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

[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.

 

[Opalg]

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 $x=0.01$.

 

[DrLiangMath]

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:

 

[I like Serena]

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.

 

[DrLiangMath]

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?

 

[I like Serena]

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.

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.

 

[DrLiangMath]

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.