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Consider the matrix $$ A=\left(\begin{array}{lll} 1 & a & a \\ a & 1 & a \\ a & a & 1 \end{array}\right) $$ Find the values of \(a\) for which \(A\) is symmetric positive definite but the Jacobi iteration does not converge.

Short Answer

Expert verified
The matrix \(A\) is symmetric positive definite but the Jacobi iteration does not converge if \(-\frac{1}{2} \leq a < 0\) or \(0 < a \leq \frac{1}{2}\).

Step by step solution

01

Check for Symmetry

Since \(A\) is given, it is trivial to check its symmetry. By inspection, it is evident that \(A=A^T\) (as all the elements are symmetric w.r.t the main diagonal). So, \(A\) is symmetric.
02

Determine Eigenvalues of A

In order to check if the matrix A is positive definite, we must make sure that all its eigenvalues are positive. To find the eigenvalues, we have to solve the characteristic equation given by $$ \text{det}(\lambda I - A) = 0, $$ which is $$ \text{det}\left(\left(\begin{array}{lll} \lambda -1 & -a & -a \\ -a & \lambda -1 & -a \\ -a & -a & \lambda -1 \\ \end{array}\right)\right) = 0. $$ Now, we calculate the determinant: $$ (\lambda-1)^3 - 2a^2(\lambda -1)-a^3 = 0. $$ Using the cubic formula, we find that the eigenvalues are $$ \lambda_1 = 1-a, \quad \lambda_2 = 1+a, \quad \lambda_3 = 1. $$
03

Check for Positive Definiteness

For \(A\) to be positive definite, all of its eigenvalues must be positive: $$ \lambda_1 > 0 \Rightarrow 1-a>0 \Rightarrow a< 1. $$ $$ \lambda_2 > 0 \Rightarrow 1+a>0 \Rightarrow a> -1. $$ $$ \lambda_3 > 0 \Rightarrow 1>0. $$ So, for \(A\) to be symmetric positive definite, we must have \(-1<a<1\).
04

Check for Diagonal Dominance

A necessary condition for the convergence of the Jacobi iteration is that \(A\) is strictly diagonally dominant. A matrix is strictly diagonally dominant if the absolute value of the diagonal element is strictly greater than the sum of the absolute values of all other elements in the row. So, for \(A\), we must have: $$ |1| > |a| + |a|. $$ However, we wish to find a condition under which Jacobi iteration does not converge. Therefore, \(A\) must not be diagonally dominant: $$ |1| \leq |a| + |a|. $$ Since \(-1 < a < 1\), it follows that \(0 \leq 1-2|a|\), leading to \(|a|\geq \frac{1}{2}\).
05

Determine the Range of Values of \(a\)

From Step 3 and Step 4, we have the following conditions: - For \(A\) to be symmetric positive definite, \(-1<a<1\). - For the Jacobi iteration not to converge, \(|a|\geq \frac{1}{2}\). Combining these conditions, we obtain the range of values for \(a\): $$ -\frac{1}{2} \leq a < 0 $$ or $$ 0<a \leq \frac{1}{2}. $$ Thus, if \(-\frac{1}{2} \leq a < 0\) or \(0<a \leq \frac{1}{2}\), the matrix \(A\) is symmetric positive definite but the Jacobi iteration does not converge.

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Key Concepts

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

Symmetric Positive Definite Matrix
In numerical linear algebra, a symmetric positive definite (SPD) matrix is a key concept. A matrix is defined as symmetric if it equals its transpose, meaning it mirrors itself across its diagonal. For example, if\[ A = \begin{pmatrix} 1 & a & a \ a & 1 & a \ a & a & 1 \end{pmatrix} \]then \( A^T = A \), ensuring symmetry because the values on either side of the diagonal are identical.

To further ensure that a symmetric matrix is positive definite, all its eigenvalues must be greater than zero. This property guarantees no zero or negative eigenvalues, which are linked to stability and meaningful solutions in many applications. In the given exercise, analyzing the matrix's eigenvalues (\(1-a\), \(1+a\), and \(1\)) confirmed that \(-1 < a < 1\) makes \( A \) positive definite as all eigenvalues satisfy the positivity condition.
Eigenvalues
Eigenvalues are pivotal in determining the characteristics of matrices in numerical linear algebra. For a given matrix like\[ A = \begin{pmatrix} 1 & a & a \ a & 1 & a \ a & a & 1 \end{pmatrix}, \]eigenvalues solve the characteristic equation \( \text{det}(\lambda I - A) = 0 \). They are critical in understanding the matrix's behavior and properties, such as its stability and transformations.

In the exercise at hand, the eigenvalues found are \( \lambda_1 = 1-a \), \( \lambda_2 = 1+a \), and \( \lambda_3 = 1 \). These eigenvalues help to determine if the matrix is positive definite. Knowing eigenvalues is fundamental since:
  • Their positivity implies positive definiteness, gauging if solutions to systems involving the matrix will be stable.
  • They reveal the nature of the matrix, helping in tasks like diagonalization or determining invertibility.
Jacobi Iteration
The Jacobi iteration is a method used to solve systems of linear equations numerically. It's an iterative algorithm that helps find solutions by making initial guesses and refining them repeatedly.

However, its convergence is not guaranteed for all matrices. Convergence depends significantly on the matrix's properties, such as diagonal dominance. If the matrix isn't strictly diagonally dominant, the Jacobi method might not converge.

Jacobi iteration leverages the idea of breaking a matrix equation into multiple parts, iteratively solving each part until the solution stabilizes or reaches an acceptable approximation. In simpler terms, it tries to "zero in" on a solution over several iterations.

For the provided exercise, the Jacobi iteration's non-convergence aligns with the case where the matrix is not strictly diagonally dominant when \( |a| \geq \frac{1}{2} \).
Diagonal Dominance
Diagonal dominance is a condition of matrices that ensures the success of iterative solutions like Jacobi iteration. It occurs when each diagonal element in a matrix is larger in absolute value than the sum of the absolute values of the other elements in its row. For example, if\[ A = \begin{pmatrix} 1 & a & a \ a & 1 & a \ a & a & 1 \end{pmatrix} \],go through each row to check if the diagonal element is greater than the other elements in the row summed.

This means ensuring \( |1| > |a| + |a| \) for all such conditions in the matrix. If a matrix satisfies this condition, methods like Jacobi often converge. However, for the matrix \( A \) in the problem, if \(|a| \geq \frac{1}{2}\), the sum of the non-diagonal terms isn't necessarily less than the diagonal term, leading to potential divergence of the Jacobi iteration. This explains why Jacobi might not converge for certain values of \( a \), particularly \( -\frac{1}{2} \leq a < 0 \) or \(0

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Most popular questions from this chapter

Show that the energy norm is indeed a norm when the associated matrix is symmetric positive definite.

Let \(A\) be a general nonsymmetric nonsingular square matrix, and consider the following two alternatives. The first is applying GMRES to solve the linear system \(A \mathbf{x}=\mathbf{b} ;\) the second is applying CG to the normal equations $$ A^{T} A \mathbf{x}=A^{T} \mathbf{b} $$ We briefly discussed this in Section \(7.5\); the method we mentioned in that context was CGLS. (a) Suppose your matrix \(A\) is nearly orthogonal. Which of the two solvers is expected to converge faster? (b) Suppose your matrix is block diagonal relative to \(2 \times 2\) blocks, where the \(j\) th block is given by $$ \left(\begin{array}{cc} 1 & j-1 \\ 0 & 1 \end{array}\right) $$ with \(j=1, \ldots, n / 2\). Which of the two solvers is expected to converge faster? [Hint: Consider the eigenvalues and the singular values of the matrices.]

Suppose CG is applied to a symmetric positive definite linear system \(A \mathbf{x}=\mathbf{b}\) where the righthand-side vector \(\mathbf{b}\) happens to be an eigenvector of the matrix \(A\). How many iterations will it take to converge to the solution? Does your answer change if \(A\) is not SPD and instead of CG we apply GMRES?

Consider the Helmholtz equation $$ -\left(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}\right)-\omega^{2} u=g(x, y) $$ defined in the unit square with homogeneous Dirichlet boundary conditions. (a) Suppose this problem is discretized on a uniform grid with step size \(h=1 /(N+1)\) using a five-point scheme as in Example \(7.1\) plus an additional term. Write down the resulting difference method. (b) Call the resulting matrix \(A\). Find a value \(\omega_{c}\) such that for \(\omega^{2}<\omega_{c}^{2}\) and \(h\) arbitrarily small, \(A\) is still positive definite, but for \(\omega^{2}>\omega_{c}^{2}\) the positive definiteness is lost. (c) Solve the problem for \(\omega=1\) and \(\omega=10\) for \(N=2^{7}-1\) using an appropriate preconditioned Krylov subspace method of your choice, or a multigrid method. Use tol = 1.e-6. Verify that the last residual norm is below tol and tell us how many iterations it took to get there.

The smoothing factor \(\mu^{*}\) for a discrete operator is defined as the worst (i.e., smallest) factor by which high frequency components are reduced in a single relaxation step. For the two-dimensional Laplacian we have discussed throughout this chapter and a basic relaxation scheme, this can be stated as follows. Suppose \(e_{0}\) is the error before a relaxation step associated with a stationary iteration matrix \(T\) and \(\mathbf{e}_{1}\) the error after that step, and write $$ \mathbf{e}_{0}=\sum_{l, m=1}^{N} \alpha_{l, m} \mathbf{v}_{l, m} $$ where \(\left\\{\mathbf{v}_{l, m}\right\\}_{l, m=1}^{N}\) are the eigenvectors of the iteration matrix. Then $$ \mathbf{e}_{1}=\sum_{l, m=1}^{N} \alpha_{l, m} \mu_{l, m} \mathbf{v}_{l, m} $$ where \(\left\\{\mu_{l, m}\right\\}_{l, m=1}^{N}\) are eigenvalues of the iteration matrix. The smoothing factor is thus given by $$ \mu^{*}=\max \left\\{\left|\mu_{l, m}\right|: \frac{N+1}{2} \leq l, m \leq N\right\\} $$ (a) Denote the discrete Laplacian by \(A\) and the iteration matrix for damped Jacobi by \(T_{\omega}\). Confirm that the eigenvectors of \(A\) are the same as the eigenvectors of \(T_{\omega}\) for this scheme. (If you have already worked on Exercise 11 , this should be old news.) (b) Show that the optimal \(\omega\) that gives the smallest smoothing factor over \(0 \leq \omega \leq 1\) for the two-dimensional Laplacian is \(\omega^{*}=\frac{4}{5}\), and find the smoothing factor \(\mu^{*}=\mu^{*}\left(\omega^{*}\right)\) in this case. Note: \(\mu^{*}\) should not depend on the mesh size. (c) Show that Jacobi (i.e., the case \(\omega=1\) ) is not an effective smoother.

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