Do eigenvalues of a linear transformation over an infinite dimensional vector space appear in conjugate pairs?

Solution 1:

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: $$ (T - \lambda I)v = 0 \implies\\ J(T - \lambda I)v = 0 \implies\\ (JT - J(\lambda I))v = 0\implies\\ (TJ - \overline{\lambda}IJ)v = 0 \implies\\ (T - \overline{\lambda} I)(Jv) = 0 \implies\\ (T - \overline{\lambda} I)\overline v = 0 $$