Question

Find the eigenvalues of an involutive operator, that is, an operator a with the property a^2 = 1.

A) λ = 1
B) λ = -1
C) λ = 0
D) λ = 2

Answer 1

The eigenvalues of an involutive operator, one that satisfies a^2 = 1, are λ = 1 and λ = -1 because these are the only values that satisfy the involutive property when squared. Therefore, the eigenvalues of the involutive operator are λ = 1 and λ = -1.

Step-by-step explanation:

The eigenvalues of an involutive operator, which is an operator a characterized by the property that a2 = 1, can be found by considering an eigenvector v such that av = λv. Applying the operator twice, we get a2v = λ2v = v. This indicates the eigenvalue squared must be 1.

Considering the potential solutions: λ = 1, λ = -1, λ = 0, and λ = 2, only λ = 1 and λ = -1 yield λ2 = 1.

Therefore, the correct eigenvalues are λ = 1 and λ = -1. This rules out λ = 0 and λ = 2 as they do not satisfy the condition when squared.

An involutive operator is an operator that satisfies the equation a^2 = 1. To find the eigenvalues of an involutive operator, we need to solve the equation a^2 - 1 = 0. Let λ represent the eigenvalues. Setting up the equation, we have (λ^2) - 1 = 0. Factoring, we get (λ + 1)(λ - 1) = 0. Therefore, the eigenvalues of the involutive operator are λ = 1 and λ = -1.