u + x = u 1 + x, u 2 + x,..., u + x). Prove that R u + x) = R u) x. =0 [Hit: Whe u is replaced by u + x, the u-board grows by x extra full) colums.] b) Now, let v = v 1, v 2,..., v ) N be such that v 1 v 2 v. Assume further that the umbers u 1 + 1, u 2 + 2,..., u + are the same as the umbers v 1 + 1, v 2 + 2,..., v +, up to order. I other words, there exists a permutatio σ S such that u i + i = v σi) + σ i) for all i [].) Prove that R u) = R v) for each N. [Hit: First, prove that R u + x) = R v + x) for each x N. The, use part a) ad the previous exercise.] To illustrate the usefuless of Exercise 3, let me express R 2 1, 2 3,..., 5, 3, 1) i a much simpler way tha before: Corollary 0.3. Let N. The, R 2 1, 2 3,..., 5, 3, 1) = ) 2!. Proof of Corollary 0.3. Defie a -tuple u = u 1, u 2,..., u ) N by u i = 2 i) + 1 for each i []). Thus, u = u 1, u 2,..., u ) = 2 1, 2 3,..., 5, 3, 1), so that u 1 u 2 u. Defie a -tuple v = v 1, v 2,..., v ) N by v i = for each i []). Thus, v = v 1, v 2,..., v ) =,,..., ),
Math 4707 Sprig 2018 Darij Griberg): homewor set 4 page 6 so that v 1 v 2 v. The umbers u 1 + 1, u 2 + 2,..., u + are the same as the umbers v 1 + 1, v 2 + 2,..., v +, up to order. I fact, the former umbers are 2, 2 1, 2 2,..., + 2, + 1, whereas the latter umbers are + 1, + 2,..., 2 2, 2 1, 2; these are just two differet ways to list all umbers from + 1 to 2.) Hece, Exercise 3 b) yields R u) = R v). I view of u = 2 1, 2 3,..., 5, 3, 1) ad v =,,..., ), this rewrites as R 2 1, 2 3,..., 5, ) 3, 1) ) = R,,..., ). But 1) applied to u = ) yields R,,..., ) =!. Hece, This proves Corollary 0.3. ) ) R 2 1, 2 3,..., 5, 3, 1) = R,,..., ) =! = ) 2!. Roo theory is the study of roo placemets ot oly i boards of the form D u), but also i more geeral subsets of Z 2. For example, the permutatios of [] ca be regarded as roo placemets i the board [] [], whereas the deragemets of [] ca be regarded as roo placemets i the board {i, j) [] [] i = j} the -chessboard without the mai diagoal ). The theory has bee developed i the 70s at the UMN by Jay Goldma, J. T. Joichi, Victor Reier ad Deis White). 0.3. The sum of the widths of all iversios of σ I the followig, umber meas real umber or complex umber or ratioal umber, as you prefer this does t mae a differece i these exercises). Exercise 4. Let N. Let σ S. Let a 1, a 2,..., a be ay umbers. Prove that [Here, the symbol ) aj a i = a i i σ i)). 1 i<j ; i=1 σi)>σj) 1 i<j ; σi)>σj) meas sum over all pairs i, j) [] 2 satisfyig i < j ad σ i) > σ j), that is, sum over all iversios of σ.] 0.4. A hollowed-out determiat I the followig, matrices are uderstood to be matrices whose etries are umbers see above). Exercise 5. Let N. Let P ad Q be two subsets of [] such that P + Q >. Let A = a i,j be a -matrix such that )1 i, 1 j The, prove that det A = 0. every i P ad j Q satisfy a i,j = 0.
Math 4707 Sprig 2018 Darij Griberg): homewor set 4 page 7 Example 0.4. Applyig Exercise 5 to = 5, P = {1, 3, 5} ad Q = {2, 3, 4}, we see that a 1 0 0 0 a 5 b 1 b 2 b 3 b 4 b 5 det c 1 0 0 0 c 5 = d 1 d 2 d 3 d 4 d 5 0 for ay umbers a 1, a 5,..., e 5. e 1 0 0 0 e 5 0.5. Arrowhead matrices Exercise 6. Let be a positive iteger. a) A permutatio σ S will be called arrowheaded if each i [ 1] satisfies σ i) = i or σ i) =. [For example, the permutatio i S 5 whose oe-lie otatio is [1, 5, 3, 4, 2] is arrowheaded.] Describe all arrowheaded permutatios σ S ad fid their umber. b) Give umbers a 1, a 2,..., a as well as 1 umbers b 1, b 2,..., b 1 ad 1 further umbers c 1, c 2,..., c 1. Let A be the -matrix a 1 0 0 c 1 0 a 2 0 c 2........ 0 0 a 1 c 1 b 1 b 2 b 1 a a i, if i = j; b j, if i = ad j = ; This is the matrix whose i, j)-th etry is for c i, if i = ad j = ; 0, if i = ad j = ad i = j all i [] ad j [].) Prove that det A = a 1 a 2 a 1 b i c i i=1 j [ 1]; j =i a j. The matrix A is called a reverse arrowhead matrix due to the shape that its ozero etries form. Reverse because the usual arrowhead matrix has its arrow poitig orthwest rather tha southeast.)