Let $T : \mathbb R^n \to \mathbb R^n$ be a dot product-preserving transformation, and let $v_1$, $v_2$ be eigenvectors with eigenvalues $\lambda_1$, $\lambda_2$. If $\lambda_1 \neq \lambda_2$, can we conclude $v_1 \cdot v_2 = 0$?
Here is some partial progress:
First, note that if $v$ is any eigenvector with eigenvalue $\lambda$, then $$||v||^2 = v \cdot v = Tv \cdot Tv = \lambda v \cdot \lambda v = \lambda^2 ||v||^2.$$
Since $v \neq 0$, we deduce $\lambda^2 = 1$. Hence $\lambda = \pm 1$.
Since $\lambda_1 \neq \lambda_2$ implies, WLOG, $\lambda_1 = 1$ and $\lambda_2 = -1$, we have
$$v_1 \cdot v_2 = Tv_1 \cdot Tv_2 = \lambda_1 v_1 \cdot \lambda_2 v_2 = \lambda_1 \lambda_2 (v_1 \cdot v_2) = -(v_1 \cdot v_2),$$
so $v_1 \cdot v_2 = 0$.
But there's no way this is that simple…
Best Answer
Your argument is correct. A real orthogonal transformation has only the possible real eigenvalues $\pm 1$ and the orthogonality follows.
Note that there are orthogonal transformations which do not have real eigenvalues, $\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}$, so in the complex case your argument becomes slightly different (as we get $|\lambda|^2 = 1$ from the invariance w.r.t. the complex inner product), but the argument for the orthogonality stay similar.
Addendum: It's not very surprising that it's simple. Do you know the proof for orthogonality of eigenvalues of symmetric matrices? This one is very similar. In fact symmetric and orthogonal matrices are closely connected (generally speaking hermitian and unitary matrices, respectively): As $z \mapsto \frac{z-i}{z+i}$ maps the real line onto the complex unit circle, the mapping $A \mapsto (A-iI)(A+iI)^{-1}$ maps hermitian matrices onto the set of unitary matrices.