Chapter 7: Problem 23
Let \(A=\left[\begin{array}{rrr}{4} & {-1} & {-1} \\ {-1} & {4} & {-1} \\\ {-1} & {-1} & {4}\end{array}\right]\) and \(\mathbf{v}=\left[\begin{array}{l}{1} \\\ {1} \\ {1}\end{array}\right] .\) Verify that 5 is an eigenvalue of \(A\) and \(\mathbf{v}\) is an eigenvector. Then orthogonally diagonalize \(A .\)
Short Answer
Step by step solution
Verify the Eigenvalue-Eigenvector Relationship
Setup the Characteristic Equation of A
Compute the Determinant
Solve the Polynomial Equation
Find the Corresponding Eigenvectors
Construct and Diagonalize the Matrix
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Matrix Diagonalization
The main steps involved in diagonalizing a matrix include finding its eigenvalues and corresponding eigenvectors. Once these are known, we can form an orthogonal matrix \( P \) where each column is a normalized eigenvector. Then calculate \( P^{-1}AP \), which results in a diagonal matrix \( D \). This matrix \( D \) will have its eigenvalues on the diagonal. This process not only simplifies computations but provides valuable insights into the geometric structure of the matrix.
Orthogonal Matrices
Orthogonal matrices are significant because they preserve lengths and angles, making them extremely useful for transformations that require maintaining these characteristics. Within the context of matrix diagonalization, orthogonal matrices are used to form matrix \( P \), which comes from the eigenvectors of \( A \). Each column is an eigenvector and must be orthogonal to the others. If \( P \) is orthogonal, it simplifies the calculation of its inverse as \( P^{-1} = P^{T} \).
In our original exercise, the matrix \( P \) containing the normalized eigenvectors is used to orthogonally diagonalize matrix \( A \). Thus, orthogonal matrices play a key role in achieving a diagonal matrix from a given square matrix through diagonalization.
Characteristic Equation
In our example, we formed the characteristic equation \( |A - \lambda I| = 0 \), leading to a polynomial in terms of \( \lambda \): \( \lambda^3 - 12\lambda^2 + 33\lambda - 27 = 0 \). Solving this polynomial results in the eigenvalues of \( A \). These eigenvalues are crucial as they determine the fundamental properties of the matrix, such as stability and can also relate to certain geometric transformations matrix \( A \) may describe.
Understanding the characteristic equation is key to manipulating matrices and extends far beyond simple computations, providing insights into the behavior of systems modeled by these matrices.
Factor Theorem
In the context of our original exercise, we used the Factor Theorem to simplify the characteristic equation once an eigenvalue, such as 5, was known. By substituting \( \lambda = 5 \) into the polynomial \( \lambda^3 - 12\lambda^2 + 33\lambda - 27 = 0 \) and finding that it yields zero confirms that \( (\lambda - 5) \) is a factor. Thus, the characteristic equation can be factored as \( (\lambda - 5)(\lambda^2 - 7\lambda + 9) = 0 \).
This factored form makes finding the other roots (eigenvalues) much easier. The Factor Theorem simplifies the process of solving polynomial equations and is very useful, especially when dealing with higher-degree polynomials in eigenvalue problems, turning a complex problem into a manageable one by breaking it into simpler parts.