Chapter 5: Problem 9
Orthogonally diagonalize the matrices by finding an orthogonal matrix \(Q\) and a diagonal matrix \(D\) such that \(Q^{T} A Q=D\) $$A=\left[\begin{array}{llll} 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 \end{array}\right]$$
Short Answer
Expert verified
The orthogonal matrix is \( Q \) and diagonal matrix is \( D \).
Step by step solution
01
Define the Matrix A
The given matrix is \( A = \begin{bmatrix} 1 & 1 & 0 & 0 \ 1 & 1 & 0 & 0 \ 0 & 0 & 1 & 1 \ 0 & 0 & 1 & 1 \end{bmatrix} \). We need to find an orthogonal matrix \( Q \) and a diagonal matrix \( D \) such that \( Q^{T} A Q = D \).
02
Find the Eigenvalues of A
To find the eigenvalues, solve the characteristic equation \( \det(A - \lambda I) = 0 \). For this matrix, the eigenvalues are \( \lambda_1 = 2 \), \( \lambda_2 = 0 \), and \( \lambda_3 = 0 \) with corresponding multiplicities.
03
Find the Eigenvectors of A
For each eigenvalue, solve \( (A - \lambda I)\mathbf{v} = 0 \) to find the eigenvectors. For \( \lambda_1 = 2 \), the eigenvectors are \( \begin{bmatrix} 1 \ 1 \ 0 \ 0 \end{bmatrix} \) and \( \begin{bmatrix} 0 \ 0 \ 1 \ 1 \end{bmatrix} \). For \( \lambda_2 = 0 \), a basis is \( \begin{bmatrix} 1 \ -1 \ 0 \ 0 \end{bmatrix} \). For \( \lambda_3 = 0 \), a basis is \( \begin{bmatrix} 0 \ 0 \ 1 \ -1 \end{bmatrix} \).
04
Form the Orthogonal Matrix Q
The columns of \( Q \) are the normalized eigenvectors. Calculate each eigenvector's norm and normalize it. Arrange these normalized eigenvectors in matrix form to get \( Q \). Here, they are \( \begin{bmatrix} 1/\sqrt{2} & 1/\sqrt{2} & 0 & 0 \ -1/\sqrt{2} & 1/\sqrt{2} & 0 & 0 \ 0 & 0 & 1/\sqrt{2} & 1/\sqrt{2} \ 0 & 0 & -1/\sqrt{2} & 1/\sqrt{2} \end{bmatrix} \).
05
Construct the Diagonal Matrix D
The diagonal matrix \( D \) is formed by placing the eigenvalues along the diagonal in the order corresponding to the order of eigenvectors in \( Q \). Thus, \( D = \begin{bmatrix} 2 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 2 \end{bmatrix} \).
06
Verify the Result
Check if \( Q^{T} A Q = D \). Calculate \( Q^{T} \), multiply \( Q^{T} A \), and then multiply the result by \( Q \). The resulting product should be \( D \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Eigenvalues
In the process of orthogonally diagonalizing a matrix, finding the eigenvalues is one of the initial steps. Eigenvalues are special numbers associated with a matrix. They describe certain linear transformations applied to the matrix that keep a vector's direction unchanged while potentially changing its magnitude.
- For a matrix \( A \), eigenvalues are found by solving the characteristic equation \( \det(A - \lambda I) = 0 \). Here, \( I \) denotes the identity matrix and \( \lambda \) represents the eigenvalue.
- In our example, the matrix has eigenvalues of \( \lambda_1 = 2 \), \( \lambda_2 = 0 \), and \( \lambda_3 = 0 \). These eigenvalues tell us that the directions associated with them are scaled by factors of 2, and in some cases, they do not change at all (0 represents a direction that doesn't change). Understanding eigenvalues is crucial as they lay the foundation for finding eigenvectors and constructing matrices like \( Q \) and \( D \).
- For a matrix \( A \), eigenvalues are found by solving the characteristic equation \( \det(A - \lambda I) = 0 \). Here, \( I \) denotes the identity matrix and \( \lambda \) represents the eigenvalue.
- In our example, the matrix has eigenvalues of \( \lambda_1 = 2 \), \( \lambda_2 = 0 \), and \( \lambda_3 = 0 \). These eigenvalues tell us that the directions associated with them are scaled by factors of 2, and in some cases, they do not change at all (0 represents a direction that doesn't change). Understanding eigenvalues is crucial as they lay the foundation for finding eigenvectors and constructing matrices like \( Q \) and \( D \).
Eigenvectors
Eigenvectors are essential when working with matrices as they provide directions that remain unchanged by certain matrix transformations, aside from scaling by their eigenvalues.
- To find an eigenvector of a matrix \( A \), solve \( (A - \lambda I) \mathbf{v} = 0 \) for each eigenvalue \( \lambda \). This equation helps identify vectors \( \mathbf{v} \) that correspond to the scaling represented by an eigenvalue.
- In our exercise, for \( \lambda_1 = 2 \), the eigenvectors are \( \begin{bmatrix} 1 \ 1 \ 0 \ 0 \end{bmatrix} \) and \( \begin{bmatrix} 0 \ 0 \ 1 \ 1 \end{bmatrix} \). For \( \lambda_2 = 0 \) and \( \lambda_3 = 0 \), the basis eigenvectors are \( \begin{bmatrix} 1 \ -1 \ 0 \ 0 \end{bmatrix} \) and \( \begin{bmatrix} 0 \ 0 \ 1 \ -1 \end{bmatrix} \), respectively.
- Each eigenvector provides a unique "perspective" on the transformation described by the matrix. They are pivotal for forming the orthogonal matrix \( Q \), as they need to be normalized and arranged into \( Q \).
- To find an eigenvector of a matrix \( A \), solve \( (A - \lambda I) \mathbf{v} = 0 \) for each eigenvalue \( \lambda \). This equation helps identify vectors \( \mathbf{v} \) that correspond to the scaling represented by an eigenvalue.
- In our exercise, for \( \lambda_1 = 2 \), the eigenvectors are \( \begin{bmatrix} 1 \ 1 \ 0 \ 0 \end{bmatrix} \) and \( \begin{bmatrix} 0 \ 0 \ 1 \ 1 \end{bmatrix} \). For \( \lambda_2 = 0 \) and \( \lambda_3 = 0 \), the basis eigenvectors are \( \begin{bmatrix} 1 \ -1 \ 0 \ 0 \end{bmatrix} \) and \( \begin{bmatrix} 0 \ 0 \ 1 \ -1 \end{bmatrix} \), respectively.
- Each eigenvector provides a unique "perspective" on the transformation described by the matrix. They are pivotal for forming the orthogonal matrix \( Q \), as they need to be normalized and arranged into \( Q \).
Orthogonal Matrix
An orthogonal matrix is a square matrix whose columns and rows are orthonormal vectors, meaning each vector is perpendicular to the others and all are unit vectors.
- Orthogonality ensures that when you multiply the matrix by its transpose, the result is the identity matrix. In terms of the mathematical operation, \( Q^{T} Q = I \) should hold where \( I \) is the identity matrix.
- To form the orthogonal matrix \( Q \) in the diagonalization process, the columns are filled with normalized eigenvectors from the matrix in focus. Normalization is a process where the length (or norm) of the vector is adjusted to be one.
- For the provided matrix, the eigenvectors are normalized as follows: dividing each component of the vectors by their respective norms, resulting in \( Q = \begin{bmatrix} 1/\sqrt{2} & 1/\sqrt{2} & 0 & 0 \ -1/\sqrt{2} & 1/\sqrt{2} & 0 & 0 \ 0 & 0 & 1/\sqrt{2} & 1/\sqrt{2} \ 0 & 0 & -1/\sqrt{2} & 1/\sqrt{2} \end{bmatrix} \). This matrix helps transform the original matrix into a diagonal matrix \( D \) when used appropriately.
- Orthogonality ensures that when you multiply the matrix by its transpose, the result is the identity matrix. In terms of the mathematical operation, \( Q^{T} Q = I \) should hold where \( I \) is the identity matrix.
- To form the orthogonal matrix \( Q \) in the diagonalization process, the columns are filled with normalized eigenvectors from the matrix in focus. Normalization is a process where the length (or norm) of the vector is adjusted to be one.
- For the provided matrix, the eigenvectors are normalized as follows: dividing each component of the vectors by their respective norms, resulting in \( Q = \begin{bmatrix} 1/\sqrt{2} & 1/\sqrt{2} & 0 & 0 \ -1/\sqrt{2} & 1/\sqrt{2} & 0 & 0 \ 0 & 0 & 1/\sqrt{2} & 1/\sqrt{2} \ 0 & 0 & -1/\sqrt{2} & 1/\sqrt{2} \end{bmatrix} \). This matrix helps transform the original matrix into a diagonal matrix \( D \) when used appropriately.
Diagonal Matrix
A diagonal matrix is a matrix in which the entries outside the main diagonal are all zero. Only the diagonal elements may be non-zero.
- Diagonal matrices are particularly valuable because they simplify the operations like matrix multiplication and finding powers of a matrix.
- In orthogonal diagonalization, the matrix \( D \) is the resulting diagonal matrix after transforming the original matrix \( A \) with the orthogonal matrix \( Q \). The diagonal of \( D \) consists of the eigenvalues of \( A \).
- For the example matrix, adjoined to the eigenvectors, the diagonal matrix is defined as \( D = \begin{bmatrix} 2 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 2 \end{bmatrix} \). Transforming \( A \) into \( D \) using \( Q \) is a cornerstone of understanding how matrices can be simplified and manipulated through orthogonal diagonalization.
- Diagonal matrices are particularly valuable because they simplify the operations like matrix multiplication and finding powers of a matrix.
- In orthogonal diagonalization, the matrix \( D \) is the resulting diagonal matrix after transforming the original matrix \( A \) with the orthogonal matrix \( Q \). The diagonal of \( D \) consists of the eigenvalues of \( A \).
- For the example matrix, adjoined to the eigenvectors, the diagonal matrix is defined as \( D = \begin{bmatrix} 2 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 2 \end{bmatrix} \). Transforming \( A \) into \( D \) using \( Q \) is a cornerstone of understanding how matrices can be simplified and manipulated through orthogonal diagonalization.