Bourbaki

Proposition 6

Tag 0BWD, permanent. Cite it as [Bourbaki, Tag 0BWD].

General Topology, chapter VI, § 1. Real number space R^n

In the ring Mn(R)\mathbf{M}_n(\mathbf{R}), the group GLn(R)\mathbf{GL}_n(\mathbf{R}) of non-singular matrices is a dense open subset, and the topology induced on this set is compatible with its group structure.

(*) If E,F,GE, F, G are three vector spaces over a field KK, a mapping ff of E×FE \times F into GG is said to be bilinear if we have identically

f(x+x′,y)=f(x,y)+f(x′,y),f(x,y+y′)=f(x,y)+f(x,y′), f(x + x', y) = f(x, y) + f(x', y), \quad f(x, y + y') = f(x, y) + f(x, y'),
f(λx,y)=f(x,λy)=λf(x,y) f(\lambda x, y) = f(x, \lambda y) = \lambda f(x, y)
for all x,x′∈Ex, x' \in E, all y,y′∈Fy, y' \in F and all λ∈K\lambda \in K.

If XX is a nonsingular square matrix, the elements of X−1X^{-1} are rational functions of the elements of XX; these functions are therefore defined and continuous in a neighbourhood of XX, so that every matrix YY in this neighbourhood is non-singular, and the mapping Y→Y−1Y \to Y^{-1} is continuous at the point XX; hence GLn(R)\mathbf{GL}_n(\mathbf{R}) is open in Mn(R)\mathbf{M}_n(\mathbf{R}) and the topology of GLn(R)\mathbf{GL}_n(\mathbf{R}) is compatible with its group structure.

Finally, GLn(R)\mathbf{GL}_n(\mathbf{R}) is the complement of the set of square matrices XX whose determinant is zero; since the determinant of XX is a polynomial in the elements of XX, Proposition 3 of no. 4 shows that GLn(R)\mathbf{GL}_n(\mathbf{R}) is dense in Mn(R)\mathbf{M}_n(\mathbf{R}).

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

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Transcribed from the printed volume. Not translated.