If $P$ is real orthogonal matrix with $\det P = -1$, prove that $-1$ is an eigenvalue of $P$. If $P$ is real orthogonal matrix with $\det P = -1$, prove that $-1$ is an eigenvalue of $P$.

can anyone help me please how can I solve this problem?
 A: Hint:
What are the possible eigenvalues of $P$ ? 
If $-1$ is not an eigenvalue, what must be the eigenvalues of $P$
? hence $\det(P)=?$
A: Belgi has already hinted at the standard way of solving this problem. More specifically, suppose $\lambda_1,\ldots,\lambda_n\in\mathbb{C}$ are the eigenvalues of $P$.


*

*Every real eigenvalue of $P$, if exists, must be equal to $-1$ or $1$.

*As $P$ is real, nonreal eigenvalues of $P$, if exist, must occur in conjugate pairs, the product of all nonreal eigenvalues of $P$ is a positive real number.

*Hence $-1$ is an eigenvalue of $P$ because $\det P$, the product of all eigenvalues of $P$, is negative.


However, there are alternative proofs as well. Here is one that makes use of Cayley transform. Suppose $-1$ is not an eigenvalue of $P$. By Cayley transform, $A=(I-P)(I+P)^{-1}$ is a well defined skew symmetric matrix such that $P=(I-A)(I+A)^{-1}$. Hence
$$
\det P=\det\left[(I-A)(I+A)^{-1}\right]=\det(I-A)^T\det(I+A)^{-1}
=\det(I+A)\det(I+A)^{-1}=1,
$$
which contradicts the condition that $\det P=-1$. Therefore $-1$ must be an eigenvalue of $P$.
A: An orthogonal matrix is diagonalizable (in an orthonormal basis, moreover) and its eigenvalues are in $\{\pm 1\}$. You see what happens if the determinant, which is the product of the eigenvalues with multiplicities, is $-1$. This means that $-1$ has odd multiplicity, in particular $\geq 1$.
