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