eigenvalues/eingenspaces of orthogonal matrices let $A\in SO(3,\mathbb{R})=\{A \in \mathbb{R}^{3\times3} | AA^{T}=I_3 \;\;\text{and}\;\;\text{det}A=1\}$ 
1.show that $1$ is an eigenvalue of multiplicity either $1$ or $3$
2.show that $\text{dim}E_1=1$ where $E_1$ is the eigenspace associated with the eigenvalue $\lambda=1$
my attempt : 
1.since $\text{det}A=1$ and $A$ is a $3$ by $3$ matrix it's got atleast $3$ complex eigenvalues
if those eigenvalues are real then clearly $1$ is an eigenvalue of multiplicity either $1$ or $3$ well if they're complex I don't know how to show it (need help here)
2.let's assume $\text{dim}E_1=3$ which means that $E_1=\mathbb{R^3}$ which is a contradiction so $\text{dim}E_1\neq3$
from here I'm stuck as well because $\text{dim}E_1$ can still be $2$ if $1$ is an eigenvalue of multiplicity $3$
any help or hints concerning both questions will be greatly appreciated. thanks! 
 A: The eigenvalues of an $n$-by-$n$ orthogonal matrix $A$ all have absolute value $1$.
The non-real ones come in complex conjugate pairs, and so there are
evenly many of them and their product is $1$. So if $a$ and $b$ are the multiplicities of the eigenvalues $1$ and $-1$, then $a+b\equiv n \pmod 2$
and $\det A=(-1)^b$.
A: This is for the first part ...
First use the property $A A^{T} = I$. For any vector $v$, we have
$$
\|A v\|^{2} = (A v)^{T} (A v)
=
v^{T} A^{T} A v
=
v^{T} v = \|v\|^{2}
$$
In particular, say $v$ is an eigenvector of $A$ with eigenvalue
$\lambda$, then
$$
Av = \lambda v
\Rightarrow
\|A v \|^{2} = \|\lambda v\|^{2} 
\Rightarrow
\|A v \|^{2}= |\lambda|^{2} \|v\|^{2}
$$
Combining both the above equations gives
$$
\|v\|^{2} = |\lambda|^{2} \|v\|^{2} \textrm{ so }
|\lambda| = 1
$$
I.e. any eigenvector will have magnitude (modulus) 1.
Now, let $p(\lambda)$ be the characteristic polynomial of $A$, it will be cubic. The roots of $p(\lambda)=0$, let's call them $\lambda_{1}$, $\lambda_{2}$, $\lambda_{3}$ are the eigenvalues of $A$.
Because $\det(A)=1$, we have $\lambda_{1} \lambda_{2} \lambda_{3} = 1$
A cubic equation either has three real roots or it has one real root and two complex roots.
For the case of three real roots, we can solve this with either 
$\lambda_{1} = \lambda_{2} = \lambda_{3} = 1$
or 
$\lambda_{1} = 1$ and $ \lambda_{2} = \lambda_{3} = -1$
For the case of one real root, assume w.l.o.g that $\lambda_{1}$ is real. The complex roots $\lambda_{2,3}$ will be conjugates of each other (they are roots of a quadratic) so we can write $\lambda_{3} = \lambda_{2}^{*}$ 
$$
\lambda_{1} \lambda_{2} \lambda_{3} = 1
\Rightarrow
\lambda_{1} \lambda_{2} \lambda_{2}^{*} = 1
\Rightarrow
\lambda_{1} | \lambda_{2} |^{2} = 1
\Rightarrow
\lambda_{1} = 1
$$
So in all cases, we have $\lambda_{1} = 1$, either with multiplicity 1 or multiplicity 3.
Can you continue from here?
