Stabilizer of Subspace stabilizes Orthogonal Complement

Theorem
Let $H$ be a finite-dimensional real or complex Hilbert space (that is, inner product space).

Let $t: H \to H$ be a normal operator on $H$.

Let $t$ stabilize a subspace $V$.

Then $t$ also stabilizes its orthogonal complement $V^\perp$.

Proof
Let $p: H \to V$ be the orthogonal projection of $H$ onto $V$.

Then the orthogonal projection of $H$ onto $V^\perp$ is $\mathbf 1 - p$, where $\mathbf 1$ is the identity map of $H$.

The fact that $t$ stabilizes $V$ can be expressed as:
 * $\left({\mathbf 1 - p}\right) t p = 0$

or:
 * $p t p = t p$

The goal is to show that:
 * $p t \left({\mathbf 1 - p}\right) = 0$

We have that $\left({a, b}\right) \mapsto \operatorname{tr} \left({a b^*}\right)$ is an inner product on the space of endomorphisms of $H$.

Here, $b^*$ denotes the adjoint operator of $b$.

Thus it will suffice to show that $\operatorname{tr} \left({x x^*}\right) = 0$ for $x = p t \left({\mathbf 1 - p}\right)$.

This follows from a direct computation, using properties of the trace and orthogonal projections: