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Subspace iteration converges to the dominant invariant subspace when a spectral gap separates the wanted and unwanted eigenvalues
Statement
Let , let , let , let be diagonalisable with eigenvalues ordered by . Let have orthonormal columns. Write in the ordered eigenbasis and write with . Assume is invertible. Then the column space of converges to that dominant invariant subspace. More precisely, in the ordered eigenbasis it is the graph of and
Facts & Assumptions
Given: The field, dimensions, diagonalisable matrix , spectral gap, initial orthonormal frame , and valid subspace iteration from the statement, for which the leading coefficient block in the statement is invertible.
A diagonalisable matrix admits an eigenbasis (A diagonalisable endomorphism is one admitting a basis of eigenvectors, equivalently a diagonal matrix representation).
Subspace iteration is the repeated QR orthonormalisation of (Subspace iteration and the dominant invariant subspace of a matrix).
Proof
By [L1], use the decomposition from the statement, where and . The hypothesis states exactly that is invertible.
The spectral gap implies , so is invertible for every . Hence has full column rank, every reduced QR step in [L2] is defined, and the column space of equals that of . Now Multiplying on the right by shows that the same column space is the graph of over the dominant invariant subspace.
The spectral gap implies Hence the graph in step 2.1 converges to the dominant invariant subspace at that rate.
Therefore the column space of converges to the dominant invariant subspace, with the precise graph-norm rate stated above.
Depends on
Used by
Dependency tree · two levels
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Sources
- Andrew Stuart and Jochen Voss, Matrix Analysis and Algorithms (standard reference, not scraped)