Stabilizer of Subspace stabilizes Orthogonal Complement

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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:

$\paren {\mathbf 1 - p} t p = 0$

or:

$p t p = t p$

The goal is to show that:

$p t \paren {\mathbf 1 - p} = 0$


We have that $\tuple {a, b} \mapsto \map \tr {a b^*}$ 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 $\map \tr {x x^*} = 0$ for $x = p t \paren {\mathbf 1 - p}$.


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

\(\ds x x^*\) \(=\) \(\ds p t \paren {\mathbf 1 - p}^2 t^* p\)
\(\ds \) \(=\) \(\ds p t \paren {\mathbf 1 - p} t^* p\)
\(\ds \) \(=\) \(\ds p t t^* p - p t p t^* p\)
\(\ds \leadsto \ \ \) \(\ds \map \tr {x x^*}\) \(=\) \(\ds \map \tr {p t t^* p} - \map \tr {p t p t^* p}\)
\(\ds \) \(=\) \(\ds \map \tr {p^2 t t^*} - \map \tr {p^2 t p t^*}\)
\(\ds \) \(=\) \(\ds \map \tr {p t t^*} - \map \tr {\paren {p t p} t^*}\)
\(\ds \) \(=\) \(\ds \map \tr {p t t^*} - \map \tr {t p t^*}\)
\(\ds \) \(=\) \(\ds \map \tr {p t t^*} - \map \tr {p t^* t}\)
\(\ds \) \(=\) \(\ds \map \tr {p \paren {t t^* - t^*t} }\)
\(\ds \) \(=\) \(\ds \map \tr 0\)
\(\ds \) \(=\) \(\ds 0\)

$\blacksquare$