Do eigenvalues of a linear transformation over an infinite dimensional vector space appear in conjugate pairs? While attempting to answer a question here (namely, the finite dimensional case of the title question: Prove that if $\lambda$ is an eigenvalue of $T$, a linear transformation whose matrix representation has all real entries, then $\overline{\lambda}$ is an eigenvalue of $T$), I noticed the asker did not specify a finite dimensional vector space. Though the person who asked the question was satisfied with a finite-dimensional response, I was wondering if the analogue was true for infinite dimensional vector spaces.
I have seen several proofs of this fact relying on $V$ being finite dimensional. One proof utilizes the roots of the characteristic polynomial; if the coefficients are real then the roots come in conjugate pairs.
The second notable proof I've seen goes something like:
$$(T-\lambda I)v = 0$$
$$\overline{(T-\lambda I)v} = 0$$
$$(\overline{T} - \overline{\lambda I})\overline{v} = 0$$
$$(T- \overline{\lambda}I)\overline{v} = 0$$
where we define $\overline{T}$ as taking the conjugate of each element of the matrix representation of $T$, and we define $\overline{v}$ as conjugating each entry in the n-tuple representation of $v$ with respect to a basis. Going backwards will give you that, given the conditions set earlier, $\lambda$ is an eigenvalue if and only if $\overline{\lambda}$ is an eigenvalue, and also, $v$ is an eigenvector with eigenvalue $\lambda$ if and only if $\overline{v}$ is an eigenvector with eigenvalue $\overline{\lambda}$.
The first thing we would have to do is have some notion that is similar to the matrix representation of $T$ having all real entries. What exactly would that be? Would we have to work with infinite matrices, or (assuming the axiom of choice) could we define $T$ such that it takes basis vectors to linear combinations of basis vectors with real coefficients and that would suffice?
If we assume the axiom of choice and take a basis of $V$, I am under the impression that the second proof I provided for the finite dimensional case could extend to the infinite dimensional case. Is it necessary to use the axiom of choice for a proof, though?
Overall, my question is: 
First, is there an analog of $T$ having all real matrix entries in an infinite dimensional case? Denote this property, if it exists, $P$.
Second: Does anyone have a proof or counterexample of the following?:
Let $V$ be an infinite dimensional complex vector space, and let $T$ be a linear transformation with $P$. $\lambda$ is an eigenvalue of $T$ if and only if $\overline{\lambda}$ is an eigenvalue of $T$.
Can we also add: $v$ is an eigenvector with eigenvalue $\lambda$ if and only if $\overline{v}$ is an eigenvector with eigenvalue $\overline{\lambda}$? Whatever $\overline{v}$ may happen to mean in this case.
If we can do this without infinite matrices, infinite basis, or assuming the axiom of choice, I would much prefer that! But I understand it may be necessary.
 A: Let $V$ be a vector space over $\Bbb R$.  As elaborated in the link in the comment, let $V_{\Bbb C} = V \oplus V$ denote the complexification of $V$, in which $v + iw = v \oplus w = (v,w)$.  We define the conjugation map by
$$
J(v + iw) = v - iw
$$
For any $v,w \in V$. Note that $J$ is $\Bbb R$-linear and that for any $\lambda = a+bi$, $x = v + iw$, we have:
$$
J(\lambda x) =
J[\lambda(v+iw)] = \overline{\lambda}(v - iw) = \overline{\lambda}J(v + iw) = \overline{\lambda}J(x)
$$
which I will let you verify.  In other words, $J$ is antilinear.
Now, if $T:V \to V$ is linear, then the unique $\Bbb C$-linear extension to $V_{\Bbb C}$ is given by
$$
\tilde T(v + iw) = T(v) + iT(w)
$$
It follows that
$$
\tilde TJ(v + iw) = T(v - iw) = T(v) - iT(w) = J\tilde T(v + iw)
$$
That is, if $\tilde T:V_{\Bbb C} \to V_{\Bbb C}$ is the $\Bbb C$-linear extension of an $\Bbb R$-linear map, then $\tilde TJ = J\tilde T$.  With that, we may proceed:

Theorem: Suppose that $\tilde T:V_{\Bbb C} \to V_{\Bbb C}$ is the $\Bbb C$-linear extension of an $\Bbb R$-linear map on $V$.  If $\lambda$ is an eigenvalue of $\tilde T$ with eigenvector $v \in \Bbb V_{\Bbb C}$, then $\overline{\lambda}$ is an eigenvalue of $\tilde T$ with eigenvector $\overline{v} = Jv$.

Proof: Note that
$$
\tilde T\overline{v} = \tilde T(Jv) = J(\tilde Tv) = J(\lambda v) = \overline{\lambda}J(v) = \overline{\lambda} \overline{v}
$$
as desired.
Or, to more closely mirror your referenced proof:
$$
(T - \lambda I)v = 0 \implies\\
J(T - \lambda I)v = 0 \implies\\
(JT - J(\lambda I))v = 0\implies\\
(TJ - \overline{\lambda}IJ)v = 0 \implies\\
(T - \overline{\lambda} I)(Jv) = 0 \implies\\
(T - \overline{\lambda} I)\overline v = 0
$$
