Stabilizer of Subspace stabilizes Orthogonal Complement

Theorem
If a normal operator $t\colon H\to H$ 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 $p\colon H\to V$ the orthogonal projection of $H$ onto $V$. Then the orthogonal projection of $H$ onto $V^\bot$ is $\boldsymbol{1}-p$, where $\boldsymbol{1}$ is the identity map of $H$.

The fact that $t$ stabilizes $V$ can be expressed as
 * $\displaystyle (\boldsymbol{1}-p)tp=0$, or $ptp=tp$.

The goal is to show that $pt(\boldsymbol{1}-p)=0$.

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 $\operatorname{tr}(xx^*)=0$ for $x=pt(\boldsymbol{1}-p)$. This follows from a direct computation, using properties of the trace and orthogonal projections: \begin{align*} xx^* &= pt(\boldsymbol{1}-p)^2t^*p = pt(\boldsymbol{1}-p)t^*p = ptt^*p - ptpt^*p,\\ \operatorname{tr}(xx^*) &= \operatorname{tr}(ptt^*p) - \operatorname{tr}(ptpt^*p) = \operatorname{tr}(p^2tt^*) - \operatorname{tr}(p^2tpt^*) = \operatorname{tr}(ptt^*) - \operatorname{tr}((ptp)t^*) \\ &= \operatorname{tr}(ptt^*) - \operatorname{tr}(tpt^*) = \operatorname{tr}(ptt^*) - \operatorname{tr}(pt^*t) = \operatorname{tr}(p(tt^*-t^*t)) = \operatorname{tr}(0) = 0. \end{align*}