Fixed points of a direct isometry

The characteristic polynomial of $f$ is a polynomial of degree 3 with real coefficients. Therefore, it admits at least one real root.

But the (complex) eigenvalues of a linear isometry have magnitude 1. So the real root can only be $+1$ or $- 1$.
There are two cases :

• Either there is a complex eigenvalue $z$, but then its complex conjugate $\bar{z}$ is also an eigenvalue (because the characteristic polynomial is with real coefficients). The determinant of $f$, which is 1 by definition of a direct isometry, is the product of the eigenvalues, but because the product of a complex number with its conjugate is $z\bar{z}=|z|^2=1$, then the only choice for the remaining eigenvalue is 1.
• Either all eigenvalues are real, so they must . In this case, again because of the determinant of $f$ being 1, there must be an even number of $-1$ eigenvalues ; as there are 3 of them, at least one of them is 1.

This shows that in any case, 1 is an eigenvalue, which means that there are eigenvectors $x$ such that $f(x)=x$, which is the definition of a fixed point. The isometry fixes at least the line $\operatorname{Span}(x)$.

By the way, this shows that direct isometries of $\mathbb{R}^3$ are rotations, with axis given as the fixed points (except for the identity which fixes all points).