Understanding a consequence of Euler's Rotation Theorem I have just finished studying the matrix proof of Euler's rotation theorem as stated in the wikipedia page here, which states: 

In three-dimensional space, any displacement of a rigid body such that
  a point on the rigid body remains fixed, is equivalent to a single
  rotation about some axis that runs through the fixed point.

The Matrix proof essentially takes an arbitrary $3 \times 3$ rotation matrix with real entries and shows using the identities $det(-A)=-det(A)$ (true for any $3 \times 3 $ matrix) and $det(R^{-1})=1$ that $det(R-I)=-det(R-I)$ therefore $\lambda=1$ is a root of the characteristic equation of $R$. 
And so $R-I$ is singular and has a non-zero kernel. Therefore there is at least one vector $n\neq 0$ with $Rn=n$. The author of the article states that this is equivalent to the first statement, which I am not sure why this is true, since it seems that all we have shown is that rotation matrices are invariant for some line, i.e their line of rotation .
In another context, I have seen it said that Euler's rotation theorem shows that any rotation may be described using three angles. I'm not sure how this statement follows from the former statement. According to Wolfram Alpha, here, a general rotation $A$ (about a line through the origin) can be written as a product of three rotation matrics $D,C,B$, i.e. $A=DCB$. Any insights much appreciated.
 A: Euler's theorem states that for any rotation about the origin, there is an axis through the origin which is unchanged by the rotation. In vector notation, any axis through the origin has the form $\{ t v\mid t \in \Bbb R\}$ for some vector $v$. So if $R$ is a rotation matrix, this means that there is some $v$ with $R(tv) = tv$ for all $t$, in particular it is true when $t = 1$: $Rv = v$. That equation means that $v$ is an eigenvector of $R$ with eigenvalue $1$.
Conversely, if $R$ is any matrix with $1$ as an eigenvalue, then there is some non-zero $v$ with $Rv = v$, and therefore $R(tv) = tv$ for all real numbers $t$. That is, $R$ leaves the line $\{tv\mid t \in \Bbb R\}$ unchanged.
So, yes, Euler's theorem in matrix form is equivalent to saying that rotation matrices have $1$ as an eigenvalue.
Now, any multiple of an eigenvector is also an eigenvector for the same eigenvalue, so we can take the vector $v$ to have norm $1$. I.e., to lie on the unit sphere. The unit sphere is a 2-dimensional surface - points on it can be identified by just 2 parameters, for example, the angles of latitude and longitude. A rotation moves all points not on the axis of rotation through the same angle with respect to that axis. So once you know the axis of rotation - the eigenvector - the rotation is completely specified by that angle of rotation.
So the rotation is completely determined by the angles of latitude and longitude, and the angle of of rotation.
