Determinants of operators and matricesLet V be a finite dimensional C-vector space and let T be an operator on V .Recall:The characteristic polynomial fTis the polynomialfT(t) :Yλ(t − λ)dλwhere dλ− dim GEλ.The Cayley-Hamilton theorem:fT(T ) = 0 as an operator on V .Goal: compute fT, without actually computing the generalized eigenspaces,or even the eigenvalues, just from MB(T ) for any arbtirary basis B for V .fT(t) :=Yλ(t − λ)dλ= tn+ a1tn−1+ · · · an,for some list of complex numbers ai.Definition 1 If T ∈ L(V ) andfT(t) = tn+ a1tn−1+ · · · an,is its characteristic polynomial, thentrace(T ) := −a1anddet(T ) := (−1)nanLast time:If A := MB(T ), thentrace(T ) = trace(A) :=Xai,i1Outline:1. Define the determinant of a matrix, det(A).2. Check that if A is uppertriangular,det(A) =Yaii.3. Show that if A and B are n × n matrices,det(AB) = det(A) det(B).4. Conclude that det(S−1AS) = det A for every invertible S.5. Conclude that if A = MB(T ), then det(A) = det(T ).Step (1) is probably the hardest. How to find the definition? Many ap-proaches. I’ll follow the book, more or less. Let’s look at cyclic spaces.Theorem 2 Suppose that V is T -cyclic, so that there is a v ∈ V withV = span(v, T v, T2v, . . . Tn−1v).Then (v, T v, . . . , Tn−1v) is a basis of V , andTnv = c0v + c1T v + · · · + cn−1Tn−1vfor a unique list (c0, . . . , cn−1) in C. Then the characteristic polynomial of T isp(t) = tn− cn−1tn−1− · · · c1t − c0.Example 3 Last time we looked at T (x1, x2) := (x2, x1). Let v := (1, 0). Then(v, T v) is a basis for V , and T2v = v. Thus c0= 1 and c1= 0, the p(t) = t2− 1.Proof: The equation for Tnv says thatp(T )(v) = 0.It follows thatp(T )(Tiv) = Tip(T )(v) = 0 for all iand since the Tiv’s span V p(T ) = 0. Since the list (v, T v, . . . Tn−1v) is indepe-nent, there is no polynomial of smaller degree that annihilates T . Thus p is theminimal polynomial of T , and since its degree is the dimension of V p is alsothe characteristic polyonmial of T .Corollary 4 In the cyclic case above, det(T ) = (−1)n+1c0.Example 5 Let 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 ofT isfT(t) = tn− 1 and det(T ) = (−1)n+1In this example, our linear transformation just permutes the basis. Our nextstep is to discuss more general cases of this.2PermutationsDefinition 6 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)(2, 4, 3, 1, 5)(2, 4, 5, 1, 3)The first of these is cycle of length 5. Note that the second doesn’t move 3or 5, and can be viewed as a cyclic permutation of the set {1, 2, 4}. The lastpermuation can be viewed as a the product (composition) of a cyclic permutationof {1, 2, 4} and a cyclic permutation of {3, 5}.Definition 7 The sign of a permuation 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 (σ(1), σ(2), . . . , σ(n))in a row below. Draw lines connecting i in the first row to i in the second. Thenm is the number of crosses.Examples:1 2 3 4 5so m = 4 and sgn = +1.2 3 4 5 11 2 3 4 5so m = 4 and sgn = +1.2 4 3 1 51 2 3 4 5so m = 5 and sgn = −1.2 4 5 1 33Example 8 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. dTheorem 9 If σ, τ ∈ Sn, thensgn(στ) = sgn(σ)sgn(τ)Omit the proof, at least for now.Definition 10 Let A be an n × n matrix. Thendet A :=X{sgn(σ)a1,σ(1)a2,σ(2)· · · an,σ(n): σ ∈ Sn}Example 11 When n = 2 there are two permuations, and we getdet(A) = a1,1a2,2− a1,2a2,1.Proposition 12 Let A and B be n × n matrices.1. If A is upper triangular, det(A) =Qiai,i.2. If B is obtained from A by interchanging two columns, then det(B) =− det(A).3. If two columns of A are equal, det(A) = 0.4. det(A) is a linear function of each column, (when all the other columnsare fixed.)5. det(AB) = det(A)