Linear Algebra

Section I. Definition 317Factor out the k to get the desired equality.= k · ∑t 1,φ(1) · · · t i,φ(i) · · · t n,φ(n) sgn(φ) = k · d(T)φFor (2) suppose that T ρ i↔ρ j−→ ˆT. We must show this is the negative of d(T).d(ˆT) =∑ ˆt 1,φ(1) · · ·ˆt i,φ(i) · · ·ˆt j,φ(j) · · ·ˆt n,φ(n) sgn(φ) (*)perm φWe will show that each term in (∗) is associated with a term in d(t), and that thetwo terms are negatives of each other. Consider the matrix from the multilinearexpansion of d(ˆT) giving the term ˆt 1,φ(1) · · ·ˆt i,φ(i) · · ·ˆt j,φ(j) · · ·ˆt n,φ(n) sgn(φ).⎛⎜⎝.ˆt i,φ(i)..ˆt j,φ(j)It is the result of the ρ i ↔ ρ j operation performed on this matrix.⎛⎜⎝..t j,φ(i).t i,φ(j)⎞⎟⎠⎞⎟⎠That is, the term with hatted t’s is associated with this term from the d(T)expansion: t 1,σ(1) · · · t j,σ(j) · · · t i,σ(i) · · · t n,σ(n) sgn(σ), where the permutationσ equals φ but with the i-th and j-th numbers interchanged, σ(i) = φ(j) andσ(j) = φ(i). The two terms have the same multiplicands ˆt 1,φ(1) = t 1,σ(1) ,. . . , including the entries from the swapped rows ˆt i,φ(i) = t j,φ(i) = t j,σ(j) andˆt j,φ(j) = t i,φ(j) = t i,σ(i) . But the two terms are negatives of each other sincesgn(φ) = − sgn(σ) by Lemma 4.3.Now, any permutation φ can be derived from some other permutation σ bysuch a swap, in one and only one way. Therefore the summation in (∗) is in facta sum over all permutations, taken once and only once.d(ˆT) =∑ ˆt 1,φ(1) · · ·ˆt i,φ(i) · · ·ˆt j,φ(j) · · ·ˆt n,φ(n) sgn(φ)perm φ= ∑t 1,σ(1) · · · t j,σ(j) · · · t i,σ(i) · · · t n,σ(n) · (−sgn(σ) )perm σThus d(ˆT) = −d(T).