Let $P_n(\mathbb{R}):=\left\{
A \in \operatorname{Mat}_{n\times n}(\mathbb{R}):\,
A=A^T \mbox{ and } (\forall i)\,
\lambda_i^A \geq 0
\right\}$, where $\left\{\lambda_i^A\right\}_i$ are the eigenvalues of $A$. Is this a manifold with boundary which can be decomposed as:
$$
P_n(\mathbb{R}) = P_n^+(\mathbb{R}) \cup Sym_n^0(\mathbb{R}),
$$
where $Sym_n^0(\mathbb{R})$ is the set of symmetric $n\times n$ matrices satisfying $x^TAx=0$ for some non-zero vector and $P_n^+(\mathbb{R})$ is the set of symmetric positive-definite matrices with real entries?
I know that $P_n^+(\mathbb{R})$ is a smooth manifold since I can explicitly write down its global chart using the matrix exponential map and (a slight variant of) the vectorization operation. However, I'm not sure if $Sym_n^0(\mathbb{R})$ is of dimension $\frac{n(n+1)}{2} -1$.
Best Answer
An explicit calculation shows that when $n = 2$, the set $\textrm{Sym}^0_n(\mathbb{R})$ is given by
$$ \textrm{Sym}^0_2(\mathbb{R}) = \left\{ \begin{pmatrix} a & b \\ b & c \end{pmatrix} \, \bigg\rvert \, ac - b^2 = 0, a + c \geq 0 \right\}. $$
Performing the change of variables $$ a = z - x, b = y, c = z + x $$ this set transforms into
$$ \left \{ \begin{pmatrix} z - x & y \\ y & z - x \end{pmatrix} \, \bigg\rvert \, z^2 = x^2 + y^2, z \geq 0 \right \}. $$
The equation $z^2 = x^2 + y^2$ is the equation of a double cone and the condition $z \geq 0$ gives us a single cone with singularity at the origin:
This shows that $\textrm{Sym}^0_n(\mathbb{R})$ is not a manifold and so $P_n(\mathbb{R})$ is not a manifold with boundary. However, if you look at the sets $S_r$ which consist of symmetric positive semi-definite matrices of rank $r$ for $0 \leq r \leq n$ then each one of those is a submanifold and the whole space of positive semi-definine matrices can be given the structure of a stratified space with strata $S_r$.