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Chapter 5: Problem 7

Orthogonally diagonalize the matrices by finding an orthogonal matrix \(Q\) and a diagonal matrix \(D\) such that \(Q^{T} A Q=D\) $$A=\left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right]$$

Short Answer

Expert verified
Orthogonally diagonalized matrix: \( Q = \begin{bmatrix} \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ 0 & 1 & 0 \\ \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \end{bmatrix} \), \( D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \).

Step by step solution

01

Verify Symmetry

First, check if the matrix \( A \) is symmetric. A matrix is symmetric if it is equal to its transpose. We calculate the transpose of \( A \) and compare it with \( A \):\[ A^T = \begin{bmatrix} 1 & 0 & -1 \ 0 & 1 & 0 \ -1 & 0 & 1 \end{bmatrix} \]Since \( A = A^T \), the matrix is symmetric and can be diagonalized orthogonally.
02

Find Eigenvalues

Next, find the eigenvalues of \( A \) by solving the characteristic equation: \[ \text{det}(A - \lambda I) = 0 \]Set up the equation:\[\begin{vmatrix}1 - \lambda & 0 & -1 \0 & 1 - \lambda & 0 \-1 & 0 & 1 - \lambda \\end{vmatrix} = 0\]Simplifying, we get:\[ (1 - \lambda)^2((1 - \lambda)^2 - 1) = 0 \]Expanding and factoring gives eigenvalues \( \lambda = 0, 0, 2 \).
03

Find Eigenvectors

For each eigenvalue, find the corresponding eigenvectors:1. **Eigenvalue \( \lambda = 0 \):**\( A - 0I = A \)Solve the system:\[ \begin{cases}x - z = 0 \y = 0 \-x + z = 0\end{cases} \]This gives eigenvectors: \( \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} \) and any scalar multiple.2. **Eigenvalue \( \lambda = 2 \):**\( A - 2I = \begin{bmatrix} -1 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & -1 \end{bmatrix} \)Solving gives the eigenvector: \( \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix} \).
04

Normalize Eigenvectors

Normalize the eigenvectors to ensure that \( Q \) is an orthogonal matrix:- For the eigenvector \( \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} \), the norm is \( \sqrt{2} \), so the normalized form is \( \begin{bmatrix} \frac{1}{\sqrt{2}} \ 0 \ \frac{1}{\sqrt{2}} \end{bmatrix} \).- For the eigenvector \( \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix} \), it is already normalized.Thus, the matrix \( Q \) is:\[ Q = \begin{bmatrix} \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \ 0 & 1 & 0 \ \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \end{bmatrix} \]
05

Form Diagonal Matrix D

Arrange the eigenvalues along the diagonal of an empty matrix to form \( D \):\[ D = \begin{bmatrix} 2 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix} \]
06

Verify Orthogonal Diagonalization

Confirm that the matrix \( Q^T A Q = D \): \[ Q^T = \begin{bmatrix} \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \ 0 & 1 & 0 \ \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \end{bmatrix}^T \]Calculate \( Q^T A Q \) to check if it equals \( D \). Simplifying will confirm that the procedure is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Orthogonal Matrix
An orthogonal matrix is a square matrix whose columns and rows are orthogonal unit vectors. This means that the columns (or rows) form an orthonormal basis. Orthonormal means each vector has a length of one and is perpendicular to the others. A key property of orthogonal matrices is that multiplying them by their transpose results in the identity matrix. Mathematically, we have:
  • If matrix \( Q \) is orthogonal, then \( Q^T Q = QQ^T = I \), where \( I \) is the identity matrix.
  • Orthogonal matrices are especially useful in preserving the dot product, which maintains vector angles and vector lengths.
  • In the context of orthogonal diagonalization, matrix \( Q \) is constructed from the normalized eigenvectors of a symmetric matrix.
Understanding how orthogonal matrices function helps in preserving numerical stability during matrix transformations, making it a crucial concept in computation and robotics.
Eigenvalues
Eigenvalues are scalars associated with a linear system of equations or a matrix, often arising in the characteristic equation \( \text{det}(A - \lambda I) = 0 \). An eigenvalue \( \lambda \) of a matrix \( A \) enables the matrix to transform a corresponding eigenvector by simply scaling it with that eigenvalue. Here's what's important about eigenvalues:
  • They provide insights into the matrix's properties, such as whether it is invertible.
  • Eigenvalues are fundamental in various applications like stability analysis and vibrations analysis in engineering.
  • For symmetric matrices, all eigenvalues are real numbers, making computations straightforward compared to general matrices.
In our current exercise, the eigenvalues \( \lambda = 0, 0, 2 \) indicate the directions in which our initial matrix has particular stretching or compressing properties.
Eigenvectors
Eigenvectors are vectors associated with a matrix, which, when transformed by the matrix, result only in a scalar multiplication (and not a direction change). For an eigenvalue \( \lambda \), a corresponding eigenvector \( \mathbf{v} \) satisfies \( A \mathbf{v} = \lambda \mathbf{v} \). Key aspects of eigenvectors include:
  • An eigenvector retains its direction after the transformation by the matrix.
  • In orthogonal diagonalization, the eigenvectors form the columns of the orthogonal matrix \( Q \).
  • Clearly, the eigenvectors must be linearly independent and orthonormalized to form a valid orthogonal matrix.
For the symmetric matrix given, the calculation of eigenvectors (\( \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} \) and \( \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix} \)) provides the necessary foundation for constructing the orthogonal matrix \( Q \), essential for the diagonalization process.
Symmetric Matrix
A symmetric matrix is a special type of square matrix that is equal to its transpose, i.e., \( A = A^T \). This property simplifies many mathematical operations, as symmetric matrices have predictable behavior:
  • All eigenvalues of symmetric matrices are real, which simplifies analysis and ensures stability in computations.
  • The eigenspaces corresponding to distinct eigenvalues are orthogonal, which helps in constructing orthogonal matrices for diagonalization.
  • Symmetric matrices often arise in physics and engineering, particularly in the study of energy and systems dynamics.
In diagonalization, starting with a symmetric matrix ensures that we can orthogonally diagonalize it, using its eigenvectors to form the orthogonal matrix \( Q \), such that \( Q^T A Q = D \), where \( D \) is a diagonal matrix of eigenvalues.

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Most popular questions from this chapter

Let \(B\) be an invertible matrix. Show that \(A=B^{T} B\) is positive definite.

Identify the quadric with the given equation and give its equation in standard form. $$x^{2}+y^{2}-2 z^{2}+4 x y-2 x z+2 y z-x+y+z=0$$

Evaluate the quadratic form \(f(\mathbf{x})=\mathbf{x}^{T} A \mathbf{x}\) for the given A and x. $$A=\left[\begin{array}{rrr} 1 & 0 & -3 \\ 0 & 2 & 1 \\ -3 & 1 & 3 \end{array}\right], \mathbf{x}=\left[\begin{array}{r} 2 \\ -1 \\ 1 \end{array}\right]$$

Let \(A\) be a real \(2 \times 2\) matrix with complex eigenvalues \(\lambda=a \pm b i\) such that \(b \neq 0\) and \(|\lambda|=1 .\) Prove that every trajectory of the dynamical system \(\mathbf{x}_{k+1}=A \mathbf{x}_{k}\) lies on an ellipse. [Hint: Theorem 4.43 shows that if \(\mathbf{v}\) is an eigenvector corresponding to \(\lambda=a-b i\), then the matrix \(P=[\operatorname{Re} \mathbf{v} \quad \operatorname{Im} \mathbf{v}]\) is invertible and \(A=P\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right] P^{-1} .\) Set \(B=\left(P P^{T}\right)^{-1} .\) Show that the quadratic \(\mathbf{x}^{T} B \mathbf{x}=k\) defines an ellipse for all \(k>0\) and prove that if \(x \text { lies on this ellipse, so does } A x .]\)

Let \(W\) be the subspace spanned by the given vectors. Find a basis for \(W^{\perp}\). $$\mathbf{w}_{1}=\left[\begin{array}{r} 4 \\ 6 \\ -1 \\ 1 \\ -1 \end{array}\right], \mathbf{w}_{2}=\left[\begin{array}{r} 1 \\ 2 \\ 0 \\ 1 \\ -3 \end{array}\right], \mathbf{w}_{3}=\left[\begin{array}{r} 2 \\ 2 \\ 2 \\ -1 \\ 2 \end{array}\right]$$
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