Problem 18
Question
In Problems 11-18, proceed as in Example 3 to construct an orthogonal matrix from the eigenvectors of the given symmetric matrix. (The answers are not unique.) $$ \left(\begin{array}{rrr} 2 & 8 & -2 \\ 8 & -4 & 10 \\ -2 & 10 & -7 \end{array}\right) $$
Step-by-Step Solution
Verified Answer
Find eigenvalues and eigenvectors, normalize eigenvectors, and form the orthogonal matrix using these vectors.
1Step 1: Find the Eigenvalues
For a symmetric matrix \(A\), the eigenvalues can be found by solving the characteristic polynomial \( \det(A - \lambda I) = 0 \). Here, \(A\) is \(\begin{pmatrix} 2 & 8 & -2 \ 8 & -4 & 10 \ -2 & 10 & -7 \end{pmatrix}\) and \(I\) is the identity matrix. Calculate the determinant of \(A - \lambda I\): \[ \det \begin{pmatrix} 2-\lambda & 8 & -2 \ 8 & -4-\lambda & 10 \ -2 & 10 & -7-\lambda \end{pmatrix} = 0 \] Expand the determinant to find the characteristic polynomial and solve for \(\lambda\).
2Step 2: Solve the Characteristic Polynomial
After expanding the determinant, you get a cubic polynomial equation in terms of \(\lambda\). Find the roots of this polynomial which are the eigenvalues of the matrix. Typically in these exercises, they are integers or simple fractions.
3Step 3: Find Corresponding Eigenvectors
For each eigenvalue \(\lambda_i\) found in the previous step, find the eigenvectors by solving the equation \((A - \lambda_i I)\mathbf{x} = \mathbf{0}\) for \(\mathbf{x}\). This involves row-reducing the matrix \(A - \lambda_i I\) and finding the null space.
4Step 4: Normalize the Eigenvectors
Ensure each eigenvector has unit length by dividing each vector by its norm. The norm of a vector \(\mathbf{v}\) is \(\|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2}\). Use this to scale each vector to unit length.
5Step 5: Construct the Orthogonal Matrix
Combine the normalized eigenvectors as columns to form the orthogonal matrix. Since the eigenvectors are orthogonal and normalized, this matrix will be an orthogonal matrix, meaning its transpose equals its inverse, \( Q^T = Q^{-1} \).
6Step 6: Verify Orthogonality
Confirm the matrix is orthogonal by checking that \(Q^T Q = I\). This ensures the correctness of the constructed orthogonal matrix from the normalized eigenvectors.
Key Concepts
EigenvectorsSymmetric MatrixEigenvaluesCharacteristic Polynomial
Eigenvectors
When dealing with matrices, you might come across the concept of eigenvectors. An eigenvector is a nonzero vector that changes at most by a scalar factor when a linear transformation is applied to it. In simpler terms, if we have a matrix \(A\) and a vector \(\mathbf{x}\), then \(\mathbf{x}\) is an eigenvector of \(A\) if it satisfies the equation \(A\mathbf{x} = \lambda \mathbf{x}\). Here, \(\lambda\) is what we call an eigenvalue, which we'll talk about in more detail later.
- Eigenvectors give us valuable insights into the properties of the transformation represented by the matrix.
- Finding eigenvectors helps in simplifying matrix computations, such as matrix diagonaliation.
- Eigenvectors are part of constructing orthogonal matrices, which have numerous applications.
Symmetric Matrix
A symmetric matrix is a special kind of square matrix. It is characterized by its symmetry about its main diagonal, meaning that it mirrors itself. Mathematically speaking, a matrix \(A\) is symmetric if \(A = A^T\) (where \(A^T\) is the transpose of \(A\)).
- One of the most interesting properties of symmetric matrices is that all of their eigenvalues are real numbers.
- When a matrix is symmetric, you can be sure there is a complete set of orthogonal eigenvectors.
- This leads to easier computations particularly in constructing orthogonal matrices, important in various branches of mathematics and physics.
Eigenvalues
Eigenvalues are intrinsic to understanding so many phenomena in mathematics. An eigenvalue associated with a square matrix \(A\) is a scalar, \(\lambda\), such that when it multiplies the identity matrix \(I\), the matrix \(A - \lambda I\) becomes singular. This means its determinant is zero:
\[ \det(A - \lambda I) = 0 \]
\[ \det(A - \lambda I) = 0 \]
- This equation is known as the characteristic equation, and solving it involves finding the roots of the resulting polynomial.
- For symmetric matrices, all eigenvalues are guaranteed to be real numbers, making them easier to handle computationally.
- The roots provide the eigenvalues needed for finding eigenvectors and constructing orthogonal matrices.
Characteristic Polynomial
Characteristic polynomials are central to finding a matrix's eigenvalues. When you subtract \(\lambda I\) (where \(I\) is the identity matrix) from your original matrix \(A\), you obtain a new matrix \((A - \lambda I)\), whose determinant expands into a polynomial in terms of \(\lambda\). This polynomial, \(p(\lambda) = \det(A - \lambda I)\), is what we call the characteristic polynomial.
- The degree of the characteristic polynomial equals the size of the matrix you started with.
- Solving \(p(\lambda) = 0\) gives you the eigenvalues of \(A\).
- For a 3x3 matrix, such as the one in the exercise, \(p(\lambda)\) will be a cubic polynomial.
Other exercises in this chapter
Problem 18
In Problems 1-20, fill in the blanks or answer true/false. The only matrices that are orthogonally diagonalizable are symmetric matrices._________
View solution Problem 18
In Problems 1-20, determine whether the given matrix \(\mathbf{A}\) is diagonalizable. If so, find the matrix \(\mathbf{P}\) that diagonalizes \(\mathbf{A}\) an
View solution Problem 18
In Problems 7-22, find the eigenvalues and eigenvectors of the given matrix. Using Theorem 8.8.2 or (6), state whether the matrix is singular or nonsingular. $$
View solution Problem 18
In Problems 15-18, evaluate the determinant of the given matrix without expanding by cofactors. $$ \mathbf{D}=\left(\begin{array}{rrr} 0 & 7 & 0 \\ 4 & 0 & 0 \\
View solution