# The set of symmetric matrices as a manifold

How would I start off proving that $S=$(the set of symmetric $n\times n$ matrices) is a manifold. I tried using the definition directly by saying $M_n =$ the space of all $n\times n$ matrices For every $A\in M_n$ there exists open sets $U=V=M_n$ and a bijection $F: U\to V$ by $F(A)= A-A^T$ Therefore we have $F(U \cap S) = F(S)$ since $S$ is a subset of $M_n=\{0\} \cap M_n$ this is where I get stuck. Also, I know that the set of all symmetric $n\times n$ matrices is $(n^2 + n)/2$, therefore that is the dimension of the manifold

Definition: A set $M$ (subset of $\Bbb{R}^n$) is a $k$-dimensional manifold if for every $x\in M$ there exists open sets $U$, $V$ and a bijection $h:U\to V$ with $x\in U$ and $H(U \cap M) = V \cap (\Bbb{R}^k \times \{c^{k+1},\ldots ,c^n\})$ for all $c$'s constants

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The set is a linear subspace... – Mariano Suárez-Alvarez Nov 3 '10 at 4:29
How would I use that? – JimJones Nov 3 '10 at 4:34
Every vector space has a basis, so there's a linear isomorphism with $\mathbb R^n$ where $n$ is the dimension of the space. This is your chart (you only need one). – Ryan Budney Nov 3 '10 at 4:37
Technically your definition of "manifold" would be called a "submanifold of Euclidean space". "Manifold" in the common math world usually means "abstract manifold". To construct your map $H$ (which I call a chart), $U$ and $V$ will be equal to $n\times n$ matrices. Choose a basis for the space of matrices which includes a basis for your subspace, and represent matrices with respect to that basis. That's the idea. – Ryan Budney Nov 3 '10 at 4:46
Oh, I'm afraid you're in for a pretty rough ride learning differential geometry without having linear algebra as background. I'd like to suggest you talk to your prof about what you're getting into. – Ryan Budney Nov 3 '10 at 5:02