Question

Let \(f(x)=\frac{a x+b}{c x+d}\) and assume \(b c-a d \neq 0\). (a) Find the formula for \(f^{-1}(x)\). (b) Why is the condition \(b c-a d \neq 0\) needed? (c) What condition on \(a, b, c\), and \(d\) will make \(f=f^{-1}\) ?

Short answer

(a) \(f^{-1}(x) = \frac{dx-b}{a-cx}\); (b) ensures function invertibility; (c) \(a = d, b = -c\).

Step-by-step solution

Understanding the problem

We are given a fractional linear transformation \(f(x)=\frac{a x+b}{c x+d}\) and need to find its inverse function, check conditions for the inverse, and determine when \(f(x) = f^{-1}(x)\).

Find the inverse function

Assume the inverse function is in the form \(f^{-1}(y) = \frac{p y + q}{r y + s}\). Start by setting \(y = f(x) = \frac{a x + b}{c x + d}\). Solving for \(x\) in terms of \(y\), we get \(y(c x + d) = a x + b\). Rearrange to form a linear equation ax - yc x = -b + yd, thus x(ac - yc) = yd - b, which gives \(x = \frac{d y - b}{a - c y}\). The inverse function is \(f^{-1}(x) = \frac{d x - b}{a - c x}\).

Why the condition \(bc - ad \neq 0\) is needed

The condition \(bc - ad eq 0\) ensures that the linear transformation is bijective, meaning it has an inverse. If \(bc - ad = 0\), the function becomes a constant or undefined, making it impossible to invert due to zero division or no unique result for each input.

Find condition for \(f = f^{-1}\)

Set \(\frac{a x + b}{c x + d} = \frac{d x - b}{a - c x}\). Cross-multiply to get \((a x + b)(a - c x) = (d x - b)(c x + d)\). Simplify both sides: \(a^2 x - ac x^2 + ab - bcx = d c x^2 + d^2 x - bcx - bd\). Equate coefficients, giving equations: 1) \(a^2 = dc\), 2) \(-ac = d^2\), 3) \(ab = -bd\). Solving these conditions reveals: a = d and b = -c, or a = 0, b = 0, c = 0, d = 0, which are truncated due to restriction \(bc - ad eq 0\).

Key concepts

Fractional Linear Transformation

Fractional linear transformation, also known as a Möbius transformation or a bilinear transformation, is a function of the form \( f(x) = \frac{a x + b}{c x + d} \). This function transforms a real or complex number \( x \) into another value, usually by means of a ratio of two linear expressions. The parameters \( a, b, c, \) and \( d \) are real or complex numbers, with the critical condition that the expression \( bc - ad eq 0 \) must hold. This condition ensures that the function does not reduce to a constant or become undefined, giving it the versatility needed for effective transformation.

Möbius transformations have significant applications in different areas of mathematics, including complex analysis and projective geometry. They can map the number line, the complex plane, or even the Riemann sphere, providing a variety of uses in transformation mechanics and symmetry analysis.

Bijective Functions

A bijective function is one that satisfies two important criteria: it is both injective (one-to-one) and surjective (onto). This means each element in the function's domain maps to a unique element in the codomain, and each element in the codomain is covered by the domain.

For a fractional linear transformation \( f(x) = \frac{a x + b}{c x + d} \), the condition \( bc - ad eq 0 \) is crucial to maintaining bijection. This condition prevents the function from degenerating into a constant, which would violate both injectivity and surjectivity, and hence fail to be bijective. This non-degenerate property ensures that there exists an inverse function \( f^{-1}(x) \), making inversion possible.

Understanding bijective functions is foundational in mathematics as they ensure every input yields a unique output, enabling equations to be solved more effectively and transformations to be reversed properly.

Function Inversion

Function inversion involves finding a function \( f^{-1}(x) \) such that applying \( f \) and then \( f^{-1} \) returns to the original input. For our given fractional linear transformation, the inverse function has a specific form: \( f^{-1}(x) = \frac{d x - b}{a - c x} \). This form is derived through algebraic manipulation, reversing the effects of the original transformation.

To find the inverse, start by setting \( y = \frac{a x + b}{c x + d} \), solve for \( x \), and rearrange. Solving the equation leads to swapping and manipulating variables to isolate \( x \) on one side. This process highlights the importance of non-zero constants in both the numerator and denominator, ensuring the equation remains valid throughout. The inverse function has meaningful applications in solving equations, modeling reversible processes, and in fields such as cryptography.

Linear Algebra

Linear algebra forms a backbone in understanding many mathematical concepts, including functions and transformations. The study of vectors, matrices, and linear mappings allows us to model and solve equations succinctly.

In the context of fractional linear transformations, linear algebra is employed to understand how transformations affect geometry and space. Consider a matrix representation \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \), where the determinant (\( ad - bc \)) signifies conditions under which transformations are invertible. When \( ad - bc eq 0 \), the corresponding matrix is termed invertible in linear algebra, mirroring our requirement for the fractional linear transformation.

This interplay between matrices and functions provides a powerful toolkit for handling a wide array of mathematical problems, empowering us to see how transformations interact in higher dimensions and across complex systems.