0
$\begingroup$

Let $f: V \to V$ be a linear map and $V = U \oplus U^\perp$ be an inner product space.

For its restriction map $f|_U$, is it invariant under $U$? And thus similarly for $f|_{U^\perp}$?

I guess one would try to see where the inner product $( ,)$ would map $f(u)$, but I don't seem to be able to get a good result.

$\endgroup$

2 Answers 2

1
$\begingroup$

If $f$ is self-adjoint, i.e. $$ \langle f(x),y\rangle=\langle x,f(y)\rangle $$ for all $x,y\in V$, then as soon as you have that $f(U)\subseteq U$ then $f(U^\perp)\subseteq U^\perp$ as well.

On the other hand, there's no condition based on the scalar product that will imply that $f(U)\subseteq U$.

$\endgroup$
2
0
$\begingroup$

When $\langle fu,v\rangle=0$ for all $u\in U$, $v\in U^\perp$ then $fu\in U$ so it leaves $U$ invariant (and similarly for the complement). Does this answer your question?

$\endgroup$
2
  • $\begingroup$ The question is when the condition $(fu, v) = 0$ would hold. $\endgroup$ Aug 24, 2016 at 21:53
  • $\begingroup$ You may have a particular example in mind? If in finite dimension you may perhaps check it on basis vectors for the splitting? $\endgroup$
    – H. H. Rugh
    Aug 24, 2016 at 21:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.