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.



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 $R$?

  • $\begingroup$ ...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

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