Question

Let \(\alpha\) be a scalar, and consider the iterative scheme $$ \mathbf{x}_{k+1}=\mathbf{x}_{k}+\alpha\left(\mathbf{b}-A \mathbf{x}_{k}\right) $$ This is the gradient descent method with a fixed step size \(\alpha\). (a) If \(A=M-N\) is the splitting associated with this method, state what \(M\) and the iteration matrix \(T\) are. (b) Suppose \(A\) is symmetric positive definite and its eigenvalues are \(\lambda_{1}>\lambda_{2}>\cdots>\lambda_{n}>0 .\) i. Derive a condition on \(\alpha\) that guarantees convergence of the scheme to the solution \(\mathbf{x}\) for any initial guess. ii. Express the condition on \(\alpha\) in terms of the spectral condition number of \(A, \kappa(A)\). iii. Show the best value for the step size in terms of maximizing the speed of convergence is $$ \alpha=\frac{2}{\lambda_{1}+\lambda_{n}} $$ Find the spectral radius of the iteration matrix in this case, and express it in terms of the condition number of \(A\). (c) Determine whether the following statement is true or false. Justify your answer. "If \(A\) is strictly diagonally dominant and \(\alpha=1\), then the iterative scheme converges to the solution for any initial guess \(\mathrm{x}_{0} .\)

Short answer

To summarize the solution: 1. For an iterative scheme using fixed step size gradient descent, we identified the iteration matrix T and the constant matrix c as follows: T = I - αA and c = αb 2. Convergence is guaranteed if the spectral radius ρ(T) < 1 or if the step size α lies in the following range: 0 < α < 2/λ_i, where λ_i are eigenvalues of A. 3. We found the condition on step size α in terms of the spectral condition number κ(A): 0 < α < 2κ(A)/λ_max 4. The best value for the step size for maximal convergence speed in terms of the eigenvalues of A is: α = 2/(λ_max + λ_min) And the spectral radius of the iteration matrix for this optimal step size is: ρ(T) = (κ(A) - 1)/(κ(A) + 1) 5. The statement that for a strictly diagonally dominant matrix A and α = 1, the iterative scheme converges to the solution for any initial guess is **false**. A counterexample demonstrates that the spectral radius ρ(T) is not necessarily less than 1 for α = 1 and strictly diagonally dominant A, and thus convergence is not guaranteed.

Step-by-step solution

(a) Identify M and T in the given splitting

Given the equation, we want to find the splitting A = M - N such that the method can be represented as the general iteration scheme x_{k+1} = Tx_k + c. We manipulate the given equation to reach that form: $$ \mathbf{x}_{k+1}=\mathbf{x}_{k}+\alpha(A\mathbf{x}_k - \mathbf{b}), $$ $$ \mathbf{x}_{k+1} = (I - \alpha A) \mathbf{x}_k + \alpha \mathbf{b}, $$ where I is the identity matrix. From this representation, we identify the iteration matrix T and the constant matrix c: $$ T = I - \alpha A, $$ $$ c = \alpha \mathbf{b}. $$ Now, we can express the splitting as: $$ A = M - N, $$ $$ M = \frac{1}{\alpha} (I - T), $$ $$ N = \frac{1}{\alpha} T - A. $$

(b. i) Derive a condition on \(\alpha\) that guarantees convergence

The spectral radius \(\rho(T)\) must be less than 1 for the method to converge. Now, we know T = I - αA and A is symmetric and positive definite. Thus, all its eigenvalues are positive and real. Let the eigenvalues of A be λ₁ , λ₂, ... , λn, then subtracting αA from I scales and shifts the eigenvalues by (1 - αλi) for i = 1,2,...,n. Convergence then requires that |1 - αλi| < 1 for all i = 1,...,n or, equivalently, $$ -1 < 1 - \alpha \lambda_i < 1, $$ which gives the condition on α: $$ 0 < \alpha < \frac{2}{\lambda_i}. $$

(b. ii) Condition on step size in terms of spectral condition number

The spectral condition number for a matrix A is given by: $$ \kappa(A) = \frac{\lambda_{max}}{\lambda_{min}}, $$ where \(\lambda_{max} = \lambda_1\) and \(\lambda_{min} = \lambda_n\). From our condition for convergence, we know that we need: $$ 0 < \alpha < \frac{2}{\lambda_i}, $$ for all i = 1, ..., n. Thus, we must have: $$ 0 < \alpha < \frac{2}{\lambda_{min}} = \frac{2}{\frac{\lambda_{max}}{\kappa(A)}} = \frac{2\kappa(A)}{\lambda_{max}}, $$ This establishes the necessary condition on \(\alpha\) in terms of the spectral condition number.

(b. iii) Best value for the step size and spectral radius of the iteration matrix

To find the best value of α for maximal convergence speed, we minimize the spectral radius of the iteration matrix T. We know that the eigenvalues of T are (1 - αλi) for i = 1, ..., n. Thus, we have: $$ \rho(T) = \max_{1 \le i \le n} |1 - \alpha \lambda_i|. $$ Differentiating ρ(T) with respect to α, we get: $$ -\lambda_1 (1 - \alpha \lambda_1)^{-1} = -\lambda_n (1 - \alpha \lambda_n)^{-1}. $$ Solving this for α, we find: $$ \alpha = \frac{2}{\lambda_1 + \lambda_n}, $$ which is the best value for the step size in terms of maximizing convergence speed. For this optimal step size, the spectral radius is: $$ \rho(T) = \left|\frac{\lambda_1 - \lambda_n}{\lambda_1 + \lambda_n}\right|. $$ Expressing this in terms of the condition number κ(A), we get: $$ \rho(T) = \frac{\kappa(A) - 1}{\kappa(A) + 1}. $$

(c) Analyze convergence for strictly diagonally dominant A and \(\alpha = 1\)

The statement is "If \(A\) is strictly diagonally dominant and \(\alpha=1\), then the iterative scheme converges to the solution for any initial guess \(\mathrm{x}_{0} .\)" We know that the spectral radius of the iteration matrix T must be less than 1 for convergence. For \(\alpha = 1\) and strictly diagonally dominant \(A\), we have: $$ T = I - A. $$ If we show that the spectral radius ρ(T) < 1, we have convergence. However, strictly diagonally dominant does not guarantee that ρ(T) < 1 for α = 1. A counterexample is a 2x2 diagonally dominant matrix A with eigenvalues λ₁ = 1.5 and λ₂ = 0.5. In this case, ρ(I - A) = |1 - 1.5| = 0.5, which satisfies the convergence criterion. However, considering a slightly different 2x2 diagonally dominant matrix B with eigenvalues ℓ₁ = 1.1 and ℓ₂ = 0.9, we have ρ(I - B) = |1 - 1.1| = 0.1 which does not guarantee convergence. Thus, the statement is **false**. Strict diagonally dominant A and α = 1 do not guarantee convergence for any initial guess.

Key concepts

Iterative Scheme

An iterative scheme is a sequence of steps to solve a problem, where each step is built upon the result of the previous step. Gradient descent is one such method commonly used in numerical optimization to find the minimum of a function. The key idea is to update the current point in the direction of the steepest descent, which is the negative of the gradient of the function at that point.

The iterative updating rule for gradient descent is expressed as: \[\mathbf{x}_{k+1} = \mathbf{x}_{k} + \alpha(\mathbf{b} - A \mathbf{x}_{k})\]where \(\alpha\) is a scalar value known as the step size or learning rate. In each iteration, the current estimate \(\mathbf{x}_{k}\) is refined to get closer to the function's minimum. The refinement is dependent on both the magnitude of \(\alpha\) and the direction provided by the term \(\mathbf{b} - A \mathbf{x}_{k}\). The choice of the step size \(\alpha\) directly impacts the effectiveness and efficiency of the scheme.

When discussing the iterative scheme's convergence, we want to ensure that the sequence of solutions will approach the true solution. To aid this process, the concept of an iteration matrix \( T \) is introduced, which forms the basis for how one estimate leads to the next.

Spectral Condition Number

The spectral condition number of a matrix, commonly denoted as \(\kappa(A)\), plays a significant role in the numerical stability of algorithms. It is defined as the ratio of the largest eigenvalue to the smallest eigenvalue of a matrix:\[\kappa(A) = \frac{\lambda_{\text{max}}}{\lambda_{\text{min}}}\]This number provides an indication of how 'ill-conditioned' a matrix is. A high spectral condition number suggests that the matrix is near singular and small perturbations in input can lead to large changes in the output, which can make numerical solutions highly sensitive and potentially unstable.

In the context of gradient descent, the spectral condition number influences the choice of step size \(\alpha\). A matrix with a smaller spectral condition number is more well-conditioned and can allow for a larger range of step sizes leading to convergence. Therefore, when optimizing the iterative scheme, we aim to select an \(\alpha\) within the bounds prescribed by the condition number to ensure the scheme converges.

Convergence Condition

The convergence condition dictates whether an iterative method will result in a sequence that approaches the actual solution. For the gradient descent method, the convergence condition requires that the spectral radius of the iteration matrix \(T\), denoted as \(\rho(T)\), is less than one:\[\rho(T) < 1\]Each eigenvalue of \(T\) derived from the original matrix \(A\) and the chosen value of \(\alpha\) must satisfy this inequality. In effect, this ensures that the iterative process is pulling the estimate towards the true solution rather than away from it.

To guarantee convergence for any initial guess, \(\alpha\) must be chosen carefully. It should be set strictly between 0 and \(\frac{2}{\lambda_{\text{max}}}\), where \(\lambda_{\text{max}}\) is the largest eigenvalue of \(A\). In the case of the gradient descent method, the best value for \(\alpha\) is often related to the inverse of the sum of the largest and smallest eigenvalues of \(A\), which corresponds to the fastest rate of convergence. This relationship is particularly important when the matrix \(A\) is symmetric positive definite, as such matrices are common in optimization problems encountered in machine learning and numerical computation.