Prove/Disprove an inner product on a complex linear space restricted to its real structure is also an inner product Let $V$ be an $n$-dimensional linear space and $(\cdot, \cdot)$ be an inner product on it. Define the conjugation map $\sigma: V \to V$ such that for any $\alpha, \beta \in V$ and $\lambda \in \mathbb{C}$, $\sigma(\alpha + \beta) = \sigma(\alpha) + \sigma(\beta)$, $\sigma(\lambda\alpha) = 
\bar{\lambda}\sigma(\alpha)$, $\sigma^2(\alpha) = \alpha$. The space
\begin{align*}
R_\sigma(V) = \{\alpha \in V: \sigma(\alpha) = \alpha\}
\end{align*}
is known as the real structure of $V$. In the link, it has been shown $R_\sigma(V)$ is an $n$-dimensional real linear space. I felt the inner product $(\cdot, \cdot)$, which is originally defined on $V$, when restricted to $R_\sigma(V)$,
is also an inner product.
While the positiveness and bilinearity are easy to prove, it seems difficult to show the symmetry. In particular, how to show $(\alpha, \beta)$ is a real number when $\alpha, \beta \in 
R_\sigma(V)$?
 A: This is not true in general.  For instance, let $V=\mathbb{C}^2$.  The vectors $v=(1,0)$ and $w=(i,1)$ are a basis for $V$, and there is a unique inner product $(\cdot,\cdot)$ on $V$ for which they are orthonormal.  Now let $\sigma:V\to V$ be given by $\sigma(a,b)=(\overline{a},\overline{b})$.  This is a conjugation, and $R_\sigma(V)=\mathbb{R}^2$.  However, the inner product is not real-valued when restricted to $\mathbb{R}^2$: for instance, $v=(1,0)$ and $w-iv=(0,1)$ are both in $\mathbb{R}^2$, but $(v,w-iv)=i$ since $v$ and $w$ are orthonormal.
A: It is entirely possible that $(\alpha,\beta)\notin\Bbb R$ for some $\alpha,\beta\in R_\sigma(V)$. Since $\sigma$ is unrelated to $(\cdot,\cdot)$ and it only provides additional structure, but no restrictions, to the $\Bbb C$-linear structure of the space, you could take any $\Bbb C$-vector space $V$ of dimension $\ge2$ already endowed with a conjugate map $\sigma$, select a $\Bbb C$-basis $\{b_1,\cdots, b_n\}$ of $V$ made of vectors in $R_\sigma(V)$, and then endow $V$ with a positive-definite Hermitian product such that $(b_1,ib_2)=2$ (which you can do as freely as you can devise positive-definite Hermitian matrices).
