I encountered the following problem and I was wondering whether it is possible to deduce even stronger conclusion that $T = \pm I$
This is the problem:
Let $T: V \to V$ be a linear transformation on a finite dimension inner-product space V and $T^*T = I$
Prove: If $W$ is $T$ invariant subspace then $W^{\bot}$ is $T$ invariant.
My approach:
Suppose $W$ is $T$ invariant.
Let $w \in W$ and let $w' \in W^{\bot}$ then $0 = <T(w),w'> =<T(w),T^*T(w')> = <T(T(w)),T(w')>$
Remember that $T^*T = I$ and $T(T(w)) \in W$
Which implies that $T(w') \in W^{\bot}$ therefore $W^{\bot}$ is $T$ invariant.
I was wondering if it's possible to conclude even further that $T = \pm I$ because I know that unitary transformations such as T rotate the whole space and if an arbitrary subspace remains T invariant then it might imply that T is the identity transformation, although I couldn't find a counterexample it seems true to m. Will appreciate any clarifications. Thanks