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!
 A: Let denote 
$$F=\ker(A-I)$$
The important idea is that $F^\perp$ is invariant by $A$ in fact, let $x\in F^\perp$ so $\forall y\in F$ we have
$$\langle Ax,y\rangle=\langle Ax,Ay\rangle=\langle x,y\rangle=0$$
so $Ax\in F^\perp$.
Now the resul follow easily: if $u\in F^\perp$ then $Au\in F^\perp$ so $Au=\alpha u$ and since $A\in SO(3)$ then $\alpha=1$ hence  $Au=u$ and if $u\in F$ then $Au=u$ and since 
$$\mathbb R^3=F\oplus F^\perp$$
then $\forall u\in\mathbb R^3, Au=u$. Conclude.
A: It's a tricky argument, but it makes sense.  We start with

$\dim(\ker(A-I))=2$.

That is, $A$ has an eigenspace of dimension $2$.  $A$ is normal, so it has an orthonormal eigenbasis

let $u$ be the vector perpendicular to $\ker(A-I)$

$u$ is in the remaining $1$-dimensional eigenspace, so $u$ must be an eigenvector.

$\det A = 1$

The determinant of a matrix is the product of its eigenvalues, up to algebraic multiplicity. Let $\lambda_u$ be the eigenvalue for $u$. 
$$\det A=\lambda_1\lambda_2\lambda_3=(1)(1)\lambda_u$$
So, $\lambda_u=1$.
Now, $A$ has a $3$ dimensional eigenspace for $\lambda = 1$.  That is, for any vector, $A v = 1$.  $A$ has to be the identity matrix.
The same cannot be done for SO(4); notice that we can't force the final eigenvalue in the same way.
A: Let $U=ker(A-I)$ and suppose $dim(U)=2$. Since $U^{\perp}\oplus U=\Bbb E^3$ we must have $dim(U^{\perp})=1$, so for any $u\in U^{\perp}, u\neq 0,$ we must have $U^{\perp}=span(u)$. For any such $u$ and any $v\in U$ we have $\left<Au,v\right>=\left<Au,Av\right>=\left<u,v\right>=0$, so $Au\in span(u)$ and $u$ is an eigenvector of $A$ with eigenvalue $\lambda$. Since $dim(U)=2$, $A$ has $2$ linearly independent eigenvectors $v_1, v_2$ with eigenvalue $1$  and so $B=\{v_1, v_2, u\}$ is a basis of eigenvectors of $A$ such that the matrix of $A$ with respect to $B$ is $D=\left(\begin{matrix}1&0&0\\0&1&0\\0&0&\lambda\end{matrix}\right).$ But then $\lambda=detD=detA=1.$ Hence, $PAP^{-1}=D=I$ and $A=P^{-1}IP=I$.
