Fixed points of a direct isometry

Let $f$ be a direct isometry of the vector space $\mathbb{R}^3$ .
Show that it fixes at least one line going through the origin in $\mathbb{R}^3$ .

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Solution by Ancient Tree

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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).

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