Chapter 8: Problem 24
In Problems 21-30, the given matrix \(\mathbf{A}\) is symmetric. Find an orthogonal matrix \(\mathbf{P}\) that diagonalizes \(\mathbf{A}\) and the diagonal matrix D such that \(\mathbf{D}=\mathbf{P}^{T} \mathbf{A P}\). $$ \left(\begin{array}{rr} 1 & -2 \\ -2 & 1 \end{array}\right) $$
Short Answer
Expert verified
Orthogonal matrix \( \mathbf{P} \) and diagonal matrix \( \mathbf{D} \) are found using eigenvectors and eigenvalues of the symmetric matrix.
Step by step solution
01
Verify Symmetry of Matrix A
Given matrix \( \mathbf{A} \) is \( \begin{pmatrix} 1 & -2 \ -2 & 1 \end{pmatrix} \). A matrix is symmetric if \( \mathbf{A} = \mathbf{A}^T \). Check the transpose: \( \begin{pmatrix} 1 & -2 \ -2 & 1 \end{pmatrix}^T = \begin{pmatrix} 1 & -2 \ -2 & 1 \end{pmatrix} \), which is equal to \( \mathbf{A} \). Thus, \( \mathbf{A} \) is symmetric.
02
Find Eigenvalues of A
The eigenvalues \( \lambda \) satisfy the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). Substitute \( \mathbf{A} \) and \( \mathbf{I} \): \[ \begin{vmatrix} 1 - \lambda & -2 \ -2 & 1 - \lambda \end{vmatrix} = 0 \]. Solving the determinant leads to \( (1-\lambda)^2 - 4 = 0 \). Simplifying gives \( \lambda^2 - 2\lambda - 3 = 0 \), which factors to \( (\lambda - 3)(\lambda + 1) = 0 \). Hence, eigenvalues are \( \lambda_1 = 3 \) and \( \lambda_2 = -1 \).
03
Find Eigenvectors for Eigenvalue 3
Calculate eigenvectors for \( \lambda_1 = 3 \): solve \( (\mathbf{A} - 3\mathbf{I})\mathbf{v} = 0 \). Substituting \( \lambda_1 \), \( \mathbf{A} - 3\mathbf{I} = \begin{pmatrix} -2 & -2 \ -2 & -2 \end{pmatrix} \). The vector \( \mathbf{v} = \begin{pmatrix} x \ y \end{pmatrix} \) satisfies \(-2x - 2y = 0 \). Solving gives \( x = -y \), thus one eigenvector is \( \mathbf{v}_1 = \begin{pmatrix} 1 \ -1 \end{pmatrix} \).
04
Find Eigenvectors for Eigenvalue -1
Calculate eigenvectors for \( \lambda_2 = -1 \): solve \( (\mathbf{A} + \mathbf{I})\mathbf{v} = 0 \). Substituting \( \lambda_2 \), \( \mathbf{A} + \mathbf{I} = \begin{pmatrix} 2 & -2 \ -2 & 2 \end{pmatrix} \). The vector \( \mathbf{v} = \begin{pmatrix} x \ y \end{pmatrix} \) satisfies \(2x - 2y = 0 \). Solving gives \( x = y \), so one eigenvector is \( \mathbf{v}_2 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \).
05
Normalize Eigenvectors
Normalize \( \mathbf{v}_1 = \begin{pmatrix} 1 \ -1 \end{pmatrix} \) to \( \mathbf{u}_1 = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} \). Normalize \( \mathbf{v}_2 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \) to \( \mathbf{u}_2 = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \).
06
Form the Orthogonal Matrix P
The orthogonal matrix \( \mathbf{P} \) is formed by the normalized eigenvectors as columns: \( \mathbf{P} = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \). Verify its orthogonality: \( \mathbf{P}^T \mathbf{P} = \mathbf{I} \).
07
Form the Diagonal Matrix D
The diagonal matrix \( \mathbf{D} \) contains eigenvalues along its diagonal: \( \mathbf{D} = \begin{pmatrix} 3 & 0 \ 0 & -1 \end{pmatrix} \).
08
Verify Diagonalization
Check that \( \mathbf{D} = \mathbf{P}^T \mathbf{A} \mathbf{P} \). Compute the product to ensure that the diagonalization holds true, confirming \( \mathbf{P}^T \mathbf{A} \mathbf{P} = \mathbf{D} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Symmetric Matrix
A symmetric matrix is a square matrix that is equal to its transpose. This means that the elements are mirrored across the main diagonal. In mathematical terms, for a matrix \( \mathbf{A} \), it is symmetric if \( \mathbf{A} = \mathbf{A}^T \). For example, consider the matrix
- \( \begin{pmatrix} 1 & -2 \ -2 & 1 \end{pmatrix} \)
- \( \begin{pmatrix} 1 & -2 \ -2 & 1 \end{pmatrix}^T \)
Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors are fundamental concepts in linear algebra. For a given matrix \( \mathbf{A} \), an eigenvector is a non-zero vector that, when multiplied by \( \mathbf{A} \), results in a vector that is a scalar multiple of itself. Mathematically, this is expressed as
To find the eigenvalues, we solve the characteristic equation
- \( \mathbf{A} \mathbf{v} = \lambda \mathbf{v} \)
To find the eigenvalues, we solve the characteristic equation
- \( \, \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \, \)
- \( \begin{pmatrix} 1 & -2 \ -2 & 1 \end{pmatrix} \)
Matrix Diagonalization
Matrix diagonalization is a process of transforming a given matrix into a diagonal form. For a symmetric matrix like
- \( \begin{pmatrix} 1 & -2 \ -2 & 1 \end{pmatrix} \)
- \( \mathbf{D} = \mathbf{P}^T \mathbf{A} \mathbf{P} \)
Normalization of Vectors
Normalization is the process of converting a vector to a unit vector, which has a magnitude of one. This is done by dividing each component of the vector by its magnitude. The magnitude of a vector \( \mathbf{v} \) is given by
- \( \| \mathbf{v} \| = \sqrt{x^2 + y^2} \)
- \( \begin{pmatrix} 1 \ -1 \end{pmatrix} \)
- \( \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} \).