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.
 A: 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 $\neq 2$.
A: We know that $\|Ax\|^2=\langle Ax,Ax\rangle=\langle AA^Tx,x\rangle=\|x\|^2$.  Since $n$ is odd, we know that there’s at least one zero of the characteristic polynomial of $A$, name it $\lambda$ and let $v$ be a correspondent eigenvector.  Then
$$\|v\|=\|Av\|=\|\lambda v\|=|\lambda|\|v\|,$$
hence $\lambda=\pm1$, that is, any real eigenvalue is either $1$ or $-1$.  
Since $\det(A)$ is the product of all eigenvalues of $A$ and that determinant equals $1$, the multiplicity of eigenvalue $-1$ must be even, say $2k$. Then the characteristic polynomial factorized over the real numbers is $(\lambda+1)^{2k}p(\lambda)$ where $p$ has an odd degree and therefore a zero, which must be $1$. That means $\det(A-1\cdot I)=0$.
A: An orthogonal $n\times n$ matrix is certainly diagonalizable over the complex numbers, being normal. However, its eigenvalues may be complex.
Your task is to prove that in the case of odd $n$, $1$ is an eigenvalue.
Since $A$ is a real matrix, its nonreal eigenvalues come in pairs $\lambda$ and $\bar{\lambda}$. So, assume the eigenvalues (possibly repeated) are
$$
\lambda_1,\dots,\lambda_k,
\lambda_{k+1},\lambda_{k+2}=\bar{\lambda}_{k+1},\dots,
\lambda_{n-1},\lambda_n=\bar{\lambda}_{n-1}
$$
where the first $k$ are real. Note that $k$ is odd. Also the module of every eigenvalue is $1$, so we have
$$
\det A=\lambda_1\dots\lambda_k
\lambda_{k+1}\lambda_{k+2}\dots\lambda_{n-1}\lambda_n=
\lambda_1\dots\lambda_k
$$
If all the real $\lambda_i$ were $-1$, for $i=1,2,\dots,k$, we'd have $\det A=(-1)^k=-1$.
