Linear Algebra (G63.2110) Professor Mel Hausner Notes on Determinants The symmetric group Sn. It is not necessary to know group theory for this discussion but it might help. Definition. The symmetric group Sn is the set of all 1-1 maps of {1,2,. ,n} into itself. Elements of Sn are called permutations. Any rearrangement {i1,i2,...,in} of {1,2,. ,n} is regarded as the element of Sn which maps k into ik. If σ ∈ Sn then σ is onto as well as 1-1, since {1,2,. ,n} is finite. Multiplication of group elements in Sn is simply composition. Thus ρ = στ is defined by the condition ρ(i)= (στ)(i)=(σ(τ(i)) for all i ∈{1, 2,...,n}. The identity map is called I, and as usual, we have σσ−1 = σ−1σ = I. It is convenient to regard Sn ⊆ Sn+1.Ifσ ∈ Sn, simply define σ(n +1)=n + 1, and the newly define σ is in Sn+1. More accurately, we have defined a 1-1 map from Sn into Sn+1 which clearly preserves multiplication. Definition. (Cycle notation) Let ii,i2,...,ik be distinct elements of {1,2,. ,n}. Then the cycle (i1 i2 ···ik) is the map σ defined by the conditions 1. σ(iν )=iν+1 for ν =1,...,k− 1. 2. σ(ik)=i1. 3. σ(x)=x if x 6∈ {ii,i2,...,ik}. The numbers ii,i2,...,ik are called elements of the cycle. The cycle σ is said to have length k 2 (or order) k. It is an easy matter to see that σ = I. In fact, σ(i1)=i2, σ (i1)=i3, ..., k−1 σ (i1)=ik. For example, if n = 5, the cycle σ = (2 5 4) of length 3 satisfies σ(1) = 1,σ(2) = 5,σ(3) = 3,σ(4) = 2,σ(5) = 4 In short, 2 7→ 5 7→ 4 7→ 2 (and 1 7→ 1, 3 7→ 3.) The elements of σ are 2, 4, and 5. Definition. A transposition (or interchange) is a cycle (ij) of length 2. Note that for transpositions τ,wehaveτ 2 = I. Theorem. Any permutation in Sn is a product of transpositions. We prove this by induction on n. It is clearly true for n = 2. Assume that it is true for n and let σ ∈ Sn+1.Ifσ(n +1)=n + 1 we have already noted that we may regard σ ∈ Sn and the result is true. Otherwise, σ(n +1)=i where 1 ≤ i ≤ n. Then τ =(n +1 i)σ maps n +1 1 into n + 1, and so we may regard (n +1 i)σ as an element of Sn. By induction (n +1 i)σ = ρ is a product of transpositions. Multiplying by (n +1i) and using (n +1i)2 = I, we find that σ =(n +1i)ρ, a product of transpositions. Note that the count of transpositions goes up by 1 in the inductive step, so we can state that every permutation in Sn is a product of at most n − 1 transpositions. Note also that we don’t really need Sn ⊆ Sn+1. The induction could have been made on the largest element x such that σ(x) =6 x, assuming σ =6 I. We’ll need the following lemma in what follows (Transpositions almost commute with per- mutations): 0 Lemma. Let σ ∈ Sn. Then if τ is a transposition in Sn, we have στ = τ σ for some transposition in Sn. Proof of Lemma. Suppose τ =(ij). Let σ(i)=i0 and σ(j)=j0, and consider the transfor- mation ρ = στσ−1. A simple check shows that ρ(i0)=j0; ρ(j0)=j0; ρ(k)=k if k =6 i0,j0 Thus ρ is a transposition and since ρ = στσ−1, we have ρσ = στ. We now prove an important and non-trivial result: Theorem: There is a transformation sign:Sn →{−1, 1} with the following properties: 1) If τ is a transposition then sign(τ)=−1. 2) For any ρ, σ ∈ Sn, sign(ρσ) = sign(ρ)sign(σ). Remark. Since the sign of a transposition is −1, and every permutation is a product of transpositions, it follows that the sign assignment is unique. The idea of this result is that if a permutation is written as a product of k transpositions, then the parity of k is unique. We prove this by induction on n, using the natural embedding of Sn in Sn+1. It is clearly true for n = 1, where sign(I) = 1, and for n = 2, where sign(I) = 1 and sign(1 2) = −1. Now assume this result true for n, and we want to show how to extend the definition of sign to Sn+1. If σ ∈ Sn+1, we have σ(n +1)= k for some k with 1 ≤ k ≤ n +1. Thus if k = n +1, σ ∈ Sn and otherwise (kn+1)σ ∈ Sn. In this case σ ∈ (kn+1)Sn. Define τk =(kn+1) for 1 ≤ k ≤ n. For convenience, also define τn+1 = I. Thus Sn+1 is partitioned as Sn+1 = τ1Sn ∪ τ2Sn ∪ ...∪ τnSn ∪ τn+1Sn We can now define sign on Sn+1.Ifρ ∈ Sn, sign(ρ) is already defined. Otherwise, ρ = τkσ, where σ ∈ Sn. In this case we naturally define sign(ρ)=−sign(σ). Under this definition, it is clear that the sign of any transposition is −1. We must prove that the sign of a product is the product of the signs. Let ρ1 = τrσ1 and let ρ2 = τsσs, where 1 ≤ r, s ≤ n + 1 and each σi is in Sn. Then ρ1ρ2 = τrσ1τsσs = τrτσ1σs 2 where, using the lemma, τ is a transpsosition. These notes present a somewhat more abstract version than the text. As usual, V is a n-dimensional vector space over a field F . Definition. Let f be an F valued function of k vector variables in V . We write f : V k → F . f is said to be k-linear if it is linear in each of its variables. Thus, for each i (1 ≤ i ≤ k), and fixed vectors v1,...,vi−1,vi+1,...,vk, and arbitrary vectors v and w, and arbitrary constants a and b, we have f(v1,...,vi−1,av+ bw, vi+1,...,vk)=af(...,v,...)+bf(...,w,...) where the dots represent the fixed vectors v1,...,vi−1,vi+1,...,vk. The following is a generalization of the corresponding result when k =1. Theorem. Let e1,...,en be a basis for V , and let k ≥ 1 be given. Let ai1i2...ik ∈ F for each k-tuple (i1,...,ik) with 1 ≤ i1,...,ik ≤ n. Then there is one and only one k-linear functional f : V k → F such that f(ei1 ,...,eik )=ai1i2...ik The proof is straightforward. Let’s first illustrate it for k = 2 and arbitrary n. In this case, we have aij ∈ F for 1 ≤ i, j ≤ n. Now if v, w ∈ V , we have, uniquely, n n v = X xiei,w= X yjej i=1 j=1 If f were bilinear in v and w with f(ei,ej)=aij, we would have n n n n f(v, w)=f(X xiei, X yjej = X xiyjf(ei,ej)= X aijxiyj i=1 j=1 i,j=1 i,j=1 Conversely, if for given aij, this formula is taken as the definition of f(v, w), it is easy to see it is bilinear (since it is linear in each of its variables) and that f(ei,ej)=aij The general case is similar. We are burdened because we can’t use dummy variables i, j,so we have to resort to the indices i1,i2,...,ik. Note the basis form of f(v1,...,vk). If the vector vk has coordinates (xk1,xk2,...,xkn) then we have n X f(v1,...,vk)= ai1...ik x1i1 ...xkik i1,...,ik=1 In addition to k-linearity, it is necessary to consider skew symmetry. 3 Definition. Let f be k-linear with f : V k → F . f is said to be skew symmetric if f =0 whenever two of the variables are equal. In this case we also have, when i<j, f(...,vi,...,vj,...)=−f(...,vj,...,vi,...) This follows from k-linearity as follows. (We take two variables for simplicity.) Since f(v + w, v + w) = 0, we have by k-linearity f(v, v)+f(v, w)+f(w, v)+f(w, w)=0 which implies f(v, w)=−f(w, v), since f(v, v)=f(w, w) = 0 by skew symmetry. The condition f(v, w)=−f(w, v) is often taken as the definition of skew symmetry. We have seen that a k-linear function f is given by a k-dimensional array ai1i2...ik . Since this can be interpreted as the value of f(ei1 ,...,eik ), where e, ...,en are a basis, we can easily see what happens if f is skew symmetric. We would have to have ai1i2...ik = 0 if the ai are not distinct and ai1i2...ik changes sign if two of indices are interchanged. It is easy to show that these conditions are sufficient for the corresponding k-linear function to be skew symmetric. Here a little algebra is very helpful. If k numbers i1,...,ik are rearranged to j1,...,jk, the rearrangement is said to be an even permutation if the rearrangement can be achieved using an even number of interchanges. It is odd if can be achieved in an odd number of interchanges. An important result in algebra is that you can’t have a permutation which is even and odd.