Question

Let each of the following matrices M describe a deformation of the \((x, y)\) plane. For each given M find: the cigenvalues and eigenvectors of the transformation, the matrix \(\mathrm{C}\) 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. $$\left(\begin{array}{rr} 2 & -1 \\ -1 & 2 \end{array}\right)$$

Short answer

Eigenvalues are 3 and 1. Eigenvectors are \(\left( \begin{array}{c} 1 \ -1 \end{array} \right)\) and \(\left( \begin{array}{c} 1 \ 1 \end{array} \right)\). The deformation is a stretch by 3 in one direction and no change in the other.

Step-by-step solution

Find the Eigenvalues

To find the eigenvalues of the matrix \( M = \left(\begin{array}{rr} 2 & -1 \ -1 & 2 \end{array}\right) \), solve the characteristic equation \( \det(M - \lambda I) = 0 \). This gives the equation: \[ \left| \begin{array}{cc} 2 - \lambda & -1 \ -1 & 2 - \lambda \end{array} \right| = 0. \]Expanding the determinant, we have: \[ (2 - \lambda)(2 - \lambda) - (-1)(-1) = 0, \]\[ (2 - \lambda)^2 - 1 = 0, \]\[ \lambda^2 - 4\lambda + 4 - 1 = 0, \]\[ \lambda^2 - 4\lambda + 3 = 0, \]\[ (\lambda - 3)(\lambda - 1) = 0. \]Thus, the eigenvalues are: \(\lambda_1 = 3\) and \(\lambda_2 = 1\).

Find the Eigenvectors

For each eigenvalue, solve the equation \( (M - \lambda I) \mathbf{v} = \mathbf{0} \).For \(\lambda_1 = 3\):\[ \left( \begin{array}{cc} 2 - 3 & -1 \ -1 & 2 - 3 \end{array} \right) \left( \begin{array}{c} x \ y \end{array} \right) = \left( \begin{array}{c} 0 \ 0 \end{array} \right), \]\[ \left( \begin{array}{cc} -1 & -1 \ -1 & -1 \end{array} \right) \left( \begin{array}{c} x \ y \end{array} \right) = \left( \begin{array}{c} 0 \ 0 \end{array} \right). \]This simplifies to \[ -x - y = 0. \]Hence, one eigenvector is \(\mathbf{v}_1 = \left( \begin{array}{c} 1 \ -1 \end{array} \right)\).For \(\lambda_2 = 1\):\[ \left( \begin{array}{cc} 2 - 1 & -1 \ -1 & 2 - 1 \end{array} \right) \left( \begin{array}{c} x \ y \end{array} \right) = \left( \begin{array}{c} 0 \ 0 \end{array} \right), \]\[ \left( \begin{array}{cc} 1 & -1 \ -1 & 1 \end{array} \right) \left( \begin{array}{c} x \ y \end{array} \right) = \left( \begin{array}{c} 0 \ 0 \end{array} \right). \]This simplifies to\[ x - y = 0. \]Hence, one eigenvector is \(\mathbf{v}_2 = \left( \begin{array}{c} 1 \ 1 \end{array} \right)\).

Form the Diagonalization Matrix and the Diagonal Matrix

The matrix \(C\) which diagonalizes \(M\) is formed using the eigenvectors as columns:\[ C = \left( \begin{array}{cc} 1 & 1 \ -1 & 1 \end{array} \right). \]The diagonal matrix \(D\) will have the eigenvalues on the diagonal:\[ D = \left( \begin{array}{cc} 3 & 0 \ 0 & 1 \end{array} \right). \]So, \( C \) and \( D \) are:\[ C = \left( \begin{array}{cc} 1 & 1 \ -1 & 1 \end{array} \right), \]\[ D = \left( \begin{array}{cc} 3 & 0 \ 0 & 1 \end{array} \right). \]

Describe the Deformation Relative to the New Axes

The new axes defined by the eigenvectors correspond to the directions along which the deformation acts as a simple scaling (stretching or compressing the plane). The deformation relative to these new axes given by matrix \(D\) indicates scaling by 3 along the direction of eigenvector \(\mathbf{v}_1\) and by 1 along the direction of eigenvector \(\mathbf{v}_2\). Thus, the deformation in the new coordinate system is a stretch by a factor of 3 along one axis and no change (identity transformation) along the other axis.

Key concepts

Matrix Diagonalization

Matrix diagonalization is the process of finding a diagonal matrix that is similar to a given square matrix. This involves finding an invertible matrix (let's call it matrix C) made up of the eigenvectors of the original matrix and a diagonal matrix D that contains the eigenvalues. The concept is crucial because it simplifies many matrix operations like finding powers of the matrix, solving differential equations, and understanding the geometric transformation properties of matrices. In the exercise, we saw that the matrix C is formed using the eigenvectors as columns, and the diagonal matrix D has the eigenvalues on the diagonal.

Diagonalizing a matrix helps in transforming complex linear transformations into simpler ones, making calculations easier. The diagonal matrix makes it clear how the transformation scales or rotates space.

Characteristic Equation

The characteristic equation of a matrix is essential in finding its eigenvalues. To derive the characteristic equation, you subtract \( \lambda I \) from the matrix M and set the determinant to zero: \[ \det(M - \lambda I) = 0. \] For our given matrix, solving the characteristic equation provided us with \( \lambda_1 = 3 \) and \( \lambda_2 = 1. \)

This process involves finding the roots of a polynomial that comes from the determinant of \( M - \lambda I. \) Each root (solution) represents an eigenvalue. By solving for these values, we identify scales of transformation along different directions in the eigenvector framework. Understanding the characteristic equation lets you predict how a matrix will stretch, compress or rotate space along its eigenvectors.

Deformation Matrix

A deformation matrix describes how a shape changes under a transformation. It's particularly useful in physics and engineering, where you need to understand how structures change under various forces. In our exercise, the given matrix M represents a deformation of the \( (x, y) \) plane.

After diagonalizing the matrix M, we obtained matrix D and C. Matrix D gives us the simplest description of the transformation in the new coordinate system: scaling by 3 along the direction of eigenvector \( \mathbf{v}_1 = \begin{array}{c} 1 \ -1 \end{array} \) and by 1 along the direction of eigenvector \( \mathbf{v}_2 = \begin{array}{c} 1 \ 1 \end{array} \).

The matrix C is the rotation to the new axes \( (x', y') \), where the deformation is given by matrix D. This scaling is crucial for understanding physical behaviors like stress, strain, and material deformation. In simpler terms, matrix D tells how much the plane stretches or compresses along the new eigen-directions, making it easier to interpret complex deformations.