Let $k$ be a field such that $\mathrm{char}\;k \neq 2$. Let $V$ be a $k$-vector space equipped with a bilinear form $p : V \times V \to k, p:(v,u) \mapsto \langle v, u\rangle$. I say that $V$ is non-singular if for every vector $v \in V$ such that $v\neq 0$ there are $u \in V$ such that $\langle v,u\rangle \neq 0$. Note, that there still can exist a non-zero $v\in V$ such that $\langle v,v\rangle = 0$. For example, consider $p$ defined by the matrix
$$ p \cong \left[ \begin{array}[cc] 00 & 1 \\ 1 & 0 \end{array} \right] . $$
Define the isometry group $\mathbf{Iso}(V)$ as a group of all invertible linear operators $T$ such that $\langle Tv, Tv \rangle = \langle v, v \rangle$. It is known that for a normal Euclidean $\mathbb{R}^n$, all eigenvalues of the operator $T \in \mathbf{Iso}(\mathbb{R}^n) = \mathbf{O}(\mathbb{R},n)$ belong to the set $\{-1,1\}$. Is this true that, in general, for every nonsingular $V$ and $T \in \mathbf{Iso}(V)$, I will have eigenvalues of $T$ in $\{-1,1\}$? Obviously, the classical proof translates to any anisotropic space.
My work so far: if $\lambda$ is an eigenvalue of $T \in \mathbf{Iso}(V)$, I can find an eigenvector $v \neq 0$, such that $T(v) = \lambda v$. It is also possible to find $w = 0$ such that $\langle v,w \rangle \neq 0$, as $V$ is non-singular. Let $u = T^{-1}w$. Then,
$$ \lambda\langle v, w \rangle = \langle Tv, w \rangle = \langle Tv, Tu \rangle = \langle v, u \rangle. $$
So, $ v \bot \lambda w - u $. But I don't know how to use this fact. I also failed to construct any counter-example.
The notation $O$ for an orthogonal group is reserved for the case that the bilinear form is symmetric; if it's skew-symmetric you get a symplectic group, and if it's neither you get a mix.
Here's a counterexample which is symmetric. Let $V$ be a finite-dimensional vector space over $k$ and consider $W = V \oplus V^{\ast}$ equipped with the bilinear form
$$B((v_1, f_1), (v_2, f_2)) = f_2(v_1) + f_1(v_2).$$
Then $GL(V)$ embeds into $O(W)$.