Let $A=A(x)$ be a square matrix and let $x^*$ be such that $A(x^* )z=0$ and $z≥0$. Let $A(x)y≫0$ for $x≠x^*$. Then $z≯0$.

How to prove this? Should I use pseudo-inverses?

  • $\begingroup$ Your question is hard to understand. It lacks appropriate quantifiers ("for all", "for some" etc.). It is not clear whether you are talking about some particular $x,y,z$ or not. And what is $y$? What does the symbol ≫ mean? Is $A$ entrywise positive or entrywise nonnegative? $\endgroup$ – user1551 Jan 11 at 11:53
  • $\begingroup$ Thank you for your comment.Let $A=A(x)$ be a square matrix and let $x∗$ be such that $A(x∗)z=0$ for some $z≥0$. Let $A(x)y≫0$ for $x≠x∗$. Then $z≯0$. the symbol $≫$ means "much bigger than". It is there to prevent $A(x)y$ tending to zero as $x$ tends to $x∗$. For $A$, I am working with an M-matrix, but I thought this could work for any matrix, at least real matrix. Not sure though. $\endgroup$ – a.giannel Jan 12 at 14:29

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