Problem 17 - Differential Equations and Linear Algebra

Question

Define \(T: P_{1}(\mathbb{R}) \rightarrow P_{1}(\mathbb{R})\) by $$T(a x+b)=(2 b-a) x+(b+a)$$ Show that \(T\) is both one-to-one and onto, and find \(T^{-1}\).

Step-by-Step Solution

Answer

\(T\) is one-to-one and onto. The inverse transformation \(T^{-1}\) is given by: \(T^{-1}(kx + l) = (2l - k)x + (l - m)\)

Step 1: Prove T is one-to-one (injective)

To prove that T is one-to-one, let's assume that \(T(p(x)) = T(q(x))\) for some polynomials \(p(x)\) and \(q(x)\) in \(P_1(\mathbb{R})\). We have: 1) \(p(x) = ax + b\) 2) \(q(x) = cx + d\) Applying the transformation T, we get: 1) \(T(p(x)) = (2b-a)x + (b+a)\) 2) \(T(q(x)) = (2d-c)x + (d+c)\) Since we assumed that \(T(p(x)) = T(q(x))\), we have: \((2b-a)x + (b+a) = (2d-c)x + (d+c)\) For the above equation to hold true, the coefficients of \(x\) and the constant terms must be equal: 1) \(2b - a = 2d - c\) 2) \(b + a = d + c\) Now we have a system of two linear equations with two variables (a, b) and (c, d). Let's solve for a and b: From the second equation, we can write: \(a = d + c - b\) Substitute this expression for a in the first equation: \(2b - (d+c-b) = 2d - c\) Solving for b, we get: \(b = d\) Substitute this result back into the equation for a: \(a = d + c - d\) \(a = c\) As we have shown that \(a=c\) and \(b=d\), we conclude that \(p(x) = q(x)\). Hence, the transformation T is one-to-one (injective).

Step 2: Prove T is onto (surjective)

To prove that T is onto, let's consider an arbitrary polynomial function \(r(x) = kx + l\) in \(P_1(\mathbb{R})\). We need to show that there exists a polynomial function \(s(x) = mx + n\) in \(P_1(\mathbb{R})\) such that \(T(s(x)) = r(x)\). Applying T to \(s(x)\), we get: \(T(s(x)) = (2n-m)x + (n+m)\) For T(s(x)) = r(x), we need to find m and n such that: 1) \(2n - m = k\) 2) \(n + m = l\) Solving this system of linear equations for m and n yields: \(m = 2l - k\) \(n = l - m\) Thus, we can construct the polynomial \(s(x) = (2l - k)x + (l - m)\), which when transformed by T gives \(r(x)\). Since \(r(x)\) was arbitrary, we have shown that the transformation T is onto (surjective).

Step 3: Find the T-inverse

Now we need to find the inverse transformation \(T^{-1}\). To do this, we will use the expressions found for m and n in the previous step. Given a polynomial \(r(x) = kx + l \in P_1(\mathbb{R})\), we found that \(s(x) = (2l - k)x + (l - m)\) is such that \(T(s(x)) = r(x)\). Therefore, we can define \(T^{-1}\) as: \(T^{-1}(kx + l) = (2l - k)x + (l - m)\) This is the inverse transformation of T.

Key Concepts

One-to-One Functions

A one-to-one function, also known as an injective function, ensures that each output of the function corresponds to exactly one unique input. This means if you have two inputs, and they give the same output, those inputs must have been equal. To determine if a function is one-to-one, we often prove that assuming the equality of outputs leads to the equality of inputs.

In this particular problem, we demonstrated that the linear transformation \(T\) is one-to-one by taking two arbitrary polynomials \(p(x)=ax+b\) and \(q(x)=cx+d\). Applying the transformation \(T\) to both, we obtained:

• \(T(p(x)) = (2b-a)x + (b+a)\)
• \(T(q(x)) = (2d-c)x + (d+c)\)

By setting \(T(p(x)) = T(q(x))\), we solved the system of equations:

• \(2b-a = 2d-c\)
• \(b+a = d+c\)

These equations led us to the conclusion that \(a=c\) and \(b=d\), meaning \(p(x) = q(x)\), thus proving \(T\) is one-to-one.

Onto Functions

An onto function, or surjective function, is one in which every element of the codomain is mapped to by at least one element of the domain. This speaks to the completeness of coverage; every possible output is achieved by some input.

In the exercise, to prove \(T\) is onto, we considered an arbitrary polynomial \(r(x) = kx + l\) in \(P_1(\mathbb{R})\). Our goal was to show there exists a polynomial \(s(x) = mx + n\) such that \(T(s(x)) = r(x)\).

• The transformation of \(s(x)\) is \(T(s(x)) = (2n-m)x + (n+m)\).

We wanted to find \(m\) and \(n\) that satisfy:

• \(2n - m = k\)
• \(n + m = l\)

Solving these equations, we found:

• \(m = 2l - k\)
• \(n = l - m\)

Hence, for any arbitrary polynomial \(r(x)\), a corresponding \(s(x)\) can be found, proving \(T\) is onto.

Inverse Functions

An inverse function is a reverse operation to the original function, such that applying the function and its inverse successively results in returning to the original input.

In our example, after establishing that \(T\) is both one-to-one and onto, we tackled finding \(T^{-1}\), the inverse of \(T\). Knowing a function that is both one-to-one and onto (bijective) guarantees an inverse exists.

• Given any \(r(x) = kx + l\), the polynomial \(s(x) = (2l - k)x + (l - m)\) was found such that \(T(s(x)) = r(x)\).

Therefore, the expression for the inverse function \(T^{-1}\) becomes \(T^{-1}(kx + l) = (2l - k)x + (l - m)\). This inverse allows us to reverse the transformation and retrieve any input from its output, demonstrating the complete operation of \(T\).