Linear Algebra

318 Chapter Four. DeterminantsFor property (1) suppose that T kρ i+ρ j−→ ˆT and consider the effect of a rowcombination.d(ˆT) =∑ ˆt 1,φ(1) · · ·ˆt i,φ(i) · · ·ˆt j,φ(j) · · ·ˆt n,φ(n) sgn(φ)perm φ= ∑ t 1,φ(1) · · · t i,φ(i) · · · (kt i,φ(j) + t j,φ(j) ) · · · t n,φ(n) sgn(φ)φDo the algebra.= ∑ φ[t1,φ(1) · · · t i,φ(i) · · · kt i,φ(j) · · · t n,φ(n) sgn(φ)+ t 1,φ(1) · · · t i,φ(i) · · · t j,φ(j) · · · t n,φ(n) sgn(φ) ]= ∑ t 1,φ(1) · · · t i,φ(i) · · · kt i,φ(j) · · · t n,φ(n) sgn(φ)φ+ ∑ t 1,φ(1) · · · t i,φ(i) · · · t j,φ(j) · · · t n,φ(n) sgn(φ)φ= k · ∑t 1,φ(1) · · · t i,φ(i) · · · t i,φ(j) · · · t n,φ(n) sgn(φ)φ+ d(T)Finish by observing that the terms t 1,φ(1) · · · t i,φ(i) · · · t i,φ(j) · · · t n,φ(n) sgn(φ)add to zero: this sum represents d(S) where S is a matrix equal to T exceptthat row j of S is a copy of row i of T (because the factor is t i,φ(j) , not t j,φ(j) )and so S has two equal rows, rows i and j. Since we have already shown that dchanges sign on row swaps, as in Lemma 2.4 we conclude that d(S) = 0. QEDWe have now shown that determinant functions exist for each size. Wealready know that for each size there is at most one determinant. Therefore, thepermutation expansion computes the one and only determinant.We end this subsection by proving the other result remaining from the priorsubsection.4.8 Theorem The determinant of a matrix equals the determinant of its transpose.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 will finish the argument by manipulating the expression on the rightto be recognizable as the determinant of the transpose. We have written allpermutation expansions with the row indices ascending. To rewrite the expressionon the right in this way, note that because φ is a permutation the row indicesφ(1), . . . , φ(n) are just the numbers 1, . . . , n, rearranged. Apply commutativityto have these ascend, giving s 1,φ −1 (1) · · · s n,φ −1 (n).= ∑ φ −1 s 1,φ −1 (1) · · · s n,φ −1 (n) sgn(φ −1 )