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I know that $\mid A\mid$ cannot be zero.

If $A$ is non-negative, then $x^{T}Ax\geq 0$. $A$ being positive is equivalent to having all its corners as positive. The largest corner $A$ can have is $A$ itself, so therefore $\mid A\mid$ must be positive too. Does it then follow that it a non negative matrix cannot have an inverse $A^{-1}$, because it would break multiplying determinants?

I am pretty lost here, help needed if possible.

minimalbob
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  • What exactly do you mean by "all its corners positive"? What is a "corner" of a matrix? Also, what is $|A|$? Does that refer to the determinant of $A$? – Ben Grossmann Mar 03 '21 at 21:08
  • In any case, one approach to this question is to note that if $A$ is non-negative, then $x^TAx = 0 \implies Ax = 0$. I give a few proofs of this fact in my post here. – Ben Grossmann Mar 03 '21 at 21:12

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