Showing subspace of linear operator is an invariant subspace

Question

I am stuck on the proof of the statement written below. How would we go about proving this fact? It seems intuitively correct. Any ideas on how to proceed? This is the assertion I'm interested in proving:

Let $V$ be a finite dimensional inner product space (real or complex) and $T ∈ L(V)$ a normal operator ($L(V)$ denotes the space of linear operators on $V$). Let $U \subseteq V$ be a $2$-dimensional invariant subspace of $T$ . Show that $U^\perp$ is also an invariant subspace of $T$.

Answer 1

If $U$ is invariant under $T$ then we have that $T$ stablizes $U$ i.e. $P_UTP_U=TP_U$ where $P_U$ is the orthogonal projection onto $U$. From this we also have that $T$ stabilizes $U^{\perp}$. So if $v\in U^{\perp}$ is arbitrary then $$T(v)=(TP_{U^{\perp}})(v)=(P_{U^{\perp}}TP_{U^{\perp}})(v)\in U^{\perp}$$