Prove the eigenvectors of a reflection transformation are orthogonal This is part of a problem to prove that all reflection transformations are diagonalizable. For $T : \mathcal{V} \to \mathcal{V} \iff T^2 = I$, I've shown (1) that cases where $T$ has only one eigenvalue, $T$ is $I$ or $-I$, and is thus already diagonal; and (2) that when it has two eigenvalues, we can construct eigenvectors for the only two possible eigenvalues, $\lambda = \pm 1$, using the equations
$$ v + Tv $$
$$ v - Tv $$
Where $v \in \mathcal{V}$. I'd like to prove that we can feed these equations an existing orthogonal basis for $\mathcal{V}$, say the standard basis $\{e_1, ..., e_n\}$, and (between the two of them) always get an orthogonal basis out, thus also showing that $\mathcal{V}$ is a direct sum of its two eigenspaces $E_1$ and $E_{-1}$, and making $T$ diagonalizable.
Now, maybe I'm overlooking a simple way to show orthogonality, but using inner products I get:
$$ \langle e_i + Te_i, \, e_i - Te_i \rangle = \langle e_i, e_i \rangle - \langle e_i, \, Te_i \rangle + \langle Te_i, \, e_i \rangle - \langle Te_i, \, Te_i \rangle$$
Over the reals, this is:
$$ \langle e_i, \, e_i \rangle - \langle Te_i, \, Te_i \rangle $$
I don't know how to handle the last term. Does this equal zero, showing orthogonality?
 A: The proposition as stated in the title of your question is patently false. All vectors are eigenvectors of the identity map, so it’s obvious that there are at least two that are not orthogonal. If $\dim\mathcal V\gt2$, the same argument applies to “proper” reflections: at least one of the eigenspaces will be at least two-dimensional.  
However, what it looks like you’re trying to prove is essentially that all reflections are orthogonally diagonalizable, which unfortunately is also false. Consider $$T:\mathbb R^2\to\mathbb R^2 \\ T: \mathbf v\mapsto\begin{bmatrix}1&-2\\0&-1\end{bmatrix}.$$ The two eigenspaces of $T$ are spanned by $(1,0)^T$ and $(1,1)^T$, respectively, which are clearly not orthogonal. $\mathbb R^2$ is the direct sum of the two eigenspaces, which is what it looks like you’re really after, but you’re not going prove that by showing that the eigenspaces are orthogonal complements relative to some fixed inner product.
As mentioned in one of the comments to your question, it would appear that there are some missing conditions (or that you’re going about it the wrong way. Hint: write $\mathbf v=\frac12(\mathbf v+T\mathbf v)+\frac12(\mathbf v-T\mathbf v)$).  
If you’ve learned about minimal polynomials, it’s quite easy to prove that reflections are diagonalizable using them: We have $T^2=I$, so the minimal polynomial must divide $\lambda^2-1$. There are only three possibilities, each of which is the product of distinct linear factors.
