For the variables $s\in\mathbb R, s > 0$, the vector $v\in\mathbb R^n$, and the symmetric positive definite matrix $M \in \mathbb R^{n \times n}$, are there cases in which
\begin{align} \frac{1}{s^2} \left( v v^T - s M \right) \end{align}
becomes positive definite?
I am aware that $v\cdot v^T$ is always positive definite, so I did (with an arbitrary vector $u\in\mathbb R^n$)
\begin{align} u^T \cdot \frac{v\cdot v^T - M\cdot s}{s^2} \cdot u \overset!> 0 \\ \Leftrightarrow u^T \cdot M \cdot u < u^T \cdot \frac{v\cdot v^T}{s} \cdot u \end{align}
where the right side would be always positive for any $u$.
For background information: $s$ is the value of a function at a specific position, $v$ the gradient at that position of that function and $M$ the respective Hessian.
If $n=1$, the answer is obviously yes. Just choose $s$ small enough. Otherwise it's no. To see this, choose any $w$ which is perpendicular to $v$. If your matrix is $A$, then $w^TAw = -\frac 1{s}w^TMw < 0$.