The Unapologetic Mathematician

Mathematics for the interested outsider

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!


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

1 Comment »

  1. […] (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 […]

    Pingback by The Real Spectral Theorem « The Unapologetic Mathematician | August 14, 2009 | Reply

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: