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.


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 $1$.

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