Let $V \subset W$. If $f : V \to V$ is an operator on inner product space where $f$ is an isometry ($(fv,fu) = (u,v)$), then is it true that $f : V^\perp \to V^\perp$? In other words is it invariant under the orthogonal complement?
I would think it requires some justification that it is unitary and self-adjoint., but the first is true.
Best Answer
If $V$ is finite-dimensional, the answer is yes.
Because $f$ is injective, it is then surjective when restricted to $V$ as both domain and codomain have the same dimension.
Let $w\in V^\perp$ and $v\in V$. As $f$ is surjective considered from $V\to V$, there exists $v_0\in V$ with $v=fv_0$. But then $$ \langle fw,v\rangle=\langle fw,fv_0\rangle=\langle w,v_0\rangle=0. $$ So $f$ maps $V^\perp$ into $V^\perp$.
If we allow $V$ to be infinite-dimensional, the answer is no. Any proper isometry would do.
Finally, there is no reason for an isometry to be selfadjoint. For instance, consider $$ f=\frac1{\sqrt2}\,\begin{bmatrix}1&1\\-1&1\end{bmatrix}\in M_2(\mathbb R). $$