# 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.