The Unapologetic Mathematician

Mathematics for the interested outsider

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!