Question about reflection+orthogonality

Question

One short (probably stupid) question, if anyone can unclear this to me:

Why for $α ∈ O(n)$ holds $r(α(v))= αr (v)α^{−1}$?

Here $r(v)$ is is reflection across the hyperplane consisting of all vectors orthogonal to v.

Thanks in advance.

Answer 1

Consider what both sides do to the vector $u = \alpha v$:

\begin{align} r(\alpha(v))(u) & = r( \alpha (v))\alpha (v)\\ & = \alpha (v) \\ &= u.\\ \\ \alpha r (v)\alpha^{−1}(u) &= \alpha r (v)\alpha^{−1}(\alpha(v)) \\ &= \alpha r (v)(v) \\ &= \alpha (v) = u \end{align}

Now suppose that $w$ is orthogonal to $\alpha(v)$. Then we have that $\alpha^{-1}(w)$ is orthogonal to $v$, because $\alpha$ preserves inner products. So:

\begin{align} r(\alpha(v))(w) & = -w\\ \\ \alpha r (v)\alpha^{−1}(w) &= \alpha r (v)(\alpha^{−1}(w)) \\ &= \alpha (-\alpha^{−1}(w)) \\ &= -\alpha (\alpha^{−1}(w)) \\ &= -w \end{align}

Thus the two sides agree both for vectors parallel and orthogonal to $v$, hence they agree everywhere.