• 日期: 2028-10-21 22:24:03
  • 标签: 常规

(a) The spectral norm of A is the maximum singular value, which is 1.9.

(b) We have ∥Ax - Akx∥ = ∥A(I - Ak)x∥ ≤ ∥A∥ ∥(I - Ak)x∥, where ∥A∥ is the spectral norm of A. Since x is a unit vector, we have ∥(I - Ak)x∥ = |1 - ak|, where ak is the kth diagonal entry of A. Therefore, we need |1 - ak| ≤ 0.2/∥A∥ = 0.2/1.9. Solving for k, we get 4 ≤ k ≤ 6.


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