# Could you prove that f(A)>=0 whenever A>0?

• MHB
• DrLiangMath
In summary, a 3x3 matrix $A$ is called positive, denoted $A>0$, if every entry of $A$ is positive, and similarly one can define $A>=0$. Let $f(x)=x^6+x^4-x^3+x^2+x$ and $f(A)=A^6+A^4-A^3+A^2+A$. The question is whether $f(A)>=0$ whenever $A>0$. After trying on different matrices, it was concluded that this result is false. A counter example was given using a matrix $A$ with zero entries and a small positive number $\alpha$. This led to the conclusion that any 6th order polynomial with at least
DrLiangMath
$A$ is a 3x3 matrix. $A$ is called positive, denoted $A>0$, if every entry of $A$ is positive. Similarly one can define $A>=0$. Let
$$f(x)=x^6+x^4-x^3+x^2+x$$
Then
$$f(A)=A^6+A^4-A^3+A^2+A$$
Could you prove that $f(A)>=0$ whenever $A>0$?

I have tried many matrices on CAS, it is true. But I don't know how to prove it.

Thanks.

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I think this result must be false. Start with the matrix $A = \begin{bmatrix}0&0&\alpha \\ \alpha&0&0 \\ 0&\alpha&0\end{bmatrix}$. Then$A^2 = \begin{bmatrix} 0&\alpha^2&0 \\ 0&0&\alpha^2 \\ \alpha^2&0&0 \end{bmatrix}$ and $A^3 = \begin{bmatrix}\alpha^3&0&0 \\ 0&\alpha^3&0 \\ 0&0&\alpha^3 \end{bmatrix} = \alpha^3I$. So $A^4 = \alpha^3A$ and $A^6 = \alpha^6I$.

The top left element of $f(A)$ is therefore $\bigl[f(A)\bigr]_{11} = \alpha^6 - \alpha^3$.

Now let $\alpha = \frac12$, so that $\bigl[f(A)\bigr]_{11} = \frac1{64} - \frac18 = -\frac7{64} < 0$.

The matrix $A$ is not positive, because it contains a lot of zero entries. But now replace all those zeros by a (small) positive number $x$ to get a new matrix $A_x = \begin{bmatrix}x&x&\alpha \\ \alpha&x&x \\ x&\alpha&x\end{bmatrix}$. This new matrix $A_x$ is positive. The $(1,1)$-entry $\bigl[f(A_x)\bigr]_{11}$ of $f(A_x)$ is a continuous function of $x$. Since that entry is negative when $x=0$, it must still be negative for all sufficiently small positive values of $x$. So $f(A_x)$ will not satisfy $f(A_x) >=0$ even though $A_x$ is positive.

You should be able verify that conclusion by taking $\alpha = 0.5$ and $x$ to be something like $x=0.01$.

Thank you so much for your insightful reply! My initial goal is to find a polynomial f(x) of degree 6, with at least one coefficient negative such that f(A)>=0 whenever A>0, where A is a 3x3 matrix. It is so hard to find and to prove:sick:

MathTutoringByDrLiang said:
Thank you so much for your insightful reply! My initial goal is to find a polynomial f(x) of degree 6, with at least one coefficient negative such that f(A)>=0 whenever A>0, where A is a 3x3 matrix. It is so hard to find and to prove:sick:

Following Opalg's proof, I believe we can find a counter example for any 6th order polynomial with at least one negative coefficient.
Consider any such polynomial of degree 6 (or less) and apply it to the matrix $A$ that Opalg suggested, and we can always find an $\alpha>0$ such that an element in the first column of the result is negative.

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Klaas van Aarsen said:
Following Opalg's proof, I believe we can find a counter example for any 6th order polynomial with at least one negative coefficient.
Consider any such polynomial of degree 6 (or less) and apply it to the matrix $A$ that Opalg suggested, and we can always find an $\alpha>0$ such that an element in the first column of the result is negative.
Good suggestion. But the situation may change if the polynomial $f(x)$ is changed. For example, if $f(x) = x^6+x^4-x^3+x^2+x+1$, then f(A)>0 for any $A$ given by Opalg as long as $A>0$. It has been proved that if f(x) of degree 6 having one negative coefficient satisfies the condition mentioned at the original post, then $f(x)$ must have the form: $f(x)=ax^6+bx^5+cx^4-dx^3+ex^2+hx+j$, where $a>0, d>0, b, c, e, h, j >=0$. It is a necessary condition, but not sufficient. So the key question is: could we find some coefficients $a, b, c, d, e, h, j$ such that $f(x)$ satisfies the desired condition?

MathTutoringByDrLiang said:
For example, if $f(x) = x^6+x^4-x^3+x^2+x+1$, then f(A)>0 for any $A$ given by Opalg as long as $A>0$.
Not every $A>0$ as given by Opalg, but specifically matrices with $\alpha^6-\alpha^3<0$. Opalg gives $\alpha=\frac 12$ as an example.

MathTutoringByDrLiang said:
It has been proved that if f(x) of degree 6 having one negative coefficient satisfies the condition mentioned at the original post, then $f(x)$ must have the form: $f(x)=ax^6+bx^5+cx^4-dx^3+ex^2+hx+j$, where $a>0, d>0, b, c, e, h, j >=0$. It is a necessary condition, but not sufficient. So the key question is: could we find some coefficients $a, b, c, d, e, h, j$ such that $f(x)$ satisfies the desired condition?
Suppose $f(x)=ax^6+bx^5+cx^4-dx^3+ex^2+hx+j$, where $a>0, d>0, b, c, e, h, j \ge 0$ satisfies the condition.

Then the top left element of $f(A)$ is $[f(A)]_{11}=a\alpha^6-d\alpha^3 = \alpha^3(a\alpha^3-d)$. Its zeros are $\alpha=0$ and $\alpha=\sqrt[3]{\frac da}>0$, and in between it is negative.
So pick e.g. $\alpha=\frac 12\sqrt[3]{\frac da}$ and we have a counter example for the given $f(x)$, which is a contradiction. It completes the proof that no such $f(x)$ exists.

Klaas van Aarsen said:
Not every $A>0$ as given by Opalg, but specifically matrices with $\alpha^6-\alpha^3<0$. Opalg gives $\alpha=\frac 12$ as an example.Suppose $f(x)=ax^6+bx^5+cx^4-dx^3+ex^2+hx+j$, where $a>0, d>0, b, c, e, h, j \ge 0$ satisfies the condition.

Then the top left element of $f(A)$ is $[f(A)]_{11}=a\alpha^6-d\alpha^3 = \alpha^3(a\alpha^3-d)$. Its zeros are $\alpha=0$ and $\alpha=\sqrt[3]{\frac da}>0$, and in between it is negative.
So pick e.g. $\alpha=\frac 12\sqrt[3]{\frac da}$ and we have a counter example for the given $f(x)$, which is a contradiction. It completes the proof that no such $f(x)$ exists.
Thank you very much for your reply. The top left element of $f(A)$ is actually $a\alpha^6-d\alpha^3 + j$ instead of $a\alpha^6-d\alpha^3$. Let $t=\alpha^3$. Then we get a quadratic function $g(t)=at^2-dt+j$. We can take j large enough such that the discriminant < 0. then $g(t)>0$ holds for any t, consequently for any $\alpha$. We can also make other elements positive. That means $f(A)>0$ whenever A>0 and A is given by Opalg.

## 1. What does f(A) represent in this statement?

F(A) represents the output or value of the function f when the input is A. In other words, it is the result of applying the function f to the value A.

## 2. What does A>0 mean in this statement?

A>0 means that the input value A is greater than zero. In other words, the function f is only being evaluated for positive values of A.

## 3. What does it mean for f(A) to be greater than or equal to 0?

When f(A) is greater than or equal to 0, it means that the output or value of the function f is either 0 or a positive number. This indicates that the function f is either non-negative or always positive when the input is positive.

## 4. Can you provide an example to illustrate this statement?

Yes, for example, if the function f(A) = A^2, then f(A) is always greater than or equal to 0 when A is positive. When A=0, f(A) = 0, and when A is a negative number, f(A) is positive. Therefore, the statement f(A)>=0 whenever A>0 is true for this function.

## 5. How can this statement be proven?

This statement can be proven using mathematical techniques such as proof by contradiction or proof by induction. It involves showing that the function f(A) produces non-negative or positive values for all positive inputs A. This can be done by analyzing the behavior of the function and its properties, and using logical reasoning to demonstrate that the statement is true.

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