6 orthonormal subspace question

[nhrock3]

6)there is normal T in unitarian final space.
$$v\neq0,v\in V$$ prove that if $$\{sp(v)\}^{\perp}$$ is T variant then
v is eigenvector of T
?
hint:prove that $$T*(v)$$ is orthogonal to $$\{sp(v)\}^{\perp}$$
what i have done:
suppose $$u\in\{sp(v)\}^{\perp}$$
we take the definition of $$T*$$
(Tu,v)=(u,T*v)
(Tu,v)=0 because $$\{sp(v)\}^{\perp}$$ is T variant so T*v is orthogonal
to u.
i know also
V=$$\{sp(v)\}^{\perp}\oplus\{sp(v)\}$$
what to do next ,how to prove the actual question?

 

[lippyka]

Thanks for help.Answer:Since \{sp(v)\}^{\perp} is T-invariant, for any u in \{sp(v)\}^{\perp}, we have Tu = u. Therefore, (Tu,v) = (u,T*v) = 0. This implies that T*v is orthogonal to \{sp(v)\}^{\perp}. Since V = \{sp(v)\}^{\perp}\oplus\{sp(v)\}, it follows that T*v must belong to \{sp(v)\}. Therefore, T*v = \lambda v for some scalar \lambda. This proves that v is an eigenvector of T with eigenvalue \lambda.