Linear Algebra, 2020a

338 Chapter Four. Determinants
Since scalars come out a row at a time we might guess that determinants are
linear a row at a time.
3.2 Definition Let V be a vector space. A map f: V n → R is multilinear if
(1) f(⃗ρ 1 ,...,⃗v + ⃗w,...,⃗ρ n )=f(⃗ρ 1 ,...,⃗v,...,⃗ρ n )+f(⃗ρ 1 ,...,⃗w,...,⃗ρ n )
(2) f(⃗ρ 1 ,...,k⃗v,...,⃗ρ n )=k · f(⃗ρ 1 ,...,⃗v,...,⃗ρ n )
for ⃗v, ⃗w ∈ V and k ∈ R.
3.3 Lemma Determinants are multilinear.
Proof Property (2) here is just Definition 2.1’s condition (3) so we need only
verify property (1).
There are two cases. If the set of other rows {⃗ρ 1 ,...,⃗ρ i−1 , ⃗ρ i+1 ,...,⃗ρ n }
is linearly dependent then all three matrices are singular and so all three
determinants are zero and the equality is trivial.
Therefore assume that the set of other rows is linearly independent. We can
make a basis by adding one more vector 〈⃗ρ 1 ,...,⃗ρ i−1 , ⃗β, ⃗ρ i+1 ,...,⃗ρ n 〉. Express
⃗v and ⃗w with respect to this basis
and add.
⃗v = v 1 ⃗ρ 1 + ···+ v i−1 ⃗ρ i−1 + v i
⃗β + v i+1 ⃗ρ i+1 + ···+ v n ⃗ρ n
⃗w = w 1 ⃗ρ 1 + ···+ w i−1 ⃗ρ i−1 + w i
⃗β + w i+1 ⃗ρ i+1 + ···+ w n ⃗ρ n
⃗v + ⃗w =(v 1 + w 1 )⃗ρ 1 + ···+(v i + w i )⃗β + ···+(v n + w n )⃗ρ n
Consider the left side of (1) and expand ⃗v + ⃗w.
det(⃗ρ 1 ,..., (v 1 + w 1 )⃗ρ 1 + ···+(v i + w i )⃗β + ···+(v n + w n )⃗ρ n , ...,⃗ρ n ) (∗)
By the definition of determinant’s condition (1), the value of (∗) is unchanged
by the operation of adding −(v 1 + w 1 )⃗ρ 1 to the i-th row ⃗v + ⃗w. The i-th row
becomes this.
⃗v + ⃗w −(v 1 + w 1 )⃗ρ 1 =(v 2 + w 2 )⃗ρ 2 + ···+(v i + w i )⃗β + ···+(v n + w n )⃗ρ n
Next add −(v 2 + w 2 )⃗ρ 2 , etc., to eliminate all of the terms from the other rows.
Apply condition (3) from the definition of determinant.
det(⃗ρ 1 ,...,⃗v + ⃗w,...,⃗ρ n )
= det(⃗ρ 1 ,...,(v i + w i ) · ⃗β,...,⃗ρ n )
=(v i + w i ) · det(⃗ρ 1 ,...,⃗β,...,⃗ρ n )
= v i · det(⃗ρ 1 ,...,⃗β,...,⃗ρ n )+w i · det(⃗ρ 1 ,...,⃗β,...,⃗ρ n )