Exercise 12, § 11

Algebra, III, 649

Bourbaki marks this one as harder than the rest.

Let $X = (\xi_{ij})$ be a square matrix of order $n$ over a commutative field $K$ ; let $H$ be a subset of $\{1, n\}$ with $p$ elements and $H'$ the complement of $H$ in $\{1, n\}$ ; suppose that, for every ordered pair of indices $h \in H, k \in H'$ ,

\sum_i \xi_{ih} \xi_{ik} = 0.

Show that, for all subsets

L, M

\{1, n\}

with

p

elements,

\rho_{M', M'} \det(X_{L, H}) \det(X_{M', H'}) - \rho_{L, L'} \det(X_{M, H}) \det(X_{L, H'}) = 0.

(Consider the columns

x_h

of index

h \in H

X

as vectors in

E = K^n

and the columns

x_k^*

of index

k \in H'

as vectors in

E^*

and show that the

(n - p)

-form

x_{H'}^*

is proportional to

\phi_p(x_H)

Solution

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