Let $H_n$ be a $(n+1)\times (n+1)$ real symmetric matrix, and let $D_0,D_1,\dots, D_n$ be the leading principal minors of $H_n$.
What I know is:
- If $H_n$ is positive definite (resp. positive semi definite), then $D_n> 0$ (resp. $D_n\geq 0$).
- If $D_k>0$ for all $0\leq k\leq n$, then $H_n$ is positive definite (by Sylvester's criterion).
What I want to know is, assuming that $H_n$ is positive semi-definite,
$\quad$ Q1. If $D_n>0$, then $H_n$ is positive definite.
$\quad$ Q2. If $H_n$ is not positive definite, then $D_n=0$.
For Q1: I believe it's done by induction over $n$. For $n=0$: If $D_0>0$, then $H_0$ is positive definite, by second point. For $n=1$: If $D_1>0$, how do you know that $D_0\neq 0$, so that we can use second point again?
For Q2: We know that $H_n$ is positive semi-definite by assumption, so $D_n\geq 0$ by first point. But, since $H_n$ is not positive semi-definite, we can't have $D_n>0$, so $D_n=0$. Is that it?
A positive semidefinite matrix is positive definite if and only if it is invertible (has non-zero determinant).
This is normally taken as a consequence of the following: a symmetric matrix is positive definite if and only if its eigenvalues are real and positive semidefinite if and only if its eigenvalues are non-negative. From there, we note that the determinant of a matrix is the product of its eigenvalues.
For a more direct proof, it suffices to note that for a (symmetric) positive semidefinite matrix $H$, we have $x^THx = 0 \iff Hx = 0$. In my post here, I prove this in a few different ways. From there, note that a matrix has zero determinant if and only if its nullspace (AKA kernel) is non-trivial.