Bourbaki

Lemma 1

Tag 05XB, permanent. Cite it as [Bourbaki, Tag 05XB].

Algebra, chapter II, § 10. Matrices

Let X=(ξij)X = (\xi_{ij}) be a matrix of type (m,n)(m, n) over a ring AA. Suppose that m⩾2m \geqslant 2 and that there exists an element ξi1\xi_{i1} in the first column of XX which is invertible in AA. Then there exist two invertible square matrices P∈Mm(A)P \in \mathbf{M}_m(A), Q∈Mn(A)Q \in \mathbf{M}_n(A) and a matrix YY of type (m−1,n−1)(m - 1, n - 1) over AA such that PP (resp. QQ) is a product of matrices of the form Bij(λ)B_{ij}(\lambda) of order mm (resp. nn) and

PXQ=(10⋯00⋮Y0). PXQ = \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & & & \\ \vdots & & Y & \\ 0 & & & \end{pmatrix}.

The matrix Bij(λ)XB_{ij}(\lambda)X is obtained by adding to the row of XX of index ii the row of index jj multiplied on the left by λ\lambda (no. 9, Example 2); if ξi1\xi_{i1} is invertible, then there exists λ∈A\lambda \in A such that, for the matrix X′=B1i(λ)X=(ξkl′)X' = B_{1i}(\lambda)X = (\xi'_{kl}), ξ11′=1\xi'_{11} = 1; multiplying X′X' on the left by suitably chosen matrices Bk1(μk)B_{k1}(\mu_k) of order mm (for 1≤k≤m1 \leq k \leq m), a matrix X′′=(ξkl′′)X'' = (\xi''_{kl}) is obtained such that ξj1′′=1\xi''_{j1} = 1, ξk1′′=0\xi''_{k1} = 0 for k≠1k \neq 1. Then the matrix obtained is multiplied successively on the right by suitable matrices B1j(vj)B_{1j}(v_j) of order nn (2≤j≤n)(2 \leq j \leq n) and a matrix is obtained of the form (57).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

Nothing.

Transcribed from the printed edition, 1998, Springer. Not translated.