5.9. Multilinearity of Determinants

Theorem 5.7.  Additivity In Each Row.

Given an n ´ n matrix A=( A 1 ,, A n )  and any n-D vector V.   Let B and C be matrices that differ from A only in the kth row, namely,

        A=( , A k , )   B=( ,V, )                  C=( , A k +V, )

Then,

        detC=detA+detB                                 (5.10)

or,

        d( , A k +V, )=d( , A k , )+d( ,V, )                (5.10a)

which is known as the additivity in the kth row.

Proof

Let

f( A )=detCdetB   =d( , A k +V, )d( ,V, )         (a)

where the right hand side of the last equality is assumed to be independent of V.  We shall prove the theorem by showing that f( A )  satisfies all 3 axioms for a determinant function so that by the uniqueness theorem, f( A )=detA .

 

For operations not involving Ak, the proof that f( A )  satisfies axioms 1 and 2 follows trivially from the fact that detC and detB are determinant functions.  For operations involving Ak, the proof for axiom 1 is as follows

        f( ,c A k , )=d( ,c A k +V, )d( ,V, )

                                =cd( , A k + V c , )cd( , V c , )

                                =c[ d( , A k + V c , )d( , V c , ) ]  

=c  f( , A k , )

where the last equality made use of the assumption that detC - detB is independent of V.  For axiom 2, we have

        f( , A k + A j ,, A j , )=d( , A k + A j +V,, A j , )d( ,V,, A j , )

                                                =d( , A k +V,, A j , )d( ,V,, A j , )

                                                =f( , A k ,, A j , )

and

        f( , A k ,, A k + A j , )=d( , A k +V,, A k + A j , )d( ,V,, A k + A j , )

Applying the theorem itself to the right hand side gives

        f( , A k ,, A k + A j , )=d( , A k +V,, A j , )d( , A k +V,, A k , )

                                                        d( ,V,, A j , )+d( ,V,, A k , )

                                                =f( , A k ,, A j , )

where we’ve used

        d( , A k +V,, A k , )=d( , A k +V A k ,, A k , )   =d( ,V,, A k , )

 

Finally, we need to prove f satisfies axiom 3, namely, f( I )=1 .  Setting A=I  in eq(a), we have

        f( I )=d( I 1 ,, I k +V,, I n )d( I 1 ,,V,, I n )

Let V=( v 1 ,, v n ) .  If k=1 , both C and B are upper triangular.  By Theorem 5.5, we have

detC=1+ v 1            and                 detB= v 1                 (b)

so that

f( I )=1+ v 1 v 1 =1                                                 (c)

and the theorem holds.  For k=2 , we have

        C=( 1 0 0 0 v 1 1+ v 2 v 3 v n 0 0 1 0 0 0 0 0 1 )   B=( 1 0 0 0 v 1 v 2 v 3 v n 0 0 1 0 0 0 0 0 1 )

To make each of these upper triangular, we need only add v 1 ×  1st row to the 2nd row.  The operation does not affect the diagonal elements so that we again recover eq(b) and hence eq(c).  Furthermore, a little reflection will show that the same holds for all k so that the theorem is proved.

Multilinearity

By repeated applications of Theorem 5.7 and Axiom 1 to the same row, we can generalize (5.10a) to

        d( ,   i=1 p t i U i , )= i=1 p t i d( , U i , )

which expresses the linearity in each row.  Since this applies to every row, the determinant function is said to be multilinear function in its rows.