Proposition 8

Tag 064G, permanent. Cite it as [Bourbaki, Tag 064G].

For every square matrix $X$ over a commutative ring, the determinant of the transpose matrix ${}^t X$ is equal to the determinant of $X$ .

Suppose that $X$ is of type (I, I). For every ordered pair of permutations $\sigma, \tau$ of $S_I$ , (since multiplication is commutative)

\prod_{i \in I} \xi_{\sigma(i), i} = \prod_{j \in I} \xi_{\sigma(\tau(j)), \tau(j)}.

In particular take $\tau = \sigma^{-1}$ ; using the fact that $\varepsilon_{\sigma^{-1}} = \varepsilon_\sigma$ , it is seen that

\det X = \sum_{\sigma \in S_I} \varepsilon_\sigma \left( \prod_{i \in I, i} \xi_{i, \sigma(i)} \right),

which proves the proposition.

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

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