Given two equations
$\displaystyle P_1 = \frac{1-X^2}{1-X^2}$
and
$\displaystyle P_2 = \frac{1-aX^2}{1-bX^2}$
I am told that there is a relationship between P1 and P2
$P_1 \geq P_2$
It is clear that by visually comparing P1 and P2, the coefficients a and b are responsible for determining the relationship between P1 and P2. That being said, I would like to express the relationship
$P_1 \geq P_2$
as
$P_1 \geq P_1f(a,b)$
$1 \geq f(a,b)$
Clearly f(a,b) can be defined as P2/P1, so disregard this solution for the time being. My question is, can I express f(a,b) as a function of a and b alone? In other words, can I factor out P1 from P2 and be left with a function of a and b alone? If so, what is the form for f(a,b)? Is it a rational expression? Is it a transform matrix? Is it another polynomial? A convolution?
My objective is to evaluate f(a,b), thus interpret the relationship between P1 and P2. In other words if f(a,b)=1 then I know P1=P2. I don't think I can make it any more clear.
If I can't extract an expression for f(a,b) as a function of a and b alone, that's fine. I'm under the impression I can't but I thought someone on here might know more than I, so I figured I would ask.
You did not say anything about the range where $a$, $b$, $x$ can be taken from. For the sake of simplicity, I will assume that $x\ne 0$, $0<1-ax^2<1$ and $0<1-bx^2<1$.
The following inequalities are equivalent to each other: \begin{align*} P_1 &\ge P_2 \\ \frac{1-x}{1-x} &\ge \frac{1-ax^2}{1-bx^2} \\ 1 &\ge \frac{1-ax^2}{1-bx^2} \\ 1-bx^2 &\ge 1-ax^2\\ ax^2 &\ge bx^2\\ a &\ge b \end{align*}
You can do similar manipulations with $P_1>P_2$, $P_1<P_2$, $P_1=P_2$, etc.
So, for example, the function $$f(a,b)= \begin{cases} 2; & a>b, \\ 1; & a=b, \\ 0; & a<b, \end{cases} $$ fulfills \begin{align*} f(a,b)>1 &\Leftrightarrow P_1>P-2\\ f(a,b)=1 &\Leftrightarrow P_1=P_2\\ f(a,b)<1 &\Leftrightarrow P_1<P_2 \end{align*}
On the other hand, if you want to have $P_1(x)=f(a,b)P_2(x)$ for every possible $x$, $a$ and $b$, this is clearly not possible, since $P_1(x)=1$ and thus the above equation gives $$f(a,b) = \frac1{P_2(x)} = \frac{1-bx^2}{1-ax^2}$$ which clearly depends on $x$.
Let me say that I have tried to answer the question assuming that $a$, $b$, $x$ are numbers. As I see that you have chose tags such as matrix-decomposition, matrices, linear-algebra, linear-transformations for your question, you might have something completely different in mind. But I cannot think of any interpretation of your question which involves matrices. So I chose the only interpretation of your question which made sense to me. If it is not what you want to ask, it's up to you to clarify your question.