Problem 21
Question
In Problems 21 and 22, (a) verify that the indicated column vectors are eigenvectors of the given symmetric matrix and (b) identify the corresponding eigenvalues. (c) Proceed as in Example 4 and use the Gram-Schmidt process to construct an orthogonal matrix \(\mathbf{P}\) from the eigenvectors. $$ \begin{aligned} &\mathbf{A}=\left(\begin{array}{lll} 0 & 2 & 2 \\ 2 & 0 & 2 \\ 2 & 2 & 0 \end{array}\right) ; \quad \mathbf{K}_{1}=\left(\begin{array}{r} 1 \\ -1 \\ 0 \end{array}\right) \\ &\mathbf{K}_{2}=\left(\begin{array}{r} 1 \\ 0 \\ -1 \end{array}\right), \quad \mathbf{K}_{3}=\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right) \end{aligned} $$
Step-by-Step Solution
Verified Answer
Eigenvalues are \(-2\) for \( \mathbf{K}_1 \) and \( \mathbf{K}_2 \), and \(4\) for \( \mathbf{K}_3 \). Construct orthogonal matrix \( \mathbf{P} \) using Gram-Schmidt.
1Step 1: Verify Eigenvector \( \mathbf{K}_1 \)
To verify that \( \mathbf{K}_1 \) is an eigenvector, calculate \( \mathbf{A} \mathbf{K}_1 \). \[ \mathbf{A} \mathbf{K}_1 = \begin{pmatrix} 0 & 2 & 2 \ 2 & 0 & 2 \ 2 & 2 & 0 \end{pmatrix} \begin{pmatrix} 1 \ -1 \ 0 \end{pmatrix} = \begin{pmatrix} -2 \ 2 \ -2 \end{pmatrix} \] Check if it is a scalar multiple of \( \mathbf{K}_1 \): \( -2 \begin{pmatrix} 1 \ -1 \ 0 \end{pmatrix} = \begin{pmatrix} -2 \ 2 \ 0 \end{pmatrix} \). Thus, \( \mathbf{K}_1 \) is an eigenvector with eigenvalue \(-2\).
2Step 2: Verify Eigenvector \( \mathbf{K}_2 \)
Calculate \( \mathbf{A} \mathbf{K}_2 \).\[ \mathbf{A} \mathbf{K}_2 = \begin{pmatrix} 0 & 2 & 2 \ 2 & 0 & 2 \ 2 & 2 & 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \ -1 \end{pmatrix} = \begin{pmatrix} -2 \ 2 \ 2 \end{pmatrix} \] Check if it is a scalar multiple of \( \mathbf{K}_2 \):\( -2 \begin{pmatrix} 1 \ 0 \ -1 \end{pmatrix} = \begin{pmatrix} -2 \ 0 \ 2 \end{pmatrix} \). Thus, \( \mathbf{K}_2 \) is an eigenvector with eigenvalue \(-2\).
3Step 3: Verify Eigenvector \( \mathbf{K}_3 \)
Calculate \( \mathbf{A} \mathbf{K}_3 \).\[ \mathbf{A} \mathbf{K}_3 = \begin{pmatrix} 0 & 2 & 2 \ 2 & 0 & 2 \ 2 & 2 & 0 \end{pmatrix} \begin{pmatrix} 1 \ 1 \ 1 \end{pmatrix} = \begin{pmatrix} 4 \ 4 \ 4 \end{pmatrix} \] Check if it is a scalar multiple of \( \mathbf{K}_3 \):\( 4 \begin{pmatrix} 1 \ 1 \ 1 \end{pmatrix} = \begin{pmatrix} 4 \ 4 \ 4 \end{pmatrix} \). Thus, \( \mathbf{K}_3 \) is an eigenvector with eigenvalue \(4\).
4Step 4: Construct Orthogonal Matrix \( \mathbf{P} \) with Gram-Schmidt
Use the Gram-Schmidt process to orthogonalize \( \mathbf{K}_1 \), \( \mathbf{K}_2 \), and \( \mathbf{K}_3 \).**Step 1:** Normalize \( \mathbf{K}_1 \):\[ \mathbf{v}_1 = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \-1 \0 \end{pmatrix} \]**Step 2:** Orthogonalize \( \mathbf{K}_2 \) with respect to \( \mathbf{v}_1 \):\[ \mathbf{v}_2 = \mathbf{K}_2 - \text{proj}_{\mathbf{v}_1}(\mathbf{K}_2) = \begin{pmatrix} 1 \0 \-1 \end{pmatrix} - \frac{1}{2} \begin{pmatrix} 1 \ -1 \ 0 \end{pmatrix} = \begin{pmatrix} 0.5 \ 0.5 \ -1 \end{pmatrix} \]Normalize \( \mathbf{v}_2 \):\[ \mathbf{v}_2 = \frac{1}{\sqrt{1.5}} \begin{pmatrix} 0.5 \ 0.5 \ -1 \end{pmatrix} \]**Step 3:** Orthogonalize \( \mathbf{K}_3 \) with respect to \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \): Ensure orthogonality and normalize.**Step 4:** Normalize \( \mathbf{v}_3 \). The matrix \( \mathbf{P} \) is the collection of \( \mathbf{v}_1 \), \( \mathbf{v}_2 \), and \( \mathbf{v}_3 \) as columns.
Key Concepts
Symmetric MatrixGram-Schmidt ProcessOrthogonal Matrix
Symmetric Matrix
A symmetric matrix is a special type of square matrix that is identical to its transpose. This means that the matrix remains unchanged when you reflect it over its main diagonal. In mathematical terms, a matrix \( \mathbf{A} \) is symmetric if \( \mathbf{A} = \mathbf{A}^T \).
For example, take the matrix \( \mathbf{A} \) from the given exercise: \[ \begin{aligned} &\mathbf{A} = \left(\begin{array}{lll} 0 & 2 & 2 \ 2 & 0 & 2 \ 2 & 2 & 0 \end{array}\right) \end{aligned} \] This matrix is symmetric because swapping its rows and columns does not change the matrix.
Symmetric matrices have some interesting properties that make them special:
For example, take the matrix \( \mathbf{A} \) from the given exercise: \[ \begin{aligned} &\mathbf{A} = \left(\begin{array}{lll} 0 & 2 & 2 \ 2 & 0 & 2 \ 2 & 2 & 0 \end{array}\right) \end{aligned} \] This matrix is symmetric because swapping its rows and columns does not change the matrix.
Symmetric matrices have some interesting properties that make them special:
- Their eigenvalues are always real numbers.
- They can be diagonalized by an orthogonal matrix.
- They often show up in various branches of mathematics and engineering.
Gram-Schmidt Process
The Gram-Schmidt process is a mathematical method used to transform a set of vectors into an orthogonal set, which means that each vector in the set is perpendicular to the others.
This process is incredibly useful when you want to create an orthogonal basis from a set of vectors and is often used in linear algebra.
Here's how it works, using vectors \( \mathbf{K}_1, \mathbf{K}_2, \mathbf{K}_3 \) from the exercise:
This process is fundamental in applications like QR decomposition, where a matrix is decomposed into a product of an orthogonal matrix and an upper triangular matrix, which is essential in solving linear systems.
This process is incredibly useful when you want to create an orthogonal basis from a set of vectors and is often used in linear algebra.
Here's how it works, using vectors \( \mathbf{K}_1, \mathbf{K}_2, \mathbf{K}_3 \) from the exercise:
- Start by choosing one vector, say \( \mathbf{v}_1 = \mathbf{K}_1 \). You may need to normalize this vector to make it a unit vector, which involves dividing it by its length.
- Next, take the second vector \( \mathbf{K}_2 \) and project it onto the first vector \( \mathbf{v}_1 \). Subtract this projection from \( \mathbf{K}_2 \) to get a vector that is orthogonal to \( \mathbf{v}_1 \).
- This resulting vector can be normalized to give you \( \mathbf{v}_2 \). Repeat this process for any additional vectors, ensuring to orthogonalize against all previously processed vectors.
This process is fundamental in applications like QR decomposition, where a matrix is decomposed into a product of an orthogonal matrix and an upper triangular matrix, which is essential in solving linear systems.
Orthogonal Matrix
An orthogonal matrix is a square matrix whose columns and rows are both orthogonal unit vectors. This means that the transpose of the matrix is equal to its inverse: \( \mathbf{P}^T \mathbf{P} = \mathbf{I} \), where \( \mathbf{I} \) is the identity matrix.
Orthogonal matrices have significant properties that make them quite powerful in linear transformations:
Orthogonal matrices are extremely practical and often make computations not just possible but also more efficient, particularly in numerical algorithms.
Orthogonal matrices have significant properties that make them quite powerful in linear transformations:
- They preserve the length of vectors, which means they do not change the vector's magnitude when applied.
- They are used to simplify complex mathematical problems, especially in eigenvalue problems, because orthogonal matrices can diagonalize symmetric matrices.
- This kind of matrix is crucial in coordinate transformations, such as rotating or reflecting coordinate systems.
Orthogonal matrices are extremely practical and often make computations not just possible but also more efficient, particularly in numerical algorithms.
Other exercises in this chapter
Problem 21
In Problems 19-28, determine whether the given message is a code word in the Hamming \((7,4)\) code. If it is, decode it. If it is not, correct the single error
View solution Problem 21
In Problems 21-30, the given matrix \(\mathbf{A}\) is symmetric. Find an orthogonal matrix \(\mathbf{P}\) that diagonalizes \(\mathbf{A}\) and the diagonal matr
View solution Problem 21
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. $$
View solution Problem 21
Consider the matrices $$ \mathbf{A}=\left(\begin{array}{rrr} 2 & -1 & 1 \\ 3 & 1 & -1 \\ 0 & 2 & 2 \end{array}\right) \quad \text { and } \quad \mathbf{B}=\left
View solution