After deflation reduces the active problem from n x n to (n-1) x (n-1), the deflated eigenvalue is stored. What is the criterion used to decide when deflation is appropriate?

Question

BlackTom AI · Accepted Answer

Deflation is triggered when a subdiagonal coupling entry becomes numerically negligible: |A^(k)(j+1,j)| < tolerance * ||A||_F (for example tolerance ≈ 10^−12).

1) Relevant concepts and knowledge points
- In QR/Francis iterations on an upper Hessenberg (or reduced) matrix, deflation isolates an eigenvalue when the subdiagonal entry that couples two blocks is effectively zero.  
- A practical, backward‑stable criterion compares the magnitude of that subdiagonal element to a scaled norm of the matrix (commonly the Frobenius norm ||A||_F) times a small tolerance.  
- Setting that element to zero decouples the matrix into smaller problems without significantly changing eigenvalues within rounding error.

2) Step‑by‑step reasoning
- At iteration k examine A^(k)(j+1,j), the subdiagonal element between row j and j+1.  
- Compute the test value: threshold = tolerance * ||A||_F. Example: tolerance = 1e−12.  
- If |A^(k)(j+1,j)| < threshold, treat the coupling as negligible and set it to zero (deflate), reducing the active block size by 1 and storing the converged eigenvalue.

3) Intermediate/calculation step (illustrative)
- Suppose ||A||_F = 5.0 and tolerance = 1e−12. Then threshold = 5.0×1e−12 = 5e−12. If |A^(k)(j+1,j)| = 3e−12 < 5e−12, deflate.

4) Why this is correct
- This test measures the subdiagonal element relative to the overall matrix scale so decisions are relative and robust to scaling. It is the standard practical criterion used in eigensolvers to decide when coupling is within round‑off and can be dropped.

5) Why the other choices are incorrect
- b: A Rayleigh shift being zero is not a general deflation test and rarely occurs exactly; it’s not the standard condition.  
- c: Deflating after a fixed number of iterations ignores actual convergence and risks incorrect eigenvalues.  
- d: Requiring a diagonal entry to "match an eigenvalue to machine precision" is circular and impractical; eigenvalues are what we seek, not known a priori.

Conclusion: The correct criterion is option a: When |A^(k)(j+1,j)| < tolerance * ||A||_F, which matches the provided answer.

After deflation reduces the active problem from n x n to (n-1) x (n-1), the deflated eigenvalue is stored. What is the criterion used to decide when deflation is appropriate?单项选择题

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