[proofplan]
First we record that an eigenvalue of a self-adjoint operator is real, using the [Hilbert space](/page/Hilbert%20Space) [inner product](/page/Inner%20Product) and the nonzero condition on the eigenvector. The commutation hypothesis then transports the eigenvalue equation from $\psi$ to $K\psi$, with conjugate-linearity of $K$ producing $\overline{\lambda}$ and self-adjointness reducing this to $\lambda$. Finally, the relation $K^2 = -I_H$ rules out the possibility that $K\psi$ is a scalar multiple of $\psi$.
[/proofplan]
[step:Show that the eigenvalue is real]
Let $(\cdot,\cdot)_H: H \times H \to \mathbb{C}$ denote the Hilbert space inner product on $H$, with the convention that it is linear in the first argument and conjugate-linear in the second argument. Since $A\psi = \lambda \psi$ and $\psi \ne 0$, we have $(\psi,\psi)_H > 0$. Because $A$ is self-adjoint and $\psi \in D(A)$,
\begin{align*}
(A\psi,\psi)_H = (\psi,A\psi)_H.
\end{align*}
Using linearity of the inner product in the first argument and conjugate-linearity in the second argument,
\begin{align*}
(A\psi,\psi)_H = (\lambda\psi,\psi)_H = \lambda(\psi,\psi)_H
\end{align*}
and
\begin{align*}
(\psi,A\psi)_H = (\psi,\lambda\psi)_H = \overline{\lambda}(\psi,\psi)_H.
\end{align*}
Therefore
\begin{align*}
\lambda(\psi,\psi)_H = \overline{\lambda}(\psi,\psi)_H.
\end{align*}
Since $(\psi,\psi)_H > 0$, it follows that $\lambda = \overline{\lambda}$, hence $\lambda \in \mathbb{R}$.
[/step]
[step:Transport the eigenvalue equation through the antiunitary symmetry]
The hypothesis $K(D(A)) = D(A)$ gives $K\psi \in D(A)$. Since $K$ is antiunitary, it is an isometry, so
\begin{align*}
(K\psi,K\psi)_H = (\psi,\psi)_H > 0.
\end{align*}
Thus $K\psi \ne 0$.
Using the commutation hypothesis with $\phi = \psi$ gives
\begin{align*}
A(K\psi) = K(A\psi).
\end{align*}
Since $A\psi = \lambda\psi$ and $K$ is conjugate-linear,
\begin{align*}
K(A\psi) = K(\lambda\psi) = \overline{\lambda}K\psi.
\end{align*}
By the previous step, $\overline{\lambda} = \lambda$. Hence
\begin{align*}
A(K\psi) = \lambda K\psi.
\end{align*}
Therefore $K\psi$ is a nonzero eigenvector of $A$ with eigenvalue $\lambda$.
[guided]
We want to prove that the symmetry $K$ sends the eigenvector $\psi$ to another eigenvector with the same eigenvalue. There are three points to check: the new vector lies in the operator domain, it is nonzero, and it satisfies the eigenvalue equation.
First, the domain condition is exactly the hypothesis $K(D(A)) = D(A)$. Since $\psi \in D(A)$, this gives $K\psi \in D(A)$, so the expression $A(K\psi)$ is well-defined.
Second, $K\psi$ is nonzero because $K$ is antiunitary. In particular, $K$ preserves Hilbert space norms, so
\begin{align*}
(K\psi,K\psi)_H = (\psi,\psi)_H.
\end{align*}
The vector $\psi$ is an eigenvector, hence $\psi \ne 0$, so $(\psi,\psi)_H > 0$. Therefore $(K\psi,K\psi)_H > 0$, which implies $K\psi \ne 0$.
Third, we compute the action of $A$ on $K\psi$. The commutation hypothesis says that for every $\phi \in D(A)$,
\begin{align*}
A(K\phi) = K(A\phi).
\end{align*}
Applying this with $\phi = \psi$ gives
\begin{align*}
A(K\psi) = K(A\psi).
\end{align*}
Because $\psi$ is an eigenvector with eigenvalue $\lambda$, we have $A\psi = \lambda\psi$. Since $K$ is anti-linear rather than linear, the scalar is conjugated:
\begin{align*}
K(A\psi) = K(\lambda\psi) = \overline{\lambda}K\psi.
\end{align*}
The previous step showed that $\lambda \in \mathbb{R}$, so $\overline{\lambda} = \lambda$. Substituting this into the last identity gives
\begin{align*}
A(K\psi) = \lambda K\psi.
\end{align*}
Thus $K\psi$ is a nonzero eigenvector of $A$ with the same eigenvalue $\lambda$.
[/guided]
[/step]
[step:Use $K^2 = -I_H$ to rule out linear dependence]
Assume, for contradiction, that $\psi$ and $K\psi$ are linearly dependent. Since $\psi \ne 0$ and $K\psi \ne 0$, there exists $c \in \mathbb{C}$ such that
\begin{align*}
K\psi = c\psi.
\end{align*}
Apply $K$ to both sides. By conjugate-linearity of $K$,
\begin{align*}
K^2\psi = K(c\psi) = \overline{c}K\psi = \overline{c}c\psi = |c|^2\psi.
\end{align*}
But $K^2 = -I_H$, so
\begin{align*}
K^2\psi = -\psi.
\end{align*}
Combining the two identities gives
\begin{align*}
|c|^2\psi = -\psi.
\end{align*}
Equivalently,
\begin{align*}
(|c|^2 + 1)\psi = 0.
\end{align*}
Since $|c|^2 + 1 > 0$ and $\psi \ne 0$, this is impossible. Hence $\psi$ and $K\psi$ are linearly independent over $\mathbb{C}$.
[/step]
