Question

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

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

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.

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 $$ \lambda_1 = 1-a, \quad \lambda_2 = 1+a, \quad \lambda_3 = 1. $$

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 \(-1<a<1\).

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 \(|a|\geq \frac{1}{2}\).

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.

Key concepts

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