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Let ##ψ_n## ,n=1,2,... be an orthonormal basis consisting of eigenstates of a Hamiltonian operator H with non-degenerate eigenvalues ##E_n##. Let A be a linear operator which acts on the energy eigenstates ##ψ_n## as

A##ψ_1##=2##ψ_1## - i ##ψ_2##

A##ψ_2##= i ##ψ_1## + 2##ψ_2##

A##ψ_n## = 0 , n=3,4,5,... Show that A is self-adjoint

The answer is - In the basis ##ψ_n## , A can be written as a matrix having zeroes everywhere except in the left upper 2 x 2 block where A = $$\begin{pmatrix} 2 & i \\ -i & 2 \end{pmatrix}$$

This makes self-adjointness obvious.

What i'm confused about is where this matrix comes from and what does it operate on ?