Suppose we have a symmetric PD or PSD matrix M which induces an inner product $\langle \cdot, \cdot \rangle_M$. If we have that $\langle x, y \rangle > 0$ for two unit vectors $x$, $y$, are there any sufficient conditions on $x, y$, and/or $M$ that ensure that $\langle{x}, {y}\rangle_M > 0$ (other than the obvious $x=y$ or $M = I$)?.


Since $M$ is PSD, it can be decomposed as $M=A^T A$, where $A$ is another matrix; see https://math.stackexchange.com/questions/1801403/decomposition-of-a-positive-semidefinite-matrix

Now $\langle x,y\rangle=x^T M y=x^T A^T Ay=\langle Ax,Ay\rangle$ and so your question is equivalent to: Which matrices $A$ preserve positive correlations between vectors? It is known that the matrices that preserve orthogonality are precisely those that are scalar multiples of orthogonal transformations: https://math.stackexchange.com/questions/2355551/linear-transformations-that-preserve-orthogonality

Certainly if a matrix does not preserve orthogonality, it does not preserve positive correlations (it can fix $x$ and map $y$ to a vector orthogonal to $x$). Conversely, scalar multiples of orthogonal matrices do preserve positive correlations -- more generally, they preserve dot products: https://math.stackexchange.com/questions/2161729/how-orthogonal-matrices-preserve-dot-product-and-volume-proof

We conclude that $A$ must be a scalar multiple of an orthogonal matrix, and hence $M$ must be a positive multiple of the identity $I$.

  • $\begingroup$ Minor edit: $M$ must be a positive multiple of $I$ (previously I had "scalar"). $\endgroup$ – Aryeh Kontorovich Sep 17 '18 at 20:35
  • $\begingroup$ Thank you! Are there any assumptions on $x$ and $y$ we could make instead? $\endgroup$ – B Merlot Sep 17 '18 at 20:46
  • $\begingroup$ For all positively correlated $x,y$, you can find a "bad" $M$ that will cause them to be orthogonal under the induced inner product. $\endgroup$ – Aryeh Kontorovich Sep 17 '18 at 20:53

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