Eigenvalues of complex special orthogonal matrix I've read that a matrix $P \in SO_3(\mathbb{C})$ must have an eigenvalue 1.  How do I prove this?
I understand the real case: the eigenvalues of $P \in SO_3(\mathbb{R})$ have complex modulus 1 and solve the characteristic polynomial of the matrix, which has real coefficients.  Since the characteristic polynomial is a real cubic, if it has complex roots they are a conjugate pair (multiplying to 1) and the third eigenvalue must be 1 to make all the eigenvalues multiply to $ 1 = \det P$.
Conceptually, this is the fixed axis of rotation.  It seems like the analogy should hold to the complex case, but I can't figure it out.
 A: The complex matrix $P$ satisfies $P^TP=I$ and $\det(P)=1$.
Let $\lambda$ be an eigenvalue of $P$ with eigenvector $x$. Then $\lambda\ne0$ and
$Px=\lambda x$ implies
$$
x=P^TPx = \lambda P^Tx,
$$
and $x$ is an eigenvector of $P^T$ to the eigenvalue $\lambda^{-1}$. As $P$ and $P^T$ have the same characteristic polynomial, it follows that $\lambda^{-1}$ is an eigenvalue of $P$ as well.
Let $\lambda_1,\lambda_2,\lambda_3$ be the three eigenvalues of $P$. 
Assume that $1$ is not an eigenvalue of $P$.
Assume $\lambda_1^2\ne1$. Then $\lambda_1^{-1}\ne \lambda_1$, and one of $\lambda_2,\lambda_3$ is equal to $\lambda_1^{-1}$. Wlog assume $\lambda_2=\lambda_1^{-1}$. Now $1=\det(P)=\lambda_1\lambda_2\lambda_3 = \lambda_3$. And $\lambda_3=1$. Contradiction.
It remains to consider the case $\lambda_1=-1$. 
If $\lambda_2=-1$ then $\lambda_3=1$ from $\det(P)=1$. 
Thus we can concentrate on the case $\lambda_2^2\ne1$. It follows
$\lambda_2\lambda_3=1$, and from $\det(P)=1$ we get $\lambda_2\lambda_3=-1$,
which is impossible.
Hence, the matrix $P$ has an eigenvalue $1$.
