# If $A$ is an $n×n$ orthogonal matrix with $\text{det}(A)=1$ and $n$ odd, then show that $\text{det}(I-A)=0$.

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?

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.