Show that a linear combination of the degenerate eigenvectors of an operator belonging to a particular eigenvalue of the operator is also an eigenvector belonging to the same eigenvalue.

In quantum mechanics, consider an operator A^\hat{A} with eigenvectors ψ1,ψ2,…,ψn\psi_1, \psi_2, \dots, \psi_n corresponding to a degenerate eigenvalue λ\lambda. This means that for each eigenvector ψi\psi_i, we have:

A^ψi=λψifori=1,2,…,n.\hat{A} \psi_i = \lambda \psi_i \quad \text{for} \quad i = 1, 2, \dots, n.

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