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Rayleigh-quotient iteration on a $2 \times 2$ symmetric matrix exhibits fast local convergence

Example

Let

$A = diag (2, 1), x (t) = \frac{1}{\sqrt{1 + t^{2}}} [\begin{array}{c} 1 \\ t \end{array}] .$

If $t \neq 0$ , one Rayleigh-quotient iteration step started from $x (t)$ has new slope $- t^{3}$ . At $t = 0$ the start vector is already an eigenvector and the Rayleigh-quotient step is not defined because the shifted matrix is singular. Thus small nonzero errors cube in one step.

Facts & Assumptions

Given: The symmetric matrix $A$ and the unit start vector $x (t)$ .

[L1]

At positive sufficiently small distance from a simple Hermitian eigendirection, whenever the next step is defined, Rayleigh-quotient iteration reduces that distance cubically (Near a simple Hermitian eigenvector, Rayleigh-quotient iteration converges cubically).

Verification

technique · computation

1.1

The Rayleigh quotient of $x (t)$ is $μ (t) = \frac{2 + t^{2}}{1 + t^{2}} .$ Therefore $A - μ (t) I = [\begin{array}{cc} \frac{t^{2}}{1 + t^{2}} & 0 \\ 0 & - \frac{1}{1 + t^{2}} \end{array}] .$

algebra

2.1

If $t = 0$ , then $μ (0) = 2$ and $A - μ (0) I = diag (0, - 1)$ is singular, so Rayleigh-quotient iteration for Hermitian matrices does not define a next iterate at that start vector. If $t \neq 0$ , solving $(A - μ (t) I) y = x (t)$ gives $y \propto [\begin{array}{c} \frac{1 + t^{2}}{t^{2}} \\ - (1 + t^{2}) t \end{array}],$ so after normalisation the new slope is $\frac{y_{2}}{y_{1}} = - t^{3} .$

step 1.1algebra

3.1

Thus every sufficiently small nonzero slope error $t$ becomes $- t^{3}$ in one step, which is the explicit two-dimensional form of the cubic behaviour from [L1].

L1step 2.1∎

Depends on

Near a simple Hermitian eigenvector, Rayleigh-quotient iteration converges cubically

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Rayleigh-quotient iteration on a 2×22×2 symmetric matrix exhibits fast local convergence