0
$\begingroup$

I have a linear operator $\Phi$ in $\mathbb{R}^n$ and have created another operator that I believe to be its adjoint (transpose) $\Phi^T$.

What is the most direct way to verify that my $\Phi^T$ is indeed correct without having access to the matrix representation of $\Phi$ or $\Phi^T$?

$\endgroup$
3
$\begingroup$

Compute $(\Phi x, y)$ and $(x, \Phi^T y)$, they should be the same.

$\endgroup$
  • $\begingroup$ Oops, I mean not classical, just regular adjoint---just the direct transpose. $\endgroup$ – Steve Dec 7 '12 at 0:24
  • $\begingroup$ I'm not sure what you mean. You're dealing with a real vector space so adjoint and transpose are the same. $\endgroup$ – Travis Dec 7 '12 at 18:56

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.