Set of symmetric matrices is a manifold.

Question

How do I prove that the set of symmetric matrices is a manifold of dimension $n(n+1)/2.$

I tried it by using Regular value theorem. For that I consider the map $$ f:M(n,\mathbb{R})\to M(n,\mathbb{R}),\ \ A\mapsto A-A^T . $$ Now I need to check following:

1. $f$ is smooth.
2. $Df$ is surjective.

Here I don't know how to calculate $Df$? Any help will be appreciated.

Thanks.

Answer 1

The set of symmetric matrices is a vector space, hence it has one global chart and its dimension is its dimension as a vector space.

Besides, your approach is doomed, $f$ is not a submersion since it is a linear non-invertible map¹, its kernel is precisely the set of symmetric matrices. More generally, if $0$ is a regular value of $f\colon\mathbb{R}^n\rightarrow\mathbb{R}^q$, then $f^{-1}(0)$ is a manifold of dimension $n-q$. Here, it would mean that the set of symmetric matrices has dimension $0$.

If you want to use the regular value theorem, you can prove that the orthogonal group is a manifold of dimension $n(n-1)/2$.

^{1: As a linear map, one has $Df\equiv f$, its differential is constant and is equal to $f$, besides $f$ being non-invertible implies that it is not-surjective since it is an endomorphism of a finite-dimensional vector space.}