# When is a linear operator invariant for a orthogonal decomposition?

Let $f: V \to V$ be a linear map and $V = U \oplus U^\perp$ be an inner product space.

For its restriction map $f|_U$, is it invariant under $U$? And thus similarly for $f|_{U^\perp}$?

I guess one would try to see where the inner product $( ,)$ would map $f(u)$, but I don't seem to be able to get a good result.

If $f$ is self-adjoint, i.e. $$\langle f(x),y\rangle=\langle x,f(y)\rangle$$ for all $x,y\in V$, then as soon as you have that $f(U)\subseteq U$ then $f(U^\perp)\subseteq U^\perp$ as well.
On the other hand, there's no condition based on the scalar product that will imply that $f(U)\subseteq U$.
When $\langle fu,v\rangle=0$ for all $u\in U$, $v\in U^\perp$ then $fu\in U$ so it leaves $U$ invariant (and similarly for the complement). Does this answer your question?
• The question is when the condition $(fu, v) = 0$ would hold. Aug 24, 2016 at 21:53