Proposition: Suppose that $ V $ is a complex vector space and $ \dim(V) < \infty $. Then $ T \in \mathcal{L}(V) $ is normal if and only if the orthogonal complement of every $ T $-invariant subspace is $ T $-invariant.
I hope that you can help me with a solution or a hint. Thanks.
My idea:
The forward implication: If $ T $ is normal, then $ T^{*} = p(T) $ for any polynomial $ p \in \Bbb{C}[X] $. Then given a $ T $-invariant subspace $ U $, we know that $ U $ is $ p(T) $-invariant. In other words, $ U $ is $ T^{*} $-invariant. As $ U $ is $ T^{*} $-invariant, it follows that $ W \stackrel{\text{df}}{=} U^{\perp} $ is $ (T^{*})^{*} $-invariant. Hence, $ W $ is $ T $-invariant.
I was unable to work out the backward implication.
Best Answer
Hint: Use for a normal operator $||Tv|| = ||T^*v||$. What does this tell you when you apply this on the matrix induced by $T$