The Wilkinson shift uses the trailing 2x2 submatrix B = A^(k-1)(n-1:n, n-1:n). The shift mu is chosen as:

Question

BlackTom AI · Accepted Answer

For a symmetric 2×2 trailing block B = [a b; b c], the eigenvalues are
lambda_{±} = (a + c ± sqrt((a - c)^2 + 4 b^2)) / 2.
Wilkinson’s shift chooses the eigenvalue of B that is closer to the bottom‑right entry c = B(2,2).

1) Relevant knowledge points and concepts
- In shifted QR for symmetric tridiagonal matrices, a scalar shift μ is chosen to accelerate convergence of the bottom subdiagonal to zero.
- The Wilkinson shift uses the trailing 2×2 block B and picks one of its two eigenvalues as μ.
- For symmetric B = [a b; b c], eigenvalues are given by the quadratic formula using trace and determinant.

2) Step‑by‑step reasoning
- Write B = [a b; b c]. Trace t = a + c, determinant d = ac − b^2.
- Eigenvalues: λ_{±} = (t ± sqrt(t^2 − 4d)) / 2 = (a + c ± sqrt((a − c)^2 + 4 b^2)) / 2.
- Compute numerical distances to c: |λ_{±} − c|. These equal |(a − c ± sqrt((a − c)^2 + 4 b^2)) / 2|.
- The Wilkinson rule selects the eigenvalue λ (either λ_+ or λ_−) with the smaller distance |λ − c|. That is, choose the eigenvalue closer to B(2,2)=c.

3) Intermediate algebra (illustrative)
- Let δ = sqrt((a − c)^2 + 4 b^2). Then λ_{±} = (a + c ± δ)/2 and λ_{±} − c = (a − c ± δ)/2.
- Choose the sign (±) that makes |(a − c ± δ)/2| minimal; that eigenvalue is μ.

4) Why this is correct
- Picking the eigenvalue nearest to the bottom diagonal entry c tends to deflate the subdiagonal element faster and promotes stable convergence of the QR iteration; this is exactly the Wilkinson shift definition.

5) Why the other options are incorrect
- (a) The average of the two eigenvalues equals trace/2 = (a + c)/2, not the Wilkinson choice (only equals one eigenvalue in special cases).
- (c) The sum of the bottom two diagonal entries (a + c) is the trace, not an eigenvalue (and not the chosen shift).
- (d) Always choosing the larger eigenvalue is not Wilkinson’s rule; Wilkinson chooses by closeness to c, which may be the smaller eigenvalue depending on a,b,c.

6) Conclusion
The correct choice is option b: the eigenvalue of B that is closer to B(2,2) = A^(k-1)(n,n). This matches the provided answer.

The Wilkinson shift uses the trailing 2x2 submatrix B = A^(k-1)(n-1:n, n-1:n). The shift mu is chosen as:单项选择题

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