The Arnoldi relation A Q_{k+1} = Q_{k+2} H-tilde_{k+1} can be rearranged to show that the residual for approximating A from the Krylov subspace is:

Question

BlackTom AI · Accepted Answer

Arnoldi relation A Q_m = Q_{m+1} 	ilde H_m implies the key residual identity.

Relevant concepts: Q_m contains m orthonormal Arnoldi basis vectors, Q_{m+1} = [Q_m, q_{m+1}] adds the next basis vector, H_m is the m×m upper Hessenberg from Arnoldi, and 	ilde H_m is the (m+1)×m matrix obtained by appending one row to H_m. In standard Arnoldi notation the appended row has a single nonzero entry h_{m+1,m} in the last column, so 	ilde H_m = [ H_m ; h_{m+1,m} e_m^T ] where e_m is the m-th unit vector.

Step-by-step algebra (set m = k for direct comparison to the answer):
1. Write Q_{k+1} = [Q_k, q_{k+1}] and 	ilde H_k = [ H_k ; h_{k+1,k} e_k^T ].
2. Multiply: Q_{k+1} 	ilde H_k = [Q_k, q_{k+1}] [ H_k ; h_{k+1,k} e_k^T ].
3. Use block multiplication: [Q_k, q_{k+1}] [ H_k ; h_{k+1,k} e_k^T ] = Q_k H_k + q_{k+1} (h_{k+1,k} e_k^T).
4. The Arnoldi relation gives A Q_k = Q_{k+1} 	ilde H_k, so substitute the product:
   A Q_k = Q_k H_k + h_{k+1,k} q_{k+1} e_k^T.
5. Rearranging yields the residual:
   A Q_k − Q_k H_k = h_{k+1,k} q_{k+1} e_k^T.

This shows the residual is a rank‑1 matrix: a scalar h_{k+1,k} times the outer product of q_{k+1} (new basis vector) and e_k^T (the last coordinate direction in the reduced space). That matches option d in content (the common form of the Arnoldi residual).

Why other options are incorrect:
- (a) identity: A Q − Q H is generally not I_n; dimensions don't match and there's no reason it equals identity.
- (b) zero: A Q = Q H only if the Krylov subspace is invariant under A (exact invariance), which is not generally true.
- (c) A Q − Q H = Q(A − H): incorrect algebraically; A acts in the large n-space while H is the small projected matrix, so you cannot factor that way.

Conclusion: the correct residual expression is the rank‑1 matrix h_{k+1,k} q_{k+1} e_k^T, confirming option d.

The Arnoldi relation A Q_{k+1} = Q_{k+2} H-tilde_{k+1} can be rearranged to show that the residual for approximating A from the Krylov subspace is:Single choice

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