What is a resolvent equation?

What is a resolvent equation?

A resolvent of an algebraic equation f(x)=0 of degree n is an algebraic equation g(y)=0, with coefficients that rationally depend on the coefficients of f(x), such that, if the roots of this equation are known, the roots of the given equation f(x)=0 can be found by solving simpler equations of degrees not exceeding n.

How do you find the spectrum of an operator?

The spectrum σ(A) of any bounded linear operator A is a closed subset of contained in |λ|≤A. Proof. Define the map F : C → B(X) by F(λ) = A − λI. We have F(λ) − F(µ) = |λ − µ|, so that F is continuous.

Are self-adjoint operators closed?

The class of self-adjoint operators is especially important in mathematical physics. Every self-adjoint operator is densely defined, closed and symmetric. The converse holds for bounded operators but fails in general.

What is the adjoint of an operator?

In mathematics, the adjoint of an operator is a generalization of the notion of the Hermitian conjugate of a complex matrix to linear operators on complex Hilbert spaces. In this article the adjoint of a linear operator M will be indicated by M∗, as is common in mathematics. In physics the notation M† is more usual.

What is resolvent of an operator?

In linear algebra and operator theory, the resolvent set of a linear operator is a set of complex numbers for which the operator is in some sense “well-behaved”. The resolvent set plays an important role in the resolvent formalism.

What is resolvent analysis?

The resolvent is a linear operator that governs how harmonic forcing inputs are amplified by the linear dynamics of a system and mapped onto harmonic response outputs. Resolvent analysis refers to the inspection of this operator to find the most responsive inputs, their gains and the most receptive outputs.

Is normal operator Diagonalizable?

theorem: every normal operator on a finite-dimensional space is diagonalizable by a unitary operator. There is also an infinite-dimensional generalization in terms of projection-valued measures.

Can the spectrum be empty?

A closed densely-defined linear operator on a Hilbert space can have empty spectrum. For example, let H=L2[0,1] and let A=ddx on the domain consisting of absolutely continuous f∈L2 for which f(0)=0 and f′∈L2.

Is a closed operator bounded?

If one takes its domain D(A) to be C1([a, b]), then A is a closed operator, which is not bounded.

Is every self-adjoint operator bounded?

Since an everywhere-defined self-adjoint operator is necessarily bounded, one needs be more attentive to the domain issue in the unbounded case. This is explained below in more detail.