0
$\begingroup$

What are the eigenvalues of an (orthogonal) projection operator that projects vectors onto some hyperplane passing through the origin?

Well, for vectors $v$ orthogonal to this hyperplane, the projection kills the vector, so that $Pv = 0$, which implies $0$ is an eigenvalue.

What are the rest of the eigenvalues? As $P$ is norm-preserving, $|\lambda_i| = 1$, that I am certain of. But is the eigenvalue exactly $1$? Anymore eigenvalues? Or the eigenvalues are simply $0$ and $1$?

For reflection onto this hyperplane, any vector on this hyperplane is its own reflection, so $Rv = v$, which shows that $1$ is an eigenvalue. For the rest of the eigenvalues, it's easy to see the symmetry from the reflected vector, so that the other eigenvalue must be $-1$, since $Rv = -v = (-1)v$ for $v$ not on the hyperplane.

Any ideas are welcome.

Thanks.

$\endgroup$
2
$\begingroup$

If $P$ is a projection operator, then $P^2 = P$. So what are the possible eigenvalues of $P$?

If $R$ is a reflection operator, then $R^2 = \operatorname{id}$. So what are the possible eigenvalues of $R$?

$\endgroup$
  • $\begingroup$ ...thanks so much for the cool hints @MichaelAlbanese, so the eigenvalues for $P$ must only be 0,1 and -1,1 for $R$, from looking at the eigenvalues of $P$, $P^2$, $R$, and $R^2$. Have a great night :-) $\endgroup$ – User001 Dec 17 '15 at 22:01

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.