Linear Algebra

Section I. Definition 311(terms with the same letters)∣ a d g∣∣∣∣∣ 0 1 0b e h∣c f i∣ = · · · + dhc · 0 0 11 0 0∣ + · · ·shows that the corresponding permutation matrices are transposes. That is,there is a relationship between these corresponding permutations. Exercise 15shows that they are inverses.4.9 Theorem The determinant of a matrix equals the determinant of itstranspose.Proof. Call the matrix T and denote the entries of T trans with s’s so thatt i,j = s j,i . Substitution gives this|T | = ∑s φ(1),1 . . . s φ(n),n sgn(φ)perms φt 1,φ(1) . . . t n,φ(n) sgn(φ) = ∑ φand we can finish the argument by manipulating the expression on the rightto be recognizable as the determinant of the transpose. We have written allpermutation expansions (as in the middle expression above) with the row indicesascending. To rewrite the expression on the right in this way, note that becauseφ is a permutation, the row indices in the term on the right φ(1), . . . , φ(n) arejust the numbers 1, . . . , n, rearranged. We can thus commute to have theseascend, giving s 1,φ −1 (1) · · · s n,φ −1 (n) (if the column index is j and the row indexis φ(j) then, where the row index is i, the column index is φ −1 (i)). Substitutingon the right gives= ∑ φ −1 s 1,φ −1 (1) · · · s n,φ −1 (n) sgn(φ −1 )(Exercise 14 shows that sgn(φ −1 ) = sgn(φ)). Since every permutation is theinverse of another, a sum over all φ −1 is a sum over all permutations φ= ∑s 1,σ ( 1) . . . s n,σ(n) sgn(σ) = ∣ ∣T trans∣ ∣perms σas required.QEDExercisesThese summarize the notation used in this book for the 2- and 3- permutations.i 1 2 i 1 2 3φ 1(i) 1 2 φ 1(i) 1 2 3φ 2 (i) 2 1 φ 2 (i) 1 3 2φ 3(i) 2 1 3φ 4 (i) 2 3 1φ 5 (i) 3 1 2φ 6(i) 3 2 1