Determinants of operators and matrices IILet V be a finite dimensional C-vector space and let T be an operator on V .Recall:The characteristic polynomial fTisfT(t) :=Yλ(t − λ)dλ= tn+ a1tn−1+ · · · an, where dλ− dim GEλ.det(T ) =Yλλdλ= (−1)nanExample 1 Cyclic permutationsLet B = (v1, . . . , vn) and let T be the operator sending v1to v2, v2to v3,and so on, but then vnto v1. Then the characterisictic polynomial of T isfT(t) = tn− 1 anddet(T ) = (−1)n+1In this example, our linear transformation just permutes the basis. Our nextstep is to discuss more general cases of this.1PermutationsDefinition 2 A permutation of the set 1, . . . , n is a bijective function σ fromthe set {1, . . . , n} to itself. Equivalently, it is a list (σ(1), . . . , σ(n)) such thateach element of {1, . . . , n} occurs exactly once. The set of all permutations oflength n is denoted by Sn.Examples in S5:(2, 3, 4, 5, 1) (cycle of length 5)(2, 4, 3, 1, 5) (cycle of length 3)(2, 4, 5, 1, 3) (cycle of length 3 and disjoint cycle of length 2)Definition 3 The sign of a permutation σ is (−1)mwhere m is the number ofpairs (i, j) where1 ≤ i < j ≤ n but σ(i) > σ(j)Here’s an easy way to count: Arrange (1, 2, . . . , n) in one row, and again ina row underneath. (σ(1), σ(2), . . . , σ(n)) in a row below. Draw lines connectingi in the first row to σ(i) in the second. Then m is the number of crosses.Examples:1 2 3 4 5so m = 4 and sgn = +1.12-3-4-5-1 2 3 4 5so m = 4 and sgn = +1.12-3?4-5?1 2 3 4 5so m = 5 and sgn = −1.12-34-5-Theorem 4 If σ and τ are elemnts of Snand στ is their composition, thensgn(στ) = sgn(σ)sgn(τ).2ExamplesLet’s just look at what happens to one typical pair. There are really fourpossibilities:1 2sgn = +11?2?sgn = +11?2?1 2sgn = −112-sgn = +11?2?1 2sgn = +11?2?sgn = −112-31 2sgn = −112-sgn = −112-A more complicated example:1 2 3 4 5so m = 4 and sgn = +1.12-3?4-5?so m = 5 and sgn = −1.12-34-5-1 2 3 4 5so m = 5 and sgn = −1.1234-5-Example 5 A cycle of length n has n − 1 crossings, and so its sign is (−1)n−1.Note that this is the same as the determinant of the corresponding linear trans-fomration.4Definition 6 Let A be an n × n matrix. Thendet A :=X{sgn(σ)a1,σ(1)a2,σ(2)· · · an,σ(n): σ ∈ Sn}Example 7 When n = 2 there are two permuations, and we getdet(A) = a1,1a2,2− a1,2a2,1.In general there are n! permuations in Sn, a very big number!It is often useful to think of a matrix A as a bunch of columns: if A is amatrix, let Ajbe its jth column. Then we can think of det as a function of ncolumns instead of a function of matrices:det(A) = det(A1, A2, . . . , An)Theorem 8 Let A and B be n × n matrices.1. If A is upper triangular, det(A) =Qiai,i.2. det(A) is a linear function of each column, (when all the other columnsare fixed, and similarly for the rows.3. If A0is obtained from A by interchanging two columns, then det(A0) =− det(A).4. More generally, if A0is obtained from A by a permutation σ of the columns,then det(A0) = sgn(σ) det(A).5. If two columns of A are equal, det(A) = 0.6. det(AB) = det(A) det(B).7. det(At) = det(A)Here are some explanations:1. If A is uppertriangular aij= 0 if j < i. Now if σ ∈ Snis not the identity,σ(i) < i for some i, and then ai,σ( i)= 0. Thus the only term is the sumdet(A) =Xσis when σ = id.2. This is fairly clear if you think about it. Imagine if a01j= ca1jfor all j,for example.3. Suppose for example that A0is obtained from A by interchanging the firsttwo columns. Let τ be the permutation interchanging 1 and 2. Then forany ja0i,j= ai,τ (j)5and for any σa0i,σ(i)= ai,τ σ(i)det A0:=Xσsgn(σ)a01,σ(1)a02,σ(2)· · · a0n,σ(n):=Xσsgn(σ)a1,τ σ(1)a2,τ σ(2)· · · an,τ σ(n):=Xσ−sgn(τσ)a1,τ σ(1)a2,τ σ(2)· · · an,τ σ(n)= − det A4. Is proved in exactly the same way.5. Follows from (3) since then det(A) = − det(A).6. Recall that in fact Bj= b1,je1+ · · · bn,jen, where eiis the jth standardbasis vector for Fnwritten as a column. Recall also that if A and B arematrices, then the jth column of AB, which we write as (AB)j, is(AB)j= ABj= AXibi,jei=Xibi,jAei=Xibi,jAiSodet(AB) = det(AB1, AB2, . . . , ABn)= det(Xibi,1Ai,Xibi,2Ai, . . . ,Xibi,nAi)Using the fact that det is linear with respect to the columns over and overagain, we can multiply this out:det(AB) =Xσbσ(1),1bσ(2),2)· · · bσ(n),n)det(Aσ(1), Aσ(2), . . . , Aσ(n))where here the sum is over all functions σ from the set {1, . . . , n} to itself.But by (5), the determinant is zero if σ is not a permutation, and if it is,we just get the determinant of A times the sign of σ. So (miracle!) weend up withdet(AB) =Xσbσ(1),1bσ(2),2)· · · bσ(n),n)sgn(σ) det(A) = det(B) det(A)7.det A :=Xσsgn(σ)a1,σ(1)a2,σ(2)· · · an,σ(n)=Xσsgn(σ−1)a1,σ−1(1)a2,σ−1(2)· · · an,σ−1(n)=Xσsgn(σ)aσ(1),1aσ(2),2· · · aσ(n),n=