Given a function $f$ and its local minima $x$. We have known that if $H(x)$ is positive definite, then $x$ is a strict local minimum for $f$. https://planetmath.org/relationsbetweenhessianmatrixandlocalextrema
But does the inverse also hold? Namely, if $x$ is a strict local minimal point, is the Hessian at $x$ a positive definite matrix?
The Hessian must be weakly positive definite (that is, $v\cdot H(x)v\ge0$ for all $v$), but need not be strictly positive definite. For example, if $f:\Bbb R\to\Bbb R$ the Hessian is just the $1\times1$ matrix $[f'']$; the function $f(x)=x^4$ has a strict global minimum at $x=0$ but $f''(0)=0$.
No, that is not true. Consider for example $f:x \to \|x\|^4$. Clearly this has a strict local minimum of in $x=0$. But $d_v^2 f(0) = 0$ for all $v$. So in fact the hessian vanishes in $0$.
