Question

In each case find new variables \(y_{1}\) and \(y_{2}\) that diagonalize the quadratic form \(q\). a. \(q=x_{1}^{2}+6 x_{1} x_{2}+x_{2}^{2}\) b. \(q=x_{1}^{2}+4 x_{1} x_{2}-2 x_{2}^{2}\)

Short answer

a. New variables are \( y_1 = x_1 + x_2 \), \( y_2 = x_1 - x_2 \); b. New variables are \( y_1 = x_1 + 2x_2 \), \( y_2 = x_1 - x_2 \).

Step-by-step solution

Write the Quadratic Form as a Matrix

Any quadratic form can be written as a matrix expression: \[ q(X) = X^T AX \]where \( X = \begin{pmatrix} x_1 \ x_2 \end{pmatrix}\) and matrix \(A\) is square and symmetric. For the quadratic form \( q = x_1^2 + 6x_1x_2 + x_2^2 \), the matrix \(A\) is:\[A = \begin{pmatrix} 1 & 3 \ 3 & 1 \end{pmatrix}\]

Find Eigenvalues of Matrix A

To diagonalize matrix \( A \), we must first find its eigenvalues. This is done by solving the characteristic equation:\[\det(A - \lambda I) = 0\]For \( A = \begin{pmatrix} 1 & 3 \ 3 & 1 \end{pmatrix} \), we have:\[\det\begin{pmatrix} 1 - \lambda & 3 \ 3 & 1 - \lambda \end{pmatrix} = (1-\lambda)(1-\lambda) - 9= 0\]\[\lambda^2 - 2\lambda - 8 = 0\]Solving this, the eigenvalues are \(\lambda_1 = 4\) and \(\lambda_2 = -2\).

Find Eigenvectors of Matrix A

For each eigenvalue, we find an eigenvector by solving \((A - \lambda I) v = 0\).For \(\lambda_1 = 4\):\[(A - 4I)v = \begin{pmatrix} -3 & 3 \ 3 & -3 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = 0\]Solving, an eigenvector is \(v_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix}\).For \(\lambda_2 = -2\):\[(A + 2I)v = \begin{pmatrix} 3 & 3 \ 3 & 3 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = 0\]Solving, an eigenvector is \(v_2 = \begin{pmatrix} 1 \ -1 \end{pmatrix}\).

Express Original Variables in Terms of New Variables

Once we have the eigenvectors, we can express the original variables in terms of the new variables:Let \( \begin{pmatrix} y_1 \ y_2 \end{pmatrix} = P^T \begin{pmatrix} x_1 \ x_2 \end{pmatrix} \), where\[P = \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix}\]The transformation is:\[\begin{pmatrix} y_1 \ y_2 \end{pmatrix} = \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix}\]So, \( y_1 = x_1 + x_2 \) and \( y_2 = x_1 - x_2 \).

Repeat Steps for Second Quadratic Form

Repeat the process for the quadratic form \(q = x_1^2 + 4x_1x_2 - 2x_2^2 \) where:\[A = \begin{pmatrix} 1 & 2 \ 2 & -2 \end{pmatrix}\]Find eigenvalues by solving: \[\lambda^2 + \lambda - 6 = 0\]The solutions are \(\lambda_1 = 2\) and \(\lambda_2 = -3\).

Calculate Eigenvectors for Second Form

For \(\lambda_1 = 2\): Solving \( (A - 2I)v = 0 \), we get the eigenvector \( v_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \).For \(\lambda_2 = -3\): Solving \( (A + 3I)v = 0 \), we get the eigenvector \( v_2 = \begin{pmatrix} 2 \ -1 \end{pmatrix} \).

Express New Variables for Second Form

With the eigenvectors, transform the variables:Let \( \begin{pmatrix} y_1 \ y_2 \end{pmatrix} = P^T \begin{pmatrix} x_1 \ x_2 \end{pmatrix} \), where:\[P = \begin{pmatrix} 1 & 2 \ 1 & -1 \end{pmatrix}\]The transformation is:\[\begin{pmatrix} y_1 \ y_2 \end{pmatrix} = \begin{pmatrix} 1 & 2 \ 1 & -1 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix}\]So, \( y_1 = x_1 + 2x_2 \) and \( y_2 = x_1 - x_2 \).

Key concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are vital when dealing with matrices, especially in the context of quadratic forms. Let's break down what they are and how they are used. An eigenvalue is a special number associated with a matrix. For a given matrix \( A \), an eigenvalue \( \lambda \) satisfies the equation \( A \mathbf{v} = \lambda \mathbf{v} \), where \( \mathbf{v} \) is an eigenvector. This means that when the matrix \( A \) acts on the eigenvector \( \mathbf{v} \), it simply scales \( \mathbf{v} \) by \( \lambda \), without changing its direction.

In the case of diagonalizing a quadratic form, we find the eigenvalues by solving the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. For example, in the given quadratic form \( q = x_1^2 + 6x_1x_2 + x_2^2 \), the corresponding matrix \( A \) has eigenvalues \( \lambda_1 = 4 \) and \( \lambda_2 = -2 \). Each eigenvalue leads to an eigenvector, such as \( v_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \) for \( \lambda_1 = 4 \). These eigenvectors are crucial for the diagonalization process.

Understanding eigenvalues and eigenvectors helps us simplify complex matrix operations and transforms them into easier, more interpretable forms.

Matrix Representation of Quadratic Forms

Quadratic forms can often be written as matrix expressions, which greatly simplifies their analysis. A quadratic form like \( q(x_1, x_2) = x_1^2 + 6x_1x_2 + x_2^2 \) can be represented in matrix form as \( q(X) = X^T A X \), where \( X \) is a column vector of variables, such as \( \begin{pmatrix} x_1 \ x_2 \end{pmatrix} \), and \( A \) is a symmetric matrix.

Writing a quadratic form this way allows us to easily identify the coefficients and relationships between variables. The symmetric matrix \( A \) for the expression above would be \( \begin{pmatrix} 1 & 3 \ 3 & 1 \end{pmatrix} \).

This representation brings clarity and helps in applying operations like finding eigenvalues and eigenvectors, which are needed for transforming the form into a diagonal structure. This diagonalized form makes calculations much simpler and reveals deeper properties of the original quadratic relationship.

Variable Transformation in Quadratic Forms

Variable transformation involves finding a new set of variables that simplifies a quadratic form, often through diagonalization. This process uses the eigenvectors from the quadratic form's matrix representation.

For our example, with the quadratic form \( q = x_1^2 + 6x_1x_2 + x_2^2 \), the eigenvectors \( v_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \) and \( v_2 = \begin{pmatrix} 1 \ -1 \end{pmatrix} \) are used. We express the new variables \( y_1 \) and \( y_2 \) as linear combinations of the original variables \( x_1 \) and \( x_2 \).

The transformation matrix \( P \), composed of the eigenvectors, allows us to convert the variables: \( y_1 = x_1 + x_2 \) and \( y_2 = x_1 - x_2 \).

This variable change often changes the form into a simpler, diagonalized state, making it more manageable to analyze and interpret. The goal is simplifying the quadratic form while preserving its essential characteristics.