TWO VECTORS AND A RECTANGULAR MATRIX 9

A short calculation yields that A + B\ = hM-i,w-i(h(^p) ® bq g a

r

) . Now that

(1.12) is established, we continue with the induction. Iff + l + wp + r and the

induction hypothesis is known to hold at w — 1, then (1.12) shows that

h [hM,w(ap ®bq ®ar)} = (-l) p /iA/

)W

(a

p

g) bx Abq ® ar);

hence,

hM,w(*P ®bi® *r) = (-l)pqbq [hMMXp ® ! ® «r)] = ( - l ) p ? C ) M * P A a

r

) = 0.

Now we prove (b). Assume that w + p -f 1 9- We prove

(i.i3) x)

(-1)(P+1)(V)

(// h

A

I^MK-)])

^

=

°-

i€Z

Let A(ap) = £ ai

j ]

8) a ^ , with

o^]

€

A*^7*

and

aji]8

G

f\p~s

F*. Fix / , with

|J| = 2. Proposition l.l.c, together with the measuring identity, gives

(/ / \ocv A [pj(&g)](ar)J J (CJF) = / / A ap (yi(bq) A ar(ujF)j

s, [J]

= ^(-1)(?"°(P"5)+^V/[4J1(M A [//(a^1, A ar)](u/F).

It follows that the left side of (1.13) is

£(-i)«e- 53(-lr^TjVM^Ml

A [ / K ^ A O , ) ] ^ ) ,

which is zero by part (a). D

The phrase "Koszul complex" has two meanings in this paper. If X: F —* R is

a map of free R—modules, then the Koszul complex associated to X is

(i.i4) . . . - ^ A ^ - ^ A ' - 1 ^ - - . ,

where

for all bq e f\q F. Of course, if bq = b[l1 A .. . A b^\ with ftW € F , then

a(fcg) = ]£(-l)

i + 1

X(bM) • *M A ... A

ftl*"1!

A

6l+1l

A ... A &M;

i=l