Linear Algebra

310 Chapter Four. Determinantswhere the sum is over all permutations σ derived from another permutation φby a swap of the i-th and j-th numbers. But any permutation can be derivedfrom some other permutation by such a swap, in one and only one way, so thissummation is in fact a sum over all permutations, taken once and only once.Thus d( ˆT ) = −d(T ).To do property (1) let T kρ i+ρ j−→ ˆT and considerd( ˆT ) = ∑= ∑ φperms φˆt 1,φ(1) · · · ˆt i,φ(i) · · · ˆt j,φ(j) · · · ˆt n,φ(n) sgn(φ)t 1,φ(1) · · · t i,φ(i) · · · (kt i,φ(j) + t j,φ(j) ) · · · t n,φ(n) sgn(φ)(notice: that’s kt i,φ(j) , not kt j,φ(j) ). Distribute, commute, and factor.= ∑ φ= ∑ 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)+ ∑ φt 1,φ(1) · · · t i,φ(i) · · · t j,φ(j) · · · t n,φ(n) sgn(φ)= k · ∑t 1,φ(1) · · · t i,φ(i) · · · t i,φ(j) · · · t n,φ(n) sgn(φ)φ+ d(T )We finish by showing 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) ).Thus, S has two equal rows, rows i and j. Since we have already shown that dchanges sign on row swaps, as in Lemma 2.3 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,the permutation expansion computes the one and only determinant value of asquare matrix.We end this subsection by proving the other result remaining from the priorsubsection, that the determinant of a matrix equals the determinant of its transpose.4.8 Example Writing out the permutation expansion of the general 3×3 matrixand of its transpose, and comparing corresponding terms∣ a b c∣∣∣∣∣ 0 0 1d e f∣g h i ∣ = · · · + cdh · 1 0 00 1 0∣ + · · ·