The Lanczos three-term recurrence is beta_k q_{k+1} = A q_k - alpha_k q_k - beta_{k-1} q_{k-1}. What are alpha_k and beta_k?

Question

BlackTom AI · Accepted Answer

The Lanczos recurrence produces a tridiagonal projection T whose diagonal and off-diagonal entries are obtained from inner products and normalisations of the Lanczos vectors q_k.

1) Relevant knowledge points and concepts
- Lanczos builds an orthonormal basis {q_1, q_2, ...} for the Krylov subspace with q_i^T q_j = δ_{ij}.
- The three-term recurrence beta_k q_{k+1} = A q_k - alpha_k q_k - beta_{k-1} q_{k-1} enforces orthogonality and defines the tridiagonal matrix T with diagonal entries alpha_k and off-diagonals beta_k.
- alpha_k is the Rayleigh quotient q_k^T A q_k; beta_k is a normalisation (2-norm) of the residual.

2) Step-by-step reasoning
Start from beta_k q_{k+1} = A q_k - alpha_k q_k - beta_{k-1} q_{k-1}.
Take inner product with q_k^T:
q_k^T (beta_k q_{k+1}) = q_k^T A q_k - alpha_k q_k^T q_k - beta_{k-1} q_k^T q_{k-1}.
Using orthonormality q_k^T q_{k+1} = 0, q_k^T q_{k-1} = 0, q_k^T q_k = 1, we get
0 = q_k^T A q_k - alpha_k,
hence alpha_k = q_k^T A q_k.

3) Intermediate steps for beta_k
Rearrange the recurrence to isolate q_{k+1}:
q_{k+1} = (1/beta_k) (A q_k - alpha_k q_k - beta_{k-1} q_{k-1}).
Since q_{k+1} must be unit norm, beta_k is chosen as the 2-norm of the residual:
beta_k = ||A q_k - alpha_k q_k - beta_{k-1} q_{k-1}||_2.
(One can also verify q_{k-1}^T A q_k = beta_{k-1} by taking inner product with q_{k-1}^T, consistent with the recurrence.)

4) Why this is correct
alpha_k = q_k^T A q_k is exactly the projection of A onto q_k (the Rayleigh quotient) and becomes the k-th diagonal of T. beta_k normalises the residual vector to produce the next orthonormal basis vector q_{k+1}; this residual’s norm is the off-diagonal element of T.

5) Why the other options are incorrect
- (a) alpha_k = 0 and beta_k = 1 is false: alpha_k depends on A and q_k and is generally nonzero; betas vary to normalise residuals.
- (c) alpha_k and beta_k are not eigenvalues of the Ritz matrix T; they are entries of T (diagonal and off-diagonal), not its eigenvalues.
- (d) alpha_k ≠ ||A q_k||_2 (alpha_k is an inner product, not the norm), and beta_k ≠ q_{k-1}^T A q_k (that inner product equals beta_{k-1}, not beta_k).

6) Conclusion
Therefore alpha_k = q_k^T A q_k and beta_k = ||A q_k - alpha_k q_k - beta_{k-1} q_{k-1}||_2, matching answer b.

The Lanczos three-term recurrence is beta_k q_{k+1} = A q_k - alpha_k q_k - beta_{k-1} q_{k-1}. What are alpha_k and beta_k?单项选择题

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