Linear Algebra and Its Applications · Problem 7.4.12

Linear Programming and Game Theory — Gilbert Strang

Mathematics · iterative-methods-singular-matrix

Problem

If AA is singular, then all splittings A=S−TA = S - T must fail. From Ax=0Ax = 0, show that S−1Tx=xS^{-1}Tx = x. So this matrix B=S−1TB = S^{-1}T has λ=1\lambda = 1 and fails.
Intuitive explanation

The key insight is that a singular matrix AA forces any splitting to inherit that singularity through the iteration matrix. When A=S−TA = S - T is singular, the null space of AA becomes a trap: if Ax=0Ax = 0 for some nonzero xx, then rearranging gives Sx=TxSx = Tx, which means S−1TS^{-1}T must map xx to itself. This eigenvalue of 1 in the iteration matrix B=S−1TB = S^{-1}T is exactly what kills convergence—the error doesn't shrink, it stays put.

Intuitive explanation

The key insight is that a singular matrix AA forces any splitting to inherit that singularity through the iteration matrix. When A=S−TA = S - T is singular, the null space of AA becomes a trap: if Ax=0Ax = 0 for some nonzero xx, then rearranging gives Sx=TxSx = Tx, which means S−1TS^{-1}T must map xx to itself. This eigenvalue of 1 in the iteration matrix B=S−1TB = S^{-1}T is exactly what kills convergence—the error doesn't shrink, it stays put.

Related theory

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