Bourbaki

Exercise 11, § 11

Algebra, III, 649

Bourbaki marks this one as harder than the rest.

Let Φ=0\Phi = 0 be an identity between minors of an invertible square matrix XX of order nn over a commutative ring AA, Φ~=0\tilde{\Phi} = 0 the complementary identity (Exercise 10) and kk an integer >0> 0. Let YY be an invertible square matrix of order n+kn + k and let Y~0\tilde{Y}_0 be the submatrix of the adjoint matrix Y~\tilde{Y} formed by suppressing in Y~\tilde{Y} the rows of index ≤k\leq k and the columns of index ≤k\leq k. If Y~0\tilde{Y}_0 is assumed to be invertible and the identity Φ~=0\tilde{\Phi} = 0 is applied to the minors of Y~0\tilde{Y}_0 and then each minor which appears in this identity (considered as a minor of Y~\tilde{Y}) is replaced by its expression as a function of the minors of YY, using identity (1) of Exercise 9, an identity Φk=0\Phi_k = 0 is obtained between minors of the matrix YY (valid if YY and Y~0\tilde{Y}_0 are invertible) which is called the extension of order kk of the identity Φ=0\Phi = 0.

In particular, Let A=(αij)A = (\alpha_{ij}) be an invertible square matrix of order n+kn + k, BB the submatrix of AA of order kk obtained by suppressing in AA the rows and columns of index >k> k and Δij\Delta_{ij} the determinant of the matrix of order k+1k + 1 obtained by suppressing in AA the rows of index >k> k except that of index k+ik + i and the columns of index >k> k except that of index k+jk + j; if CC denotes the matrix (Δij)(\Delta_{ij}) of order nn, show the identity

det⁡C=(det⁡A)(det⁡B)n−1 \det C = (\det A)(\det B)^{n-1}
valid when AA and BB are invertible (show that it is the extension of order kk of the total expansion of a determinant of order nn).

State the identities obtained by extending the Laplace expansion (§ 8, no. 6) and the identity of Exercise 2 of § 8.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.