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.

  • $\begingroup$ Please explain "becomes positive definite". Becomes when what? Are you tuning $s$ until the matrix becomes positive definite? What is given? What is to be determined? $\endgroup$ – Rodrigo de Azevedo Oct 19 '19 at 14:35
  • $\begingroup$ @RodrigodeAzevedo: You are able to turn all those parameters under the conditions specified. Given is what is mentioned with "Given..." It is to be determined whether the matrix $(v\cdot v^T-M\cdot s) / s^2$ can become pd. I am confused... How is my question not clear? $\endgroup$ – Make42 Oct 19 '19 at 16:39
  • $\begingroup$ When I read "given", I assume that what follows are values of various types. From that viewpoint, the matrix is either PD or not. There's no tuning involved whatsoever. If there is tuning, then that which is tuned is a variable and, thus, is not given. $\endgroup$ – Rodrigo de Azevedo Oct 19 '19 at 16:43
  • $\begingroup$ @RodrigodeAzevedo: Understand, that also makes sense. I try to improve the question. $\endgroup$ – Make42 Oct 19 '19 at 16:58

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 $w^TAw = -\frac 1{s}w^TMw < 0$.

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  • $\begingroup$ Not sure if we misunderstand each other: I do not mean that this is supposed to always be positive definite, just that is could be positive definite. $\endgroup$ – Make42 Oct 16 '19 at 16:09
  • $\begingroup$ Whatever $v$ you take, whatever $s$ you take, your matrix will never be positive definite. That's what my answer was supposed to tell you. $\endgroup$ – amsmath Oct 16 '19 at 16:38

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