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.\(\left(\begin{array}{ll}3 & 2 \\ 2 & 3\end{array}\right)\)

Short answer

Eigenvalues: 5, 1.Eigenvectors: \( \begin{pmatrix} 1 \ 1 \end{pmatrix} \), \( \begin{pmatrix} -1 \ 1 \end{pmatrix} \).Matrix C: \( \begin{pmatrix} 1 & -1 \ 1 & 1 \end{pmatrix} \).Matrix D: \( \begin{pmatrix} 5 & 0 \ 0 & 1 \end{pmatrix} \).

Step-by-step solution

Find the Eigenvalues of Matrix M

Given matrix \[ M = \begin{pmatrix} 3 & 2 \ 2 & 3 \end{pmatrix} \]. To find the eigenvalues, solve the characteristic equation \[ \text{det}(M - \lambda I) = 0 \]. Compute the determinant: \[ \text{det} \left( \begin{pmatrix} 3 & 2 \ 2 & 3 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \right) = 0 \]. This simplifies to: \[ \text{det} \left( \begin{pmatrix} 3 - \lambda & 2 \ 2 & 3 - \lambda \end{pmatrix} \right) = 0 \]. Solving \[ (3 - \lambda)^2 - 4 = 0 \]. This simplifies to \[ \lambda^2 - 6\lambda + 5 = 0 \].The roots of the quadratic equation are: \[ \lambda_1 = 5 \] and \[ \lambda_2 = 1 \].

Find the Eigenvectors of Matrix M

For \( \lambda_1 = 5 \), solve \[ (M - 5I) \mathbf{v} = 0 \]. This yields \[ \begin{pmatrix} 3 - 5 & 2 \ 2 & 3 - 5 \end{pmatrix} \mathbf{v} = \begin{pmatrix} -2 & 2 \ 2 & -2 \end{pmatrix} \mathbf{v} = 0 \]. This gives eigenvector \[ \mathbf{v}_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \].For \( \lambda_2 = 1 \), solve \[ (M - 1I) \mathbf{v} = 0 \]. This yields \[ \begin{pmatrix} 3 - 1 & 2 \ 2 & 3 - 1 \end{pmatrix} \mathbf{v} = \begin{pmatrix} 2 & 2 \ 2 & 2 \end{pmatrix} \mathbf{v} = 0 \]. This gives eigenvector \[ \mathbf{v}_2 = \begin{pmatrix} -1 \ 1 \end{pmatrix} \].

Form Matrix C using Eigenvectors

Matrix \( C \) is formed by placing the eigenvectors as columns: \[ C = \begin{pmatrix} 1 & -1 \ 1 & 1 \end{pmatrix} \].

Form Diagonal Matrix D using Eigenvalues

Matrix \( D \) is formed by placing the eigenvalues on the diagonal: \[ D = \begin{pmatrix} 5 & 0 \ 0 & 1 \end{pmatrix} \].

Describe the Deformation Relative to the New Axes

The deformation relative to the new axes can be observed by analyzing matrix \( D \). Along the new axes corresponding to the eigenvectors, the transformation scales vectors by 5 in the direction of eigenvector \( \mathbf{v}_1 \) and by 1 in the direction of eigenvector \( \mathbf{v}_2 \). Hence, the transformation stretches the plane by a factor of 5 along \( \mathbf{v}_1 \) and leaves vectors along \( \mathbf{v}_2 \) unchanged.

Key concepts

Characteristic Equation

To understand the deformation described by a matrix, we first find its eigenvalues. This is done using the characteristic equation. For a given matrix \( M \), the characteristic equation is found by solving \( \text{det}(M - \lambda I) = 0 \).
In our example, given matrix \( M = \begin{pmatrix} 3 & 2 \ 2 & 3 \end{pmatrix} \), we compute the determinant of \( \begin{pmatrix} 3 - \lambda & 2 \ 2 & 3 - \lambda \end{pmatrix} \). This simplifies to \( (3 - \lambda)^2 - 4 = 0 \), which gives the characteristic equation \( \lambda^2 - 6\lambda + 5 = 0 \).
Solving this quadratic equation, we find the roots, or eigenvalues: \( \lambda_1 = 5 \) and \( \lambda_2 = 1 \). These eigenvalues are crucial for determining the behavior of the transformation described by the matrix.

Diagonalization of Matrices

Diagonalizing a matrix involves transforming it into a diagonal matrix, which simplifies understanding its effects on a vector space. After finding the eigenvalues, we proceed to find the eigenvectors.
For \( \lambda_1 = 5 \), solving \( (M - 5I) \mathbf{v} = 0 \) gives us the eigenvector \( \mathbf{v}_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \).
For \( \lambda_2 = 1 \), solving \( (M - 1I) \mathbf{v} = 0 \) gives us the eigenvector \( \mathbf{v}_2 = \begin{pmatrix} -1 \ 1 \end{pmatrix} \).
Forming matrix \( C \) using these eigenvectors, we place them as columns: \( C = \begin{pmatrix} 1 & -1 \ 1 & 1 \end{pmatrix} \).
Matrix \( D \) is formed by placing the eigenvalues on the diagonal: \( D = \begin{pmatrix} 5 & 0 \ 0 & 1 \end{pmatrix} \).
This diagonalization process simplifies understanding the transformation, making it clear how each axis (aligned with the eigenvectors) is scaled by the corresponding eigenvalues.

Matrix Deformation

Matrix deformation describes how a transformation matrix alters the plane. The matrix \( D \), which is diagonal, shows the deformation relative to the new axes.
In the example, the given matrix \( M \) deforms the plane by stretching it differently along the eigenvectors. The eigenvalues \( \lambda_1 = 5 \) and \( \lambda_2 = 1 \) indicate the scaling factors along these new axes.
Specifically, with eigenvector \( \mathbf{v}_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \), the matrix stretches the plane by a factor of 5. With eigenvector \( \mathbf{v}_2 = \begin{pmatrix} -1 \ 1 \end{pmatrix} \), it leaves the plane unchanged (scaling factor of 1).
This information is crucial as it breaks down the complex transformation into simpler, understandable components, showing how each part of the plane is impacted.

Scaling Transformation

Scaling transformation refers to changing the size of objects in a vector space without altering their shape. In linear algebra, matrices often describe these transformations.
The matrix \( M = \begin{pmatrix} 3 & 2 \ 2 & 3 \end{pmatrix} \) scales the vector space along the directions of the eigenvectors \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \).
From our diagonal matrix \( D = \begin{pmatrix} 5 & 0 \ 0 & 1 \end{pmatrix} \), the scaling factors are clear: a factor of 5 along \( \mathbf{v}_1 \) and a factor of 1 along \( \mathbf{v}_2 \).
This means any vector aligned with \( \mathbf{v}_1 \) will be stretched 5 times longer, while vectors aligned with \( \mathbf{v}_2 \) will remain unchanged in length.
Such transformations help in numerous practical applications, from computer graphics to engineering design, where understanding how an object scales in different dimensions is critical.