Question

Let each of the following matrices \(M\) describe a deformation of the \((x, y)\) plane. For each given \(M\) find: the eigenvalues and eigenvectors of the transformation, the matrix \(C\) which diagonalizes \(M\) and specifies the rotation to new axes \(\left(x^{\prime}, y^{\prime}\right)\) along the eigenvectors, and the matrix \(D\) which gives the deformation relative to the new axes. Describe the deformation relative to the new axes.\(\left(\begin{array}{ll}3 & 1 \\ 1 & 3\end{array}\right)\)

Short answer

Eigenvalues: 4, 2. Eigenvectors: \(\begin{pmatrix} 1 \ 1 \end{pmatrix}\), \(\begin{pmatrix} 1 \ -1 \end{pmatrix}\). Matrices: \( C = \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix}\) and \( D = \begin{pmatrix} 4 & 0 \ 0 & 2 \end{pmatrix}\).

Step-by-step solution

Find the Eigenvalues

To find the eigenvalues of matrix M, solve the characteristic equation \[ \det(M - \lambda I) = 0 \] where \( M = \begin{pmatrix} 3 & 1 \ 1 & 3 \end{pmatrix} \) and \( I \) is the identity matrix.

Set Up and Solve the Determinant Equation

Compute \(\det( \begin{pmatrix} 3 & 1 \ 1 & 3 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}) = 0 \). This results in \(\det( \begin{pmatrix} 3 - \lambda & 1 \ 1 & 3 - \lambda \end{pmatrix}) = 0 \) which simplifies to \[((3-\lambda)^2 - 1 = 0) \].

Solve for Eigenvalues

Solve the quadratic equation \((3 - \lambda)^2 - 1 = 0 \) to find the eigenvalues. This yields \((\lambda - 4)(\lambda - 2) = 0 \), so the eigenvalues are \(\lambda_1 = 4\) and \(\lambda_2 = 2\).

Find the Eigenvectors

For each eigenvalue, substitute \(\lambda\) back into \((M - \lambda I)\mathbf{v} = 0 \) to solve for the corresponding eigenvectors.

Eigenvector for \(\lambda_1 = 4\)

Substitute \(\lambda = 4\) into \((\begin{pmatrix} 3 & 1 \ 1 & 3 \end{pmatrix} - 4 \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix})\mathbf{v} = 0 \) giving \(\begin{pmatrix} -1 & 1 \ 1 & -1 \end{pmatrix}\mathbf{v} = 0 \). Solving this, we get the eigenvector \(\mathbf{v}_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \).

Eigenvector for \(\lambda_2 = 2\)

Substitute \(\lambda = 2\) into \((\begin{pmatrix} 3 & 1 \ 1 & 3 \end{pmatrix} - 2 \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix})\mathbf{v} = 0 \) giving \(\begin{pmatrix} 1 & 1 \ 1 & 1 \end{pmatrix}\mathbf{v} = 0 \). Solving this, we get the eigenvector \(\mathbf{v}_2 = \begin{pmatrix} 1 \ -1 \end{pmatrix} \).

Construct the Matrix C

Place the eigenvectors as columns in the matrix C. So, \( C = \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} \).

Verify the Diagonalization

Check that \( C^{-1} * M * C = D \), where D is a diagonal matrix with the eigenvalues on the diagonal. Calculate \( C^{-1} \) and verify the relation. \( D = \begin{pmatrix} 4 & 0 \ 0 & 2 \end{pmatrix} \).

Describe the Deformation Relative to New Axes

The deformation in the new \(x', y'\) axes is represented by \( D \), which scales along the eigenvectors by 4 and 2 respectively.

Key concepts

Matrix Diagonalization

Matrix diagonalization is a powerful method to simplify a square matrix into a diagonal form. This technique is useful because diagonal matrices are easier to work with, especially for computations involving matrix powers or exponentials. Given a square matrix \(M\), we want to find another matrix \(C\) such that \(C^{-1}MC\) is a diagonal matrix \(D\). The matrix \(C\) is formed from the eigenvectors of \(M\), while \(D\) contains the eigenvalues of \(M\) on its main diagonal. Diagonalization is significant in solving systems of differential equations, in performing matrix exponentiation, and in understanding the geometry of linear transformations.

Linear Transformation

A linear transformation is a function between two vector spaces that preserves addition and scalar multiplication. For instance, in the exercise, matrix \(M\) describes a deformation of the 2D plane, mapping coordinates \((x, y)\) to new coordinates. This transformation can stretch, rotate, or otherwise change the shape of objects in the plane. By analyzing the eigenvectors and eigenvalues of \(M\), we gain insights into the nature of this transformation. For example, eigenvectors indicate directions that remain invariant under the transformation, and their corresponding eigenvalues show how much these directions are scaled.

Characteristic Equation

The characteristic equation is a determinant-based polynomial equation used to find the eigenvalues of a matrix. For a matrix \(M\), the characteristic equation is formed by \(\text{det}(M - \text{λ}I) = 0\), where \(\text{λ}\) represents the eigenvalues and \(I\) is the identity matrix. In our solved exercise, for matrix \(M = \begin{pmatrix}3 & 1 \ 1 & 3\begin{pmatrix}\), we get \(\text{det}( \begin{pmatrix} 3 & 1 \ 1 & 3 \begin{pmatrix} - \text{λ}\begin{pmatrix} 1 & 0 \ 0 & 1\begin{pmatrix}) = 0\) , which simplifies to \((3 - \text{λ})^2 - 1 = 0\). Solving this gives the eigenvalues \(\text{λ} = 4\) and \(\text{λ} = 2\).

Eigenvalues

Eigenvalues are special scalars associated with a matrix that provide deep insights into its properties. For a given square matrix \(M\), an eigenvalue \(\text{λ}\) satisfies the equation \(M \text{v} = \text{λ} \text{v}\) for some non-zero vector \(\text{v}\) (the eigenvector). These values tell us about the scaling factor in the directions of the eigenvectors. In our featured solution, the eigenvalues 4 and 2 show how the matrix \(M\) compresses or stretches the plane along the direction of its eigenvectors. High eigenvalues suggest greater stretching, while low eigenvalues indicate compression.

Eigenvectors

Eigenvectors are non-zero vectors that only scale during a matrix transformation. They do not change direction, making them crucial for understanding the behavior of the transformation. For the matrix \(M\), eigenvectors are found by solving \((M - \text{λ} I) \text{v} = 0\) for each eigenvalue \(\text{λ}\). In our exercise, the eigenvectors corresponding to \(\text{λ} = 4\) and \(\text{λ} = 2\) are \(\text{v}_1 = \begin{pmatrix}1 \ 1 \begin{pmatrix}\) and \(\text{v}_2 = \begin{pmatrix}1 \ -1\begin{pmatrix}\), respectively. These vectors define the new axes, revealing the intrinsic directions that remain invariant under the linear transformation.