Trouble understanding relation between determinant and eigenvectors

Trouble understanding relation between determinant and eigenvectors

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!