Problem 18 - Differential Equations and Linear Algebra

Question

Define \(T: P_{2}(\mathbb{R}) \rightarrow P_{1}(\mathbb{R})\) by $$T\left(a x^{2}+b x+c\right)=(a-b) x+c$$ Determine whether \(T\) is one-to-one, onto, both, or neither. Find \(T^{-1}\) or explain why it does not exist.

Step-by-Step Solution

Answer

The transformation \(T: P_{2}(\mathbb{R}) \rightarrow P_{1}(\mathbb{R})\) is both one-to-one and onto. Its inverse, \(T^{-1}: P_{1}(\mathbb{R}) \rightarrow P_{2}(\mathbb{R})\), exists and is given by: $$T^{-1}(r(x)) = T^{-1}(gx + h) = x^2 + (1-g)x + h$$

Step 1: Check if the transformation is one-to-one.

To check if \(T\) is one-to-one, we must show that if \(T(p(x)) = T(q(x))\), then \(p(x) = q(x)\). Let \(p(x) = ax^2 + bx + c\) and \(q(x) = dx^2 + ex + f\). Suppose that \(T(p(x)) = T(q(x))\). Then, we have: $$T(p(x)) = (a - b)x + c$$ $$T(q(x)) = (d - e)x + f$$ Since \(T(p(x)) = T(q(x))\), we get: $$(a - b)x + c = (d - e)x + f$$ Now, we have to show that \(a = d\), \(b = e\), and \(c = f\). Comparing the coefficients of the polynomials, we have: 1. Coefficient of \(x^1\): \(a - b = d - e \Rightarrow a - d = e - b\). 2. Coefficient of \(x^0\): \(c = f\). From the first equation, we get that \(a - d = b - e\), which implies that \(a - d = e - b\) and \(a = d\). So, the transformation is one-to-one.

Step 2: Check if the transformation is onto.

To check if \(T\) is onto, we must show that for any polynomial \(r(x)\) in \(P_{1}(\mathbb{R})\), there exists a polynomial in \(p(x)\) in \(P_{2}(\mathbb{R})\) such that \(T(p(x)) = r(x)\). Let \(r(x) = gx + h\) be any polynomial in \(P_{1}(\mathbb{R})\). We want to find a polynomial \(p(x) = ax^2 + bx + c\) such that: $$(a - b)x + c = gx + h$$ Comparing the coefficients, we get the following system of equations: 1. Coefficient of \(x^1\): \(a - b = g\). 2. Coefficient of \(x^0\): \(c = h\). There are no constraints for \(a, b, c\), so we can always find a polynomial in \(P_{2}(\mathbb{R})\) that maps to \(r(x)\). This implies that the transformation is onto.

Step 3: Find the inverse transformation.

Since \(T\) is both one-to-one and onto, we can find its inverse \(T^{-1}\). Let \(r(x) = gx + h\) be any polynomial in \(P_{1}(\mathbb{R})\) and \(p(x) = ax^2 + bx + c\) be its pre-image in \(P_{2}(\mathbb{R})\). From the previous step, we know that: 1. \(a - b = g\). 2. \(c = h\). We can rewrite the first equation as \(b = a - g\), and substitute this into the expression for \(p(x)\): $$p(x) = ax^2 + (a - g)x + h$$ Thus, the inverse transformation \(T^{-1}: P_{1}(\mathbb{R}) \rightarrow P_{2}(\mathbb{R})\) is given by: $$T^{-1}(r(x)) = T^{-1}(gx + h) = x^2 + (1-g)x + h$$

Key Concepts

Polynomial Functions

Polynomial functions are a key concept in mathematics, especially in algebra. A polynomial consists of variables and coefficients, combined using addition, subtraction, multiplication, and non-negative integer exponents. For example, a polynomial such as \(ax^2 + bx + c\) is called quadratic because its highest power of the variable \(x\) is 2. Other types, such as cubic, quartic, etc., follow based on the highest exponent.
Here's a quick breakdown of polynomials:

• Terms: These can be constants, variables, or a product of both.
• Degree: The highest power of \(x\) in the polynomial. For \(ax^2 + bx + c\), the degree is 2.
• Coefficients: The numbers before the variables, like \(a\), \(b\), and \(c\) in the example.

Polynomials, like \(P_{2}(\mathbb{R})\), are often used in linear transformations. Understanding the degree and structure of polynomials helps in visualizing how transformations occur and their properties.

One-to-One Mapping

One-to-one mapping (injective mapping) is a function where every element in the domain maps to a unique element in the range. In simpler terms, no two different inputs produce the same output.
Checking if a transformation like \(T\) is one-to-one involves ensuring that if \(T(p(x)) = T(q(x))\), then it must follow that \(p(x) = q(x)\).
In the context of the example given:

• If applying the transformation \((a - b)x + c\) resulted in identical outputs \((a-b)x+c = (d-e)x+f\), then it must conclude \((a = d\), \(b = e\), and \(c = f\)).
• This property ensures the uniqueness of the mapping, confirming the linear transformation \(T\) as one-to-one.

Recognizing one-to-one transformations is crucial for understanding their inverses and how data relations can be rearranged without overlap.

Onto Mapping

Onto mapping (surjective mapping) is a fundamental concept where each element in the target space is mapped by at least one element from the domain. This assures complete coverage of the range.
To verify if \(T\) is onto, we must confirm that for every resulting polynomial \(r(x) = gx + h\) in the range, there is a corresponding original polynomial \(p(x) = ax^2 + bx + c\) in the domain.
Regarding our example:

• For any \(gx + h\), assert that equations \(a - b = g\) and \(c = h\) can be satisfied.
• Since there are no restrictions on \(a, b,\) or \(c\), any linear polynomial can be reached, proving \(T\) as onto.

This characteristic is vital for ensuring that the transformation does not leave elements unmapped in the range, making each output truly representative of an initial input.

Inverse Transformations

An inverse transformation is a function that reverses the effects of another function. If a transformation \(T\) is both one-to-one and onto, it is invertible, meaning there exists a transformation \(T^{-1}\) that undoes \(T\).
For \(T\) given by \((a-b)x+c\), the corresponding \(T^{-1}\) allows you to retrieve the original polynomial from any transformed polynomial \(gx + h\).
To find \(T^{-1}\):

• Start from the equation given by the transformation, such as \((a-b = g)\) and \(c = h)\).
• Reexpress \(b = a - g\) to substitute into the polynomial \(p(x) = ax^2 + (a-g)x + h\).
• The inverse \(T^{-1}\) is then \(x^2 + (1-g)x + h\).

Knowing how to establish inverse functions provides greater control in navigating between transformed data and their original forms, enabling a complete understanding of the transformation's effect.