Invariant Subspaces of Self-Adjoint Transformations

Okay, today I want to nail down a lemma about the invariant subspaces (and, in particular, eigenspaces) of self-adjoint transformations. Specifically, the fact that the orthogonal complement of an invariant subspace is also invariant.

So let’s say we’ve got a subspace $W\subseteq V$ and its orthogonal complement $W^\perp$ . We also have a self-adjoint transformation $S:V\rightarrow V$ so that $S(w)\in W$ for all $w\in W$ . What we want to show is that for every $v\in W^\perp$ , we also have $S(v)\in W^\perp$

Okay, so let’s try to calculate the inner product $\langle S(v),w\rangle$ for an arbitrary $w\in W$ .

$\displaystyle\langle S(v),w\rangle=\langle v,S(w)\rangle=0$

since $S$ is self-adjoint, $S(w)$ is in $W$ , and $v$ is in $W^\perp$ . Then since this is zero no matter what $w\in W$ we pick, we see that $S(v)\in W^\perp$ . Neat!

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August 11, 2009 - Posted by John Armstrong | Algebra, Linear Algebra

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[...] (how?). The subspace is then invariant under the action of . But then the orthogonal complement is also invariant under . So we can restrict it to a transformation [...]

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