Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31
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If μ is not an eigenvalue and one simple eigenvalue is uniquely nearest to μ, shifted inverse iteration converges to its eigendirection

Statement

Let A be diagonalisable with eigenpairs (λi,vi), let μ{λ1,,λn}, and suppose one simple eigenvalue λj satisfies

λjμ<λiμ(ij).

If the start vector has nonzero vj-component, then shifted inverse iteration with shift μ converges projectively to the eigendirection of vj.

Facts & Assumptions

Given: A diagonalisable matrix A, a shift μ not equal to any eigenvalue, and a start vector with nonzero vj-component.

[L1]

Shifted inverse iteration is power iteration for (AμI)1 (Inverse iteration and shifted inverse iteration).

Proof

technique · direct
1.1

The eigenvectors of (AμI)1 are the same vi, and the corresponding eigenvalues are (λiμ)1. The hypothesis λjμ<λiμ means 1λjμ>1λiμ(ij).

givenalgebra
2.1

Therefore (AμI)1 has a simple eigenvalue of strictly largest modulus in the eigendirection vj. The start vector has nonzero vj-component by hypothesis, so [L2] applies.

L2step 1.1
3.1

Since [L1] identifies shifted inverse iteration with that power iteration, the iterates converge projectively to the eigendirection of vj.

L1step 2.1

Depends on

Used by

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources