(a) Prove that the dimension of a Krylow subspace is bounded...

(a) Prove that the dimension of a Krylow subspace is bounded by the degree of the minimal 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?

Key Concepts

Krylov Subspace

A Krylov subspace is generated by the successive applications of a matrix to a vector, forming the span {v, Av, A²v, ...}. This concept is central in numerical linear algebra, particularly in iterative methods for solving systems of equations and eigenvalue problems. The structure of Krylov subspaces allows one to capture the action of the matrix in a progressively richer set of directions, which eventually leads to insights about the invariant subspaces and the spectrum of the matrix.

Minimal Polynomial

The minimal polynomial of a matrix is the unique monic polynomial of smallest degree such that when applied to the matrix, it results in the zero matrix. This polynomial encapsulates fundamental properties of the matrix, including its eigenvalues and their algebraic multiplicities. It also provides an upper bound on the dimension of any Krylov subspace because any power of the matrix higher than the degree of the minimal polynomial can be expressed as a linear combination of lower powers, thus preventing the Krylov sequence from being indefinitely linearly independent.

Cyclic Vector

A cyclic vector for a matrix is one from which the entire space (or a large invariant subspace) can be generated by the repeated action of the matrix, meaning that the resulting Krylov subspace is as large as possible. The existence of a cyclic vector ensures that the dimension of the Krylov subspace reaches the degree of the minimal polynomial. However, not every vector is cyclic; hence, while the dimension of any Krylov subspace is bounded by the minimal polynomial's degree, not all choices of the starting vector will achieve this maximal dimension.

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