0

Primarily, here $A$ can be either real or complex matrix and as $A$ is orthogonal, hence all the eigenvalues of $A$ are of the form $e^{i\Delta}$ for $i^2=-1$ and real $\Delta$.

I tried to show it by induction but I couldn't prove for the $n=2$ case that $1$ is an eigenvalue of $A$. Can someone please provide me with a short hint?

Rabi Kumar Chakraborty
  • 2,193
  • Can someone please edit the text using Latex (as I am new in learning it) ??? – Rabi Kumar Chakraborty Jul 25 '19 at 18:24
  • 5
    Worse than lacking LaTeX, you seem to mean "odd" when you say "even", as otherwise there are simple counterexamples. See https://math.stackexchange.com/questions/3114533/if-q-is-a-proper-orthogonal-transformation-matrix-deduce-that-det1-q-0 . – darij grinberg Jul 25 '19 at 18:27

1 Answers1

0

Every $n \times n$ orthogonal matrix $M$ with determinant $1$ defines an orientation-preserving rotation on the $n-1$ sphere $S^{n-1} \subset \mathbb{R}^n$, and this rotation has a fixed point at $x$ iff $x$ is an eigenvector of $M$ with eigenvalue $1$. The problem statement is thus a consequence of the well-known topological fact that every diffeomorphism from an even-dimensional sphere to itself of degree different to $-1$ has a fixed point.

Connor Harris
  • 7,070