Linear-algebra first course problem about orthogonal matrices I am trying to demonstrate next assert about matrices:
$A$ is a matrix of $n$ order, with $n$ odd, that obeys $A A^T =I$ and $\det\, A=1$. Then $\det\,(A-I)=0$.
I have tried a number of things but none of them work. That $n$ is odd seems to indicate to the trace of the matrix an its properties but I have also tried to find a product of matrix useful but it did not work out.
All ideas well be apreciated.

We have $\left(-1\right)^n = -1$ (since $n$ is odd). But
\begin{equation}
\left(A-I\right)A^T = \underbrace{AA^T}_{=I} - A^T = I-A^T = \left(I-A\right)^T .
\end{equation}
Taking determinants of both sides of this equality, we obtain
\begin{align}
\det\left(\left(A-I\right)A^T\right)
&= \det\left(\left(I-A\right)^T\right) = \det\left(\underbrace{I-A}_{=-\left(A-I\right)}\right) \\
& = \det\left(-\left(A-I\right)\right) = \left(-1\right)^n \det\left(A-I\right) = - \det\left(A-I\right)
\end{align}
(since $\left(-1\right)^n = -1$).
Thus,
\begin{align}
-\det\left(A-I\right) = \det\left(\left(A-I\right) A^T\right) = \det\left(A-I\right) \cdot \underbrace{\det \left(A^T\right)}_{= \det A = 1} = \det\left(A-I\right) ,
\end{align}
so that $0 = 2 \cdot \det\left(A-I\right)$ and thus $\det\left(A-I\right) = 0$, assuming that your matrices are over a field of characteristic $\neq 2$.
