Problem 31 - A History of Mathematics: An Introduction

Question

Show that if the substitution \(x=\alpha x^{\prime}+\beta y^{\prime}, y=\gamma x^{\prime}+\) \(\delta y^{\prime}\) with \(\alpha \delta-\beta \gamma=1\) transforms the quadratic form \(F=\) \(a x^{2}+2 b x y+c y^{2}\) into the form \(F^{\prime}=a^{\prime} x^{\prime 2}+2 b^{\prime} x^{\prime} y^{\prime}+\) \(c^{\prime} y^{\prime 2}\), then there is an "inverse" substitution of the same form that transforms \(F^{\prime}\) into \(F\).

Step-by-Step Solution

Answer

Question: Given a quadratic form F and a substitution which transforms it into another quadratic form F', show that there exists an inverse substitution that transforms F' back into F. Answer: Yes, the existence of an inverse substitution is demonstrated when applying the given substitution and its inverse to their respective quadratic forms, F and F', showing that terms in both equations correspond to each other. This means that if the given substitution transforms F into F', the inverse substitution will transform F' back into F.

Step 1: Analyze the given substitution

The given substitution is: \(x = \alpha x' + \beta y',\ y = \gamma x' + \delta y'\) with \(\alpha \delta - \beta \gamma = 1\)

Step 2: Find the inverse substitution

To find the inverse substitution, we should find expressions for \(x'\) and \(y'\) in terms of \(x\) and \(y\). Using the condition \(\alpha \delta - \beta \gamma = 1\), we can find the inverse of the matrix associated with the linear transformation: \(\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \delta & -\beta \\ -\gamma & \alpha \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}\) Thus, the inverse substitution is: \(x' = \delta x - \beta y,\ y' = -\gamma x + \alpha y\)

Step 3: Apply the given substitution to F

Substitute \(x = \alpha x' + \beta y'\) and \(y = \gamma x' + \delta y'\) into F: \(F' = a(\alpha x' + \beta y')^2 + 2b(\alpha x' + \beta y')(\gamma x' + \delta y') + c(\gamma x' + \delta y')^2\)

Step 4: Apply the inverse substitution to F'

Substitute \(x' = \delta x - \beta y\) and \(y' = -\gamma x + \alpha y\) into F': \(F = a'(\delta x - \beta y)^2 + 2b'(\delta x - \beta y)(-\gamma x + \alpha y) + c'(-\gamma x + \alpha y)^2\)

Step 5: Compare F and F'

By comparing the expressions for F and F', we can notice that the terms in both equations correspond to each other. This means that the transformation applied by the given substitution can indeed be reversed by its inverse transformation, and the quadratic form F can be obtained from F' by applying the inverse substitution. Thus, we have shown that if the given substitution transforms F into F', there exists an inverse substitution that transforms F' back into F.

Key Concepts

Linear Transformations

Linear transformations are mathematical operations that change variables through a linear function. In our example, we have a pair of equations: - \(x = \alpha x' + \beta y'\)- \(y = \gamma x' + \delta y'\)where coefficients \(\alpha, \beta, \gamma, \) and \(\delta\) are constants. These constants form a matrix, and the transformation can be represented using matrix operations.Linear transformations are crucial because they allow us to explore how a shape or space behaves as we change variables. They maintain the "linearity," meaning that lines remain lines and planes remain planes after transformation.This property is particularly handy when dealing with quadratic forms as it helps simplify complex expressions by changing the reference frame.

Inverse Substitutions

Inverse substitutions are, essentially, reversing a transformation to retrieve the original variables from the transformed ones. In the problem, we aim to find the inverse of the transformation defined by:- \(x = \alpha x' + \beta y'\)- \(y = \gamma x' + \delta y'\)To achieve this, we find expressions for \(x'\) and \(y'\) in terms of \(x\) and \(y\). This involves finding the inverse of the transformation matrix. The original matrix is:\[ \begin{pmatrix} \alpha & \beta \ \gamma & \delta \end{pmatrix} \]Its inverse is given by:\[ \begin{pmatrix} \delta & -\beta \ -\gamma & \alpha \end{pmatrix} \]The condition \(\alpha \delta - \beta \gamma = 1\) ensures that this matrix is invertible. By using these inverse matrix components, new variables \(x'\) and \(y'\) can be expressed as:- \(x' = \delta x - \beta y\)- \(y' = -\gamma x + \alpha y\)This inverse operation is crucial when wanting to reverse a transformation, ensuring that we can switch back to our original form.

Matrix Algebra

Matrix algebra provides a powerful mechanism for solving linear equations and transformations. It is especially helpful in the context of quadratic forms, as matrices can represent transformations compactly.In matrix algebra, we use matrices and vectors to perform operations that effect transformations on spatial coordinates. For instance, given a matrix \(M\) that represents a transformation, the multiplication of \(M\) with a vector \(\mathbf{v}\) transforms \(\mathbf{v}\) as \(M\mathbf{v}\). When finding an inverse, as needed in this exercise, the matrix \(M\) is replaced by its inverse \(M^{-1}\), so that \(M^{-1}M\mathbf{v} = \mathbf{v}\). This step effectively "undoes" the original transformation. - For our exercise, the original matrix \(\begin{pmatrix} \alpha & \beta \ \gamma & \delta \end{pmatrix}\) has the inverse \(\begin{pmatrix} \delta & -\beta \ -\gamma & \alpha \end{pmatrix}\).This process shows how matrix algebra can simplify the manipulation of systems of equations, allowing us to navigate transformations and their inverses seamlessly. With quadratic forms, this is particularly useful because it efficiently solves problems that involve multi-variable transformations.