Exercise 1966Topology › Topology of Mn(ℝ)L1Added 4 Nov 2026

A homeomorphism of $M_n(\mathbb{R})$

#homeomorphism#matrix map#inverse map

Statement

Consider $\Phi:M_n(\mathbb{R})\to S_n(\mathbb{R})\times A_n(\mathbb{R})$ defined by $\Phi(A)=(A+A^T,A-A^T)$ , where $S_n$ (resp. $A_n$ ) is the set of symmetric (resp. antisymmetric) matrices. Show $\Phi$ is a homeomorphism.

Hint

Explicitly find the inverse; both maps are polynomial.

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Solution

Inverse. Given $(S,A)$ with $S$ symmetric and $A$ antisymmetric, set $M=\frac{S+A}{2}$ . Then $M+M^T=S$ , $M-M^T=A$ ; indeed $M=\frac{S+A}{2}$ . So $\Phi^{-1}(S,A)=\frac{S+A}{2}$ .

Continuity. Both $\Phi$ and $\Phi^{-1}$ are linear maps between finite-dimensional spaces, hence continuous.

Conclusion. $\Phi$ is a homeomorphism.

A homeomorphism of $M_n(\mathbb{R})$

Statement

Solution

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