I've found a vague note in my geometry textbook, it has more to do with linear algebra however so that is why I post it with this tag.

Suppose we are working in $\mathbb{E}^3$. And that $A \in SO(3)$ is the linear part of the isometry $F(x)= Ax +b$. We now want to show that if $A \neq I$ - where $I$ is the identical matrix- then $dim( ker (A-I))=1$. This is the argument that is given in my textbook.
"Suppose $dim (ker (A- I) )= 2$, and let $u$ be a vector perpendicular to $ker(A-I)$. Then $u$ also is an eigenvector of A and since $det A =1$, the eigenvalue associated with $u$ is 1 (Why is this the case?). Wich implies that $A=I$. (I don't get this either). "

I was wondering can something similar be done for $A \in SO(4)$?

I hope you guys can help because I really can't see it... Thanks!