A matrix has singular values , ,

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Numerical and computational mathematics · Numerical and computational mathematics

Error, Chance and the Machine, Ch Linear systems, least squares and the singular value decomposition

Problem

A 3×33\times3 matrix A\mathsf{A} has singular values σ1=10\sigma_1 = 10, σ2=1\sigma_2 = 1, σ3=10−12\sigma_3 = 10^{-12}. Use ∣det⁡A∣=σ1σ2σ3|\det\mathsf{A}| = \sigma_1\sigma_2\sigma_3, which holds because the orthogonal factors U\mathsf{U} and V\mathsf{V} of the decomposition have determinant ±1\pm1. ((a)) Show that det⁡A≠0\det\mathsf{A} \ne 0, and hence that A\mathsf{A} is, in the strict mathematical sense, invertible. ((b)) Someone argues: 'the determinant is non-zero, so A\mathsf{A} is safely invertible and solving Ax=b\mathsf{A}\mathbf{x} = \mathbf{b} numerically is no cause for concern.' State the value of κ2(A)\kappa_2(\mathsf{A}) and use it to say whether this claim is correct, explaining what actually happens to the solution's accuracy in binary64 (u=2−53≈1.1×10−16u = 2^{-53} \approx 1.1\times10^{-16}).

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