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Ignoring eigenvectors, the shifted QR algorithm for computing eigenvalues in the symmetric tridiagional case costs $O(n)$ per iteration, converges globally, and converges cubically near the end. What is its worst case complexity (which presumably depends on the desired error)?

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up vote 4 down vote accepted

With the right shift stategy (mixing RQ and W shift appropriately), it is globally cubically convergent, with a guaranteed linear convergence factor, which makes its complexity of the order of $N$ times the logarithm of the off-diagonal entries. Unfortunately,

See http://www.springerlink.com/content/f1w76x84463l52t7/ for local cubic convergence.

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Does that mean there is no entry-independent worst case upper bound? – Geoffrey Irving Aug 13 '12 at 20:18
Probably no known one. But my knowledge about this is about 10 years old, so you should check more recent literature. – Arnold Neumaier Aug 14 '12 at 7:02

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