DETERMINANTS Math121,11/14/2005O.Knill COROLLARY: det(AT ) = det(A) Determinants are characterized as the only n-linear alternating and normalized function from Mn(F ) to F . −1 T −1 −1 Proof. For any permutation π define a inverse permutation by π by a Pπ = Pπ . We have σ(π)= σ(π ) because the inverse permutation has the same signature: interchanging two elements in the permutation changes DEFINITION. The determinant of a n × n matrix M over the field F is defined recursively by the Laplace sign. If one needs l such transpositions to unscramble the permuation then σ(π)=(−1)l. But then, one needs (= cofactor) expansion also l transpositions to scramble it. Replacing each π with π−1 in the permutation formula does not change the n 1+j ˜ value of the formula. det(M)= j=1(−1) M1jdet(M1j P PROPERTIES. The determinant is a normalized n-form: where A˜ij is the (n − 1) × (n − 1) matrix obtained from M by deleting i’th row and j’th column. N-LINEARITY: The determinant is linear in a column (row), if the other columns (rows) fixed. 1 2 2 ALTERNATING: The determinant changes sign, if two columns (rows) are interchanged. EXAMPLE. The determinant of the 3 × 3 matrix A = 1 2 1 in the field F3 of characteristic 3 is NORMALIZED: The determinant of the identity matrix is 1. 1 1 2 Proof: induction with repect to n. The statements for rows become statements for columns for the transpose 2 1 1 1 1 2 M T 1det( ) − 2det( ) + 2det( ) = 1 · 3 − 2 · 1 + 2 · (−1) = −1 = 2 . matrix so that it suffices to prove the statment for columns. 1 2 1 2 1 1 For n = 2, we have det(A) = A11A22 − A12A21 and all properties are true. Assume the property is true for n − 1. To prove (i), distinguish the case, where the column is the first column and the case, where we have an other column. In the first case, we see the linearity directly, in the second case, we use induction. (ii) follows REMARK. The usefulnes to consider also other fields F besides R or C had been demonstrated again recently in from the permutation formula. The third statement is obvious. cryptology: on November 8’th, the number RSA-640 had been factored using an algorithm, which uses matrices ROW REDUCTION. The determinant changes sign, if two rows are interchanged. The determinant is multplied defined over F2. by a factor a if a row is multiplied with a. The determinant does not change if a multiple of one row is added M p q a 0 1 0 to an other row. Therefore, det( )=(−1) j=1 j where p is the number of row TYPE 1 ELEMENTARY MATRICES have determi- Example: 1 0 0 which switches the first and transpositions and aj are the scaling factors appliedQ during the row reductions. nant −1. 0 0 1 REMARK: Row reduction is the most convenient algorithm to compute determinants. Instead of adding n! second row. terms, the row reduction algorithm only needs O(n3) steps since there are O(n2) row reduction steps each 1 0 0 involving O(n) operations. TYPE 2 ELEMENTARY MATRICES have determi- Example: 0 a 0 which multplies the second nant a. 0 0 1 COROLLARY. Any invertible matrix M is the product M = E · ... · E of elementary matrices. 1 k row with a. Proof. This is a reformulation of what happens during row reduction. 1 0 b TYPE 3 ELEMENTARY MATRICES have determi- Example: 0 1 0 which adds b times the third nant 1. 0 0 1 COROLLARY. det(AB) = det(A) · det(B). row to the first row. Proof. If A is not invertible, its rank is smaller than n and so the rank of AB so that AB is not invertible and PERMUTATION MATRIX. A permutation π = {π(1), ..., π(n)} of {1, ..., n} defines a permutation matrix both sides of the formula are the same. In the invertible case, det(M)= i det(Ei). Because the multiplicative Pπej = eπ(j). Define the function σ(π) = det(Pπ). on the group Sn of all permutations. If a permutation formula is true if the first matrix is an elementary the formula is true inQ general by inducuction. matrix M can be written as a product of k type 1 elementary matrices, then det(M)=(−1)k. In other words, σ(PQ)= σ(P )σ(Q) for permutation matrices. COROLLARY. det(A) 6= 0 if and only if A is invertible. Proof. If A is invertible, then AB = 1 for some matrix B. Because det is multiplicative det(A)det(B) = 1 so that det(A) 6= 0. If det(A) 6= 0, then each of the elementary matrices building up A is invertible and so also A. THEOREM. det(M)= π∈Sn σ(π)M1π(1)A2π(2) ··· Anπ(n) P THEOREM. If δ : M (F ) → F is n-linear, alternating and normalized, then δ = det. PROOF. Induction in n: group all permutations according to which element k is first. Each group is the set n of all permutations of n − 1 elements. So π = [k, ρ(2), ..., ρ(n)] where ρ is a permutation of {1, ..., n}\{k}. PROOF. step (i): If E is an elementary matrix of type 1, then δ(E )=(−1) The signatures satisfy σ(π)=(−1)k+1σ(ρ). Grouping the terms in the sum appearing in the statement of the 1 1 if E is an elementary matrix of type 2, then δ(E )= a theorem gives the Laplace expansion. 2 2 If E3 is an elementary matrix of type 3, then δ(E3)=1. step (ii): it follows that δ is multiplicative δ(AB) = δ(A)δ(B): Proof: check this first if A is an elementary REMARK. Without the σ(π) part, the sum is called the permanant of M. It is not known whether there is matrix. Because every inverible matrix is a product of elementary matrices, it is true in general. a polynomial algorithm to computer permanents. step (iii): if A is invertible, then δ(A) 6= 0. Proof: there exists then a matrix B such that AB = 1. Because of the multiplicative property (ii) and the assumption δ(1) = 1, it can not be that δ(A)=0. n i+1 ˜ COROLLARY. We can also do Laplace expansion with respect to first column: det(A) = i=1(−1) Ai1 or step (iv): it follows that if M = E1 · ... · Ek is the product of k elementary matrices, then det(M)= i det(Ei). n i+j ˜ n i+j ˜ any column det(A)= i=1(−1) Aij or any row det(A)= j=1(−1) Aij. P step (v): it follows that if M is invertible, then each Ei is invertible and since for invertible elementaryQ matrices P P δ(E) 6= 0, also δ(M) 6= 0. step (vi): if A is not invertible, then δ(A) = 0, the same as det(A). step (vi): for invertible matrices, we can r r r r write M = i=1 Ei. Because δ(Ei) = det(Ei), we get delta(M)= δ( i=1 Ei)= i=1 δ(Ei)= i=1 det(Ei)= r δ( i=1 Ei)=Q δ(M). Q Q Q Q DETERMINANTSII Math121,11/16/2005O.Knill WHY DO WE CARE ABOUT DETERMINANTS? check invertibility of matrices allow to define orientation in any dimensions A summary of all properties about determinants. • • have geometric interpretation as volume appear in change of variable formulas in higher • • dimensional integration. PERMUTATIONS. A permutation of 1, 2,...,n is a rearrangement explicit algebraic expressions for inverting a ma- { } • of 1, 2,...,n . There are n! = n (n 1)... 1 different permutations trix proposed alternative concepts are unnatural, { } · − · of 1, 2,...,n : fixing the position of first element leaves (n 1)! • hard to teach and and harder to understand { } − as a natural functional on matrices it appears in possibilities to permute the rest. • formulas in particle or statistical physics determinants are fun • EXAMPLE. There are 6 permutations of 1, 2, 3 : (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1). { } TRIANGULAR AND DIAGONAL MATRICES. PARTITIONED MATRICES. PATTERNS AND SIGN. The matrix A with zeros everywhere except The determinant of a diagonal or triangular The determinant of a partitioned matrix matrix is the product of its diagonal elements. A 0 Ai,π(i) = 1 is called a permutation matrix or the pattern of π. An is the product det(A)det(B). inversion is a pair k < l such that σ(k) < σ(l). The sign of a permuta- 0 B 1 0 0 0 tion π, denoted by ( 1)σ(π) is ( 1) for an odd number of inversions in 3 4 0 0 − − 4 5 0 0 1 2 0 0 the pattern, otherwise, the sign is 1. (To get the sign in the permutations to Example: det( ) = 20. Example det( ) = 2 12 = 24. 2 3 4 0 0 0 4 2 · the right, count the number of pairs of black squares, where the upper square − 1 1 2 1 0 0 2 2 is to the right). EXAMPLES. σ(1, 2) = 0, σ(2, 1) = 1. σ(1, 2, 3) = σ(3, 2, 1) = σ(2, 3, 1) = 1. σ(1, 3, 2) = σ(3, 2, 1) = σ(2, 1, 3) = 1. LINEARITY OF THE DETERMINANT. If the columns of A and B are the same except for the i’th column, − det([v , ..., v, ...v ]] + det([v , ..., w, ...v ]] = det([v , ..., v + w, ...v ]] DETERMINANT The determinant of a n n matrix A is defined as the sum 1 n 1 n 1 n σ(π)A A A , where π is× a permutation of 1, 2,...,n and σ(π) is its sign.