# Eigenvalue of a vector in a subspace [closed]

Consider, a quantum system has a hamiltonian with eigenstates $$\{|\phi_1\rangle,|\phi_2\rangle,|\phi_3\rangle\}$$ and associated eigenvalues $$\{\lambda_a,\lambda_a,\lambda_b\}$$. My notes state that any vector in the subspace $$\{|\phi_1\rangle,|\phi_2\rangle\}$$ has the corresponding eigenvalue $$\lambda_a$$.

This seems like an obvious statement, but would like to know how I could prove it (if there is a way to do so).

Suppose we have an arbitrary state $$|\phi\rangle$$ in the subspace $$\{|\phi_1\rangle,|\phi_2\rangle\}$$. What this means is that we can write $$|\phi\rangle$$ as a superposition (linear combination) of these two states. For example, $$|\phi\rangle = a|\phi_1\rangle + b|\phi_2\rangle$$ for which $$|a|^2 + |b|^2 = 1$$.
Now, we can find the eigenvalue of $$|\phi\rangle$$ by applying the Hamiltonian, as follows:
$$H|\phi\rangle = H(a|\phi_1\rangle + b|\phi_2\rangle) = aH|\phi_1\rangle + bH|\phi_2\rangle = a\lambda_a|\phi_1\rangle + b\lambda_a|\phi_2\rangle = \lambda_a(a|\phi_1\rangle + b|\phi_2\rangle) = \lambda_a|\phi\rangle$$
So, we have shown that any state in the subspace $$\{|\phi_1\rangle,|\phi_2\rangle\}$$ has eigenvalue $$\lambda_a$$.