Question on Arnoldi Iteration Let $p_n$ be a monic polynomial of degree n (a polynomial of degree n with the coefficient of the term of degree n equal to 1). 

Let A be a square matrix and $A=Q^*HQ$ where H is a Hessenberg matrix. 
Given a vector b, why can the polynomial $p(A)b$ be written as $A^nb - Qy$ for some vector y?

This is the first step in proving the following property of the arnoldi iteration, which I am trying to understand. 

"The matrix Hn can be characterized by the following optimality condition. The characteristic polynomial of Hn minimizes ||p(A)q1||2 among all monic polynomials of degree n. This optimality problem has a unique solution if and only if the Arnoldi iteration does not break down."
The question should be edited in this way: any monic polynomial p(A) can be written as $A^n-Qy$ where y is an arbitrary vector and Q is the orthonormal basis for $K_n$ (The $n_{th}$ Krylov subspace of A). Hence, the answer to this question becomes straightforward: The term with the highest degree of the monic polynomial would be by definition $A^n$ and the rest of polynomial can be formed by Q which is the orthonormal basis for $K_n:=&lt;b,Ab,A^2b,...,A^{n-1}b&gt;$
For a more comprehensive elaboration on Arnoldi iteration and Krylov subspaces you can look into numerical linear algebra handbooks e.g. Chapter 34 "Numerical Linear Algebra" by Lloyd N. Trefethen.
The question should be edited in this way: any monic polynomial p(A) can be written as $A^n-Qy$ where y is an arbitrary vector and Q is the orthonormal basis for $K_n$ (The $n_{th}$ Krylov subspace of A). Hence, the answer to this question becomes straightforward: The term with the highest degree of the monic polynomial would be by definition $A^n$ and the rest of polynomial can be formed by Q which is the orthonormal basis for $K_n:=&lt;b,Ab,A^2b,...,A^{n-1}b&gt;$
For a more comprehensive elaboration on Arnoldi iteration and Krylov subspaces you can look into numerical linear algebra handbooks e.g. Chapter 34 "Numerical Linear Algebra" by Lloyd N. Trefethen.