Permutations, the Parity Theorem, and Determinants


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1 1 Permutatios, the Parity Theorem, ad Determiats Joh A. Guber Departmet of Electrical ad Computer Egieerig Uiversity of Wiscosi Madiso Cotets 1 What is a Permutatio 1 2 Cycles Traspositios 4 3 Orbits 5 4 The Parity Theorem Decompositio of Permutatios ito Cycles with Disjoit Supports 7 5 Determiats 9 Refereces 12 Abstract The Parity Theorem says that wheever a eve (resp. odd) permutatio is expressed as a compositio of traspositios, the umber of traspositios must be eve (resp. odd). The purpose of this article is to give a simple defiitio of whe a permutatio is eve or odd, ad develop just eough backgroud to prove the parity theorem. Several examples are icluded to illustrate the use of the otatio ad cocepts as they are itroduced. We the defie the determiat i terms of the parity of permutatios. We establish basic properties of the determiat. I particular, we show that detba = detb deta, ad we show that A is osigular if ad oly if deta 0. If you fid this writeup useful, or if you fid typos or mistakes, please let me kow at 1. What is a Permutatio A permutatio is a ivertible fuctio that maps a fiite set to itself. 1 If we specify a order for the elemets i the fiite set ad apply a give permutatio to each poit i order, the the fuctio values we geerate simply list all the poits 1 To say that a fuctio is ivertible meas that it is both oetooe ad oto. Oetooe meas that o pair of poits ca map to a commo destiatio poit. Oto meas that every poit is the image of some poit.
2 2 of the set i a ew order. I this way, a permutatio specifies a reorderig of the elemets of a fiite set. Without loss of geerality, it suffices to take as our fiite set {1,...,} for some positive, fiite iteger. A permutatio ϕ o {1,...,} ca be described explicitly with the 2 matrix ( 1 ϕ1 ϕ The top row lists the first itegers i their usual order, ad the bottom row lists them i a ew order. Example 1. O {1,2,3,4,5}, is the permutatio such that ϕ := ). ( ) ϕ1 = 4, ϕ2 = 3, ϕ3 = 5, ϕ4 = 1, ad ϕ5 = 2. Similarly, is the permutatio such that ψ := ( ) ψ1 = 4, ψ2 = 5, ψ3 = 3, ψ4 = 1, ad ψ5 = Cycles A cycle is a especially simple kid of permutatio. Give k distict elemets i {1,...,}, say x 1,...,x k, we write ϕ = (x 1,...,x k ) if ϕ takes x 1 to x 2, x 2 to x 3,..., x k 1 to x k, ad x k to x 1, while leavig all other iputs uchaged. I other words, ad (x 1,...,x i,...,x k )x i = x i+1, where x k+1 := x 1, (1) (x 1,...,x i,...,x k )x = x, if x / {x 1,...,x k }. (2) Such a permutatio is called a kcycle. The support of a kcycle ϕ = (x 1,...,x k ) is suppϕ := {x 1,...,x k }. Formula (2) says that all other values of x are fixed poits of the cycle. Remarks. (i) From (1), it is apparet that for x suppϕ, ϕ k x = x. Of course, for x / suppϕ, (2) also implies ϕ k x = x. Hece, for a kcycle ϕ, we always have that ϕ k
3 3 is the idetity, which we deote by I. I particular, for k = 2, (x 1,x 2 )(x 1,x 2 ) = I, which says that a 2cycle is its ow iverse. (ii) The cycle otatio is ot uique i the sese that ay circular shift of the sequece x 1,...,x k yields the same permutatio; i.e., (x 1,...,x k ), (x 2,...,x k,x 1 ),..., ad (x k,x 1,...,x k 1 ) all defie the same permutatio. I particular, ote that (x 1,x 2 ) = (x 2,x 1 ). Example 2. O {1,2,3,4,5,6,7,8,9}, cosider the 3cycle ψ := (1,2,6) ad the 4cycle ϕ := (3, 5, 7, 6). The followig calculatios illustrate how to compute with cycle otatio: ϕ7 = (3,5,7,6)7 = 6 ϕ6 = (3,5,7,6)6 = 3 ψ6 = (1,2,6)6 = 1 ψϕ6 = (1,2,6)(3,5,7,6)6 = (1,2,6)3 = 3 ϕψ6 = (3,5,7,6)(1,2,6)6 = (3,5,7,6)1 = 1 ϕ4 = (3,5,7,6)4 = 4 ψ4 = (1,2,6)4 = 4 ψϕ4 = (1,2,6)(3,5,7,6)4 = (1,2,6)4 = 4 ϕψ4 = (3,5,7,6)(1,2,6)4 = (3,5,7,6)4 = 4. I particular, otice that ψϕ6 = 3 1 = ϕψ6. So i geeral, cycles do ot commute. Propositio 3. If cycles ψ 1,...,ψ m have pairwise disjoit supports, the ψ i x, x suppψ i, ψ 1 ψ m x = m x x / suppψ i. Furthermore, the ψ i commute ad ca be applied i ay order. Proof. Fix a value of i i the rage from 1 to m. If x suppψ i, the ψ i x also belogs to suppψ i. Therefore, x ad ψ i x do ot belog to the supports of the other cycles; i.e., x ad ψ i x are fixed poits of the other cycles. Hece, we ca write ψ 1 ψ m x = ψ 1 ψ m 1 x, sice x is a fixed poit of ψ m, i=1 = ψ 1 ψ m 2 x, sice x is a fixed poit of ψ m 1, =. = ψ 1 ψ i 1 ψ i x
4 4 = ψ 1 ψ i 2 ψ i x, sice ψ i x is a fixed poit of ψ i 1, = ψ 1 ψ i 3 ψ i x, sice ψ i x is a fixed poit of ψ i 2, =. = ψ 1 ψ i x = ψ i x. We ca similarly argue that ψ m ψ 1 x = ψ i x. I fact, applyig the cycles i ay order to x suppψ i always results i ψ i x. Now suppose x does ot belog to the support of ay ψ i. The x is a fixed poit of every ψ i ad we ca write ψ 1 ψ m x = ψ 1 ψ m 1 x = = ψ 1 x = x. Agai, we ca apply the cycles i ay order; e.g., ψ m ψ 1 x = x. Propositio 3 is importat, because we will see later that every permutatio ca be decomposed ito a compositio of cycles with pairwise disjoit supports Traspositios A traspositio is a 2cycle such as (x,y), where x y. Thus, (x,y)x = y ad (x,y)y = x, while for all z x,y, we have (x,y)z = z. As metioed i the Remarks above, a traspositio is its ow iverse, ad (x,y) = (y,x). Lemma 4. Let ψ be a kcycle, ad let τ be a traspositio whose support is a subset of the support of ψ. The ψτ is equal to the compositio of two cycles with disjoit supports. Proof. Suppose ψ = (x 1,...,x i,...,x j,...,x k ) ad τ = (x i,x j ). It is easy to check that ψτ = (x 1,...,x i,x j+1,...,x k )(x i+1,...,x j ), which is the compositio of two cycles with disjoit supports. Lemma 5. Let ψ ad η be cycles with disjoit supports, ad let τ be a traspositio whose support itersects the supports of both ψ ad η. The ψητ is a sigle cycle whose support is suppψ suppη. Proof. Suppose ψ = (x 1,...,x i,...,x k ), η = (y 1,...,y j,...,y m ), ad τ = (x i,y j ). It is easy to check that ψητ = (x 1,...,x i,y j+1,...,y m,y 1,...,y j,x i+1,...,x k ), which is a sigle cycle whose support is suppψ suppη.
5 5 3. Orbits Let ϕ be a permutatio o {1,...,}. Give x {1,...,}, cosider the sequece x,ϕx,ϕ 2 x,...,ϕ x. These + 1 values all belog to {1,...,}. So there must be at least two values that are the same, say ϕ k x = ϕ m x for some 0 k < m. Now apply ϕ 1 to both sides k times to get x = ϕ m k x. Hece, for ay x, there is a smallest positive iteger l (depedig o x) for which ϕ l x = x. The orbit of x (uder ϕ) is the set O x := {x,ϕx,...,ϕ l 1 x}. If l = 1, the orbit is just the sigleto set {x}. A sigleto orbit is a fixed poit. Example 6. Cosider the permutatio ϕ of Example 1. We see that O 1 = {1,4} ad O 2 = {2,3,5}. However, O 3 = O 2, O 4 = O 1, ad O 5 = O 2. The orbits of the permutatio ψ of Example 1 are O 1 = {1,4}, O 2 = {2,5}, ad O 3 = {3}, while O 4 = O 1 ad O 5 = O 2. For each x {1,...,} we ca determie its orbit O x, ad sice each x belogs to its ow orbit; i.e., x O x, we ca write {1,...,} = O x. x=1 Cosider two orbits O y ad O z for y z. We show below that if they are ot disjoit, the they are the same. If O y = O z, the the above uio ca be simplified to {1,...,} = O x. x=1 x z Proceedig i this way, after a fiite umber of steps, we obtai a sequece of distict poits y 1,...,y with {1,...,} = O yi, (3) where ad the orbits O y1,...,o y are pairwise disjoit. If is eve, we say that the permutatio ϕ is eve, ad we write sgϕ = 1. If is odd, we say that the permutatio ϕ is odd, ad we write sgϕ = 1. The quatity sgϕ is called the sig, sigature, or parity of the permutatio ϕ. Example 7. Cosider the idetity permutatio, which we deote by I. For each x, I x = x, ad so the orbit of x uder iperm is {x}. Hece, the umber of disjoit orbits of I is =. Sice = 0 is eve, the idetity is a eve permutatio, ad sg I = 1. i=1
6 6 Example 8. Let us determie the disjoit orbits of a kcycle ϕ = (x 1,...,x k ). If we start with x 1 ad apply ϕ over ad over usig (1), we fid that O x1 = {x 1,...,x k } = suppϕ. If fact, startig with x 2 or x 3,..., or x k, we fid they all have the same orbit, suppϕ. O the other had if we start with a x / suppϕ, we have by (2) that ϕx = x. Hece, the orbit of such a x is the sigleto set {x}. We have thus show that for a kcycle, the umber of disjoit orbits is = 1 + ( k). Sice this formula is equivalet to = k 1, the sig of a kcycle is 1 if k is odd ad 1 if k is eve. We ow show that if two orbits are ot disjoit, they are the same. Suppose O y O z, ad let x deote a poit i O y O z. Sice x O y, we must have x = ϕ r y for some oegative iteger r. Sice x O z, we must have x = ϕ s z for some oegative iteger s. Hece, ϕ r y = ϕ s z, which implies y = ϕ s r z. This further implies ϕ m y = ϕ m+s r z O z for all m; hece, O y O z. Similarly, writig z = ϕ r s y implies O z O y. 4. The Parity Theorem It is easy to write a kcycle as a compositio of traspositios. Cosider the formula (x 1,...,x k ) = (x 1,x k )(x 1,x k 1 ) (x 1,x 3 )(x 1,x 2 ). (4) Notice that each of x 2,...,x k appears i oly oe factor, while x 1 appears i every factor. If we apply the righthad side to x 1, we get (x 1,x k )(x 1,x k 1 ) (x 1,x 3 )(x 1,x 2 )x 1 = (x 1,x k )(x 1,x k 1 ) (x 1,x 3 )x 2 = x 2, sice x 2 is ot i the supports of ay of the remaiig traspositios. If we start with x 2, we get (x 1,x k )(x 1,x k 1 ) (x 1,x 4 )(x 1,x 3 )(x 1,x 2 )x 2 = (x 1,x k )(x 1,x k 1 ) (x 1,x 4 )(x 1,x 3 )x 1 = (x 1,x k )(x 1,x k 1 ) (x 1,x 4 )x 3 = x 3. Cotiuig i this way, we see that (4) holds. Notice that i (4), there are k 1 traspositios. We saw earlier that a kcycle ϕ has = 1 + ( k) orbits so that k 1 = determies the sig of ϕ. Hece, it is possible to write a kcycle ϕ as a compositio of traspositios such that the umber of traspositios is eve if sgϕ = 1, ad the umber of traspositios is odd if sgϕ = 1. However, there are may ways to write a kcycle as compositios of differet umbers of traspositios.
7 7 Example 9. O {1,2,3,4,5}, we ca use (4) to write the 3cycle (1,2,3) as (1,2,3) = (1,3)(1,2). Sice k = 3 is odd, k 1 = 2 is eve. Hece, it is o surprise that we ca write a 3cycle as the compositio of 2 (a eve umber) traspositios. However, sice (4, 5)(4, 5) = I, we ca also write (1, 2, 3) = (1, 3)(1, 2)(4, 5)(4, 5), which is aother way to write this 3 cycle as a eve umber of traspositios Decompositio of Permutatios ito Cycles with Disjoit Supports For a arbitrary permutatio ϕ, oce we have idetified its disjoit orbits we ca associate each orbit with a cycle i the followig way. The disjoit orbit decompositio (3) suggests that we put { ϕx, x Oyi, ψ i x := x, otherwise. (5) Note that the ψ i are cycles ad have disjoit supports. We ca further write ϕ = ψ 1 ψ, (6) which expresses ϕ as the compositio of cycles with disjoit supports. The fact that (6) holds follows from Propositio 3 ad the defiitio (5). Parity Theorem. Wheever a eve (resp. odd) permutatio is expressed as a compositio of traspositios, the umber of traspositios must be eve (resp. odd). Proof. Cosider a permutatio ϕ with disjoit orbits ad correspodig represetatio as cycles with disjoit supports as i (6). Suppose also that ϕ = τ 1 τ m, where each τ i is a traspositio. Recallig that a traspositio is its ow iverse, write I = ϕϕ 1 = (ψ 1 ψ )(τ 1 τ m ) 1 = (ψ 1 ψ )(τ m τ 1 ) = ψ 1 ψ τ m τ 1. Cosider the expressio ψ 1 ψ τ m. Sice the supports of the ψ i partitio {1,...,}, the two poits i the support of τ m must belog to the supports of the ψ i. If τ m = (u,v), there are two cases to cosider. First, there is the case that u ad v both belog to the support of a sigle ψ i. Secod, u belogs to the support of some ψ i, while v belogs to the support of some other ψ j with j > i. 2 I the first case, we use Propositio 3 ad Lemma 4 to write ψ 1 ψ τ m = ψ 1 ψ i 1 ψ i+1 ψ (ψ i τ m ), 2 There is o loss of geerality i assumig j > i because (u,v) is equal to (v,u).
8 8 where ψ i τ m is equal to the compositio of two cycles with disjoit supports created from suppψ i. Hece, the above expressio is a permutatio with + 1 pairwise disjoit orbits. I the secod case, we use Propositio 3 ad Lemma 5 to write ψ 1 ψ τ m = ψ 1 ψ i 1 ψ i+1 ψ j ψ j+1 ψ (ψ i ψ j τ m ), where ψ i ψ j τ m is equal to a sigle cycle whose support is the uio of the supports of ψ i ad ψ j. Hece, the above expressio is a permutatio with 1 pairwise disjoit orbits. Sice we do ot kow which of the two cases τ m falls ito, let us deote the ew umber of disjoit orbits by + σ m, where σ m = ±1. Now that we have determied the umber of disjoit orbits of ψ 1 ψ τ m, we ca determie the umber of disjoit orbits of ψ 1 ψ τ m τ m 1 as + σ m + σ m 1, ad so o. The umber of disjoit orbits of I = ψ 1 ψ τ m τ 1 ca be writte as + m k=1 where each σ k = ±1. The sig of I = ψ 1 ψ τ m τ 1 is determied by whether ( m ) + σ k = ( m ) k=1 σ k, σ k k=1 is eve or odd. However, we kow from Example 7 that the idetity is eve. This meas that ( ) mius the above sum has to be eve. Therefore, if ( ) is eve, the above sum must be eve, while if ( ) is odd, the above sum must be odd. Now observe that if m is eve, the possible values for the above sum are m,m 2,...,4,2,0, 2, 4,..., m, which are all eve, while if m is odd the possible values are m,m 2,...,3,1, 1, 3,..., m, which are all odd. Hece, if ( ) is eve, i.e., if the origial permutatio ϕ is eve, the m must be eve, while if ϕ is odd, the m must be odd. Corollary 10. If ϕ ad ψ are permutatios, the sg(ϕψ) = sg(ϕ) sg(ψ), ad is therefore equal to sg(ψϕ). I particular, if ψ is itself a traspositio (so that sg(ψ) = 1), sg(ϕψ) = sg(ϕ). Proof. We oly prove sg(ϕψ) = sg(ϕ)sg(ψ) sice the other parts of the corollary follow immediately from this. Suppose ϕ = τ 1 τ m ad ψ = θ 1 θ k, where the τ i ad θ j are traspositios. By the Parity Theorem, sg(ϕψ) = ± accordig to whether m + k is eve or odd. Similarly for sg(ϕ) ad m ad sg(ψ) ad k. It is easy to verify sg(ϕψ) = sg(ϕ)sg(ψ) for each of the four possibilities of m ad k beig eve/odd.
9 9 5. Determiats If A is a matrix with colums a(1),...,a(), the the ith row of the colum vector a( j), deoted by a i ( j), is the i, j etry of A. The determiat of A is deta := sg(ϕ)a 1 (ϕ1) a (ϕ), (7) ϕ where the sum is over all possible! permutatios ϕ of the itegers {1,...,}. Propositio 11. The determiat of a triagular matrix is the product of its diagoal etries. Proof. To begi, recall that A is upper (resp. lower) triagular if the elemets below (resp. above) the mai diagoal are zero. Without loss of geerality, assume A is upper triagular so that a i ( j) = 0 for i > j. The result will follow if we ca show that all terms i (7) with ϕ I have zero as a factor. If ϕ I, the for some i, we must have i > ϕi. 3 ad for such i, sice A is upper triagular, a i (ϕi) = 0. Let I deote the idetity matrix, whose i, j etry is 1 if i = j ad 0 for i j. Our first observatio is that the determiat of the idetity matrix is oe. To see this, cosider a typical term i (7). If ϕ I, the for some i, ϕi i, ad so the factor a i (ϕi) = 0 whe A is the idetity matrix. The oly term i (7) that is ot zero is the term with ϕ = I. That term is equal to oe. Fix a particular colum idex u, ad cosider a typical term i (7). This term cotais the factor a i (u) for some i, 4 ad o other factor i that term ivolves a etry from the colum a(u). Hece, whe all the colums of A are fixed except for colum u, deta is liear i colum u. Let u ad v be two distict itegers from {1,...,}, ad let τ := (u,v) be the traspositio that iterchages u ad v. Let B deote the matrix with colums b( j) := a(τ j). The a( j), j / {u,v}, b( j) := a(τ j) = a(v), j = u, a(u), j = v. I other words, B is obtaied from A by iterchagig colums u ad v. We claim that detb = deta. To see this, write detb = sg(ϕ)b 1 (ϕ1) b (ϕ) ϕ 3 To see this, cosider the decompositio (6). If ϕ I, some ψ i must be a kcycle with k 2, say ψ i = (x 1,...,x k ). These x j are all distict, ad oe is the largest, say x l. The ϕx l = ψ i x l < x l. 4 The value of i is ϕ 1 u.
10 10 = sg(ϕ)a 1 (τϕ1) a (τϕ) ϕ = sg(τ 1 ψ)a 1 (ττ 1 ψ) a (ττ 1 ψ) ψ = sg(τ 1 ψ)a 1 (ψ) a (ψ) ψ = sg(τ 1 )sg(ψ)a 1 (ψ) a (ψ) ψ = sg(ψ)a 1 (ψ) a (ψ) ψ = deta, where the third equality follows by substitutig ϕ = τ 1 ψ; the fifth equality follows because the sig of a compositio of permutatios is the product of their sigs by Corollary 10; ad the sixth equality follows because traspositios are odd. Now suppose A has two equal colums, ad B is obtaied by iterchagig those two colums. The by the precedig paragraph, det B = det A. But sice B = A, detb = deta. Therefore, deta = 0. We showed above that if b( j) = a(τ j) for some traspositio τ, the detb = deta. What if b( j) = a(ϕ j) for some arbitrary permutatio ϕ? Writig ϕ i terms of traspositios, say ϕ = τ 1 τ k, we see that detb = ( 1) k deta = sg(ϕ)deta. Theorem 12. detba = detb deta. Proof. It is coveiet to write deta i terms of its colums. We use the otatio (a(1),...,a()) = deta. If B is aother matrix with colums b(1),...,b(), the the colums of BA are Ba(1),...,Ba(), ad detba = (Ba(1),...,Ba()). Now recall that Write a( j) = i=1 ([ detba = B = i=1 i=1 a i ( j)e(i), j = 1,...,. ] ) a i (1)e(i),Ba(2),...,Ba() a i (1) (Be(i),Ba(2),...,Ba()).
11 11 Repeatig this calculatio for a(2),...,a() yields detba = = i 1 =1 i 1 =1 i =1 i =1 a i1 (1) a i () (Be(i 1 ),...,Be(i )) a i1 (1) a i () (b(i 1 ),...,b(i )). Amog the tuples (i 1,...,i ), if the i k are ot distict, the (b(i 1 ),...,b(i )) is the determiat of a matrix with two or more equal colums ad is therefore zero. Otherwise, (i 1,...,i ) = (ϕ1,...,ϕ) for some permutatio ϕ. Hece, detba = a ϕ1 (1) a ϕ () (b(ϕ1),...,b(ϕ)) ϕ = a ϕ1 (1) a ϕ ()sg(ϕ)detb ϕ = detbsg(ϕ)a 1 (ϕ 1 1) a (ϕ 1 ) ϕ = detbsg(ψ 1 )a 1 (ψ1) a (ψ) ψ = detbsg(ψ)a 1 (ψ1) a (ψ) ψ = detb deta. Corollary 13. deta = 0 if ad oly if A is osigular. Proof. If A is osigular, the 1 = deti = det(a 1 A) = (deta 1 )(deta) implies deta 0. Now suppose A is sigular. The the colums of A are liearly depedet. Without loss of geerality, suppose a() + 1 j=1 c ja( j) = 0 for some coefficiets c j. Hece, 0 = ( a(1),...,a( 1),0 ) ( 1 ) = a(1),...,a( 1),a() + c j a( j) j=1 = ( a(1),...,a( 1),a() ) 1 + = deta + 0, j=1 c j ( a(1),...,a( 1),a( j) ) sice i the sum over j, the jth term ivolves the determiat of a matrix whose colum j ad colum are the same. Such a determiat is zero.
12 12 Refereces [1] T. L. Barlow, A historical ote o the parity of permutatios, The America Mathematical Mothly, vol. 79, o. 7, pp , [2] W. Phillips, O the defiitio of eve ad odd permutatios, The America Mathematical Mothly, vol. 74, o. 10, pp , [3] W. Rudi, Priciples of Mathematical Aalysis, 3rd ed. New York: McGrawHill, 1976.
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