(a) Prove that the dimension of a Krylov subspace is bounded by the degree of theminimal polynomial of the matrix A, as defined in Exercise 8.6.23. 

(b) Is there always a Krylov subspace whose dimension equals the degree of the minimal polynomial?

Data From Exercise 8.6.23

Let A be an n × n matrix. By definition, the minimal polynomial  of A is the monic polynomial m_A(t) = t^k + c_k−1t^k−1 + · · · + c₁t + c₀ of minimal  degree k that annihilates A, so m_A(A) = A^k + c_k−1A^k−1 + · · · + c₁A + c₀I = O.