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Sep 22, 2016 at 10:11 comment added Nik Weaver @ChristianRemling: ah, you're right. I was unconsciously assuming $2b - a - c \neq 0$ to first order in $\max(a,b,c)$.
Sep 22, 2016 at 5:21 comment added Christian Remling @NikWeaver: I think you're not paying enough attention to the details here. Did you see my answer, which I believe refutes what you claim?
Sep 22, 2016 at 5:06 comment added Nik Weaver I think it's right. Try working out the $2 \times 2$ and $3\times 3$ cases if you don't believe me.
Sep 22, 2016 at 0:36 comment added NullOfMatrix Thank you for your reply! But I do not think things could be that easy since there maybe interactions of $det(A)$ and $\sum A_{ij}$, and $A_{ij}$ have different signs which make the numerator term more complicated. But that is a good point!
Sep 21, 2016 at 2:12 history answered Nik Weaver CC BY-SA 3.0