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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?
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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
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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
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Spoiler counterexample \(\displaystyle A=\left( \begin{array}{ccc} u & u & u \\ u & u & u \\ u & u & u \end{array} \right)\) where \(\displaystyle 0
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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.
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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 \(\displaystyle +1~?\)
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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. Hopefully I have answered your question.
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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!?
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