Stabilizer of Subspace stabilizes Orthogonal Complement

Theorem
If a normal operator $T$ on a finite-dimensional real or complex Hilbert space (i.e. inner product space) $H$ stabilizes a subspace $V$, then it also stabilizes its orthogonal complement $V^\bot$.

Proof
Denote by $P_V$ the orthogonal projection of $H$ onto $V$. Then the orthogonal projection of $H$ onto $V^\bot$ is $\boldsymbol{1}_H-P_V$.

The fact that $T$ stabilizes $V$ can be expressed as
 * $\displaystyle (\boldsymbol{1}_H-P_V)TP_V=0$, or $TP_V=P_VTP_V$.

The goal is to show that
 * $P_VT(\boldsymbol{1}_H-P_V)=0$.

Let $X=P_VT(\boldsymbol{1}_H-P_V)$.

Since $(A,B)\mapsto\operatorname{tr}(AB^*)$ is an inner product on the space of endomorphisms of $H$ (here $B^*$ denotes the adjoint operator of $B$), it will suffice to show that $\operatorname{tr}(XX^*)=0$. This follows from a direct computation, using properties of the trace and orthogonal projections: \[ XX^* = P_VT(\boldsymbol{1}_H-P_V)^2T^*P_V = P_VT(\boldsymbol{1}_H-P_V)T^*P_V = P_VTT^*P_V - P_VTP_VT^*P_V, \] \begin{align*} \operatorname{tr}(XX^*) &= \operatorname{tr}(P_VTT^*P_V) - \operatorname{tr}(P_VTP_VT^*P_V) = \operatorname{tr}(P_V^2TT^*) - \operatorname{tr}(P_V^2TP_VT^*) = \operatorname{tr}(P_VTT^*) - \operatorname{tr}(P_VTP_VT^*) \\ &= \operatorname{tr}(P_VTT^*) - \operatorname{tr}(TP_VT^*) = \operatorname{tr}(P_VTT^*) - \operatorname{tr}(P_VT^*T) = \operatorname{tr}(P_V(TT^*-T^*T)) = 0. \end{align*}