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A is a 3x3 matrix. A is called positive, denoted A>0, if every entry of A is positive. Let $$f(x)=x^6 - x^3 + 1$$ Could you prove that f(A)>0 whenever A>0?

Idea said: Spoiler counterexample $$\displaystyle A=\left( \begin{array}{ccc} u & u & u \\ u & u & u \\ u & u & u \end{array} \right)$$ where $$\displaystyle 00 whenever A>0. Hopefully you could help me figure one such polynomial. Thank you in advance by 0 like 0 dislike Idea said: Spoiler counterexample \(\displaystyle A=\left( \begin{array}{ccc} u & u & u \\ u & u & u \\ u & u & u \end{array} \right)$$ where $$\displaystyle 0 by 0 like 0 dislike Spoiler counterexample \(\displaystyle A=\left( \begin{array}{ccc} u & u & u \\ u & u & u \\ u & u & u \end{array} \right)$$ where $$\displaystyle 0 by 0 like 0 dislike Plato said: Actually \(\displaystyle f(A)=A^6 - A^3 + 1$$ is an abuse of notation. If it were $$\displaystyle A^6 - A^3$$ that is possible. SEE THIS How do you understand how to use $$\displaystyle +1~?$$ Click to expand... It was my mistake/typo. It should be $A^6-A^3+I$ where $I$ is the 3x3 identity matrix.
MathTutoringByDrLiang said: Yes, A stands for an arbitrary 3x3 matrix. We say A>0 (A is then called positive) if every entry of A is positive, but I didn't say A has a numerical value. Furthermore, f(A) means A^6 - A^3 + 1. Click to expand... Actually $$\displaystyle f(A)=A^6 - A^3 + 1$$ is an abuse of notation. If it were $$\displaystyle A^6 - A^3$$ that is possible. SEE THIS How do you understand how to use $$\displaystyle +1~?$$
MathTutoringByDrLiang said: A is a 3x3 matrix. A is called positive, denoted A>0, if every entry of A is positive. Let $$f(x)=x^6 - x^3 + 1$$ Could you prove that f(A)>0 whenever A>0? Click to expand... "A" is supposed to be the name of a 3 x 3 matrix. How can it also have a numerical value at the same time!?