Bourbaki

Example 6

Tag 00QC, permanent. Cite it as [Bourbaki, Tag 00QC].

Algebra, VIII, p. 454, Appendix 2. Determinants over a Noncommutative Field

Suppose n=2n= 2. Let A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} be an element of GL2(D)\mathbf{G}\mathbf{L}_2(D). The elements aa and cc of D are not both zero. We will give the determinant of A explicitly.

If aa is not zero, then we have

(24) A=(10ca−11)(a00d−ca−1b)(1a−1b01)A = \begin{pmatrix} 1 & 0 \\ ca^{-1} & 1 \end{pmatrix}\begin{pmatrix} a & 0 \\ 0 & d-ca^{-1}b \end{pmatrix}\begin{pmatrix} 1 & a^{-1}b \\ 0 & 1 \end{pmatrix}.

Consequently, ad−aca−1b≠0ad-aca^{-1}b\not= 0 and

(25) det(A) =π(ad−aca−1b)=\pi (ad-aca^{-1}b).

b) If aa is zero, then we have c≠0c\not= 0 and

(26) A=(0bcd)=(0110)(cd0b)A = \begin{pmatrix} 0 & b \\ c & d \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} c & d \\ 0 & b \end{pmatrix}.

Consequently, by a) and Example 2, we have cb≠0cb\not= 0 and

(27) det(A) =π(−cb)=\pi (-cb).

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Transcribed from the printed edition, 2023, Springer Nature. Not translated.