Need help with proof of $SO(3)$ is path connected I am working on an exercise in Tapp's matrix groups for undergraduates. It is a proof that $SO(3)$ is path-connected. $SO(3)$ is the group
$$ SO(3) = \{A \in O(n)\mid \det A = 1 \}$$
where $O(n)$ is the group of orthogonal matrices. 
My work so far:
An element in $SO(3)$ is a matrix with columns $(p|v|p\times v)$ where $p \bot v$ and $\|p\|=\|v\|=1$. Let $A=(p_a|v_a|p_a \times v_a)$ and $B=(p_b|v_b|p_b\times v_b)$ be two elements in $SO(3)$. The goal is to find a path.
Let $R$ be the rotation of $\mathbb R^3$ that takes $p_a$ to $p_b$. Let $\varphi_R$ be a parametrisation of $R$ such that $\varphi_R(0)$ is the identity map and $\varphi_R({1\over 2})$ has rotated $p_a$ to $p_b$.
Similarly, if $R'$ is the rotation that rotates $v_a$ to $v_b$ around the axis $p_b = R(p_a)$ and $\varphi_{R'}:[{1\over 2}, 1]\to \mathbb R^3$ its parametrisation
then $\varphi_{R'}\circ \varphi_R$ is a path from $A$ to $B$.
My problem is: this proof is missing all the details but I don't know how to write it out. Could someone please show me how to write out this proof rigourously and in detail?
 A: One possible approach is to use that (1) every element has $1$ as eigenvalue,  (2) any element of $SO(3)$ that fixes a given vector $v\neq0$ also fixes the plane $v^\perp$ and is determined by its restriction to $v^\perp$, and (3) that restriction is an orientation-preserving orthogonal linear transformation of that plane, i.e., a rotation of the plane. Now given $M\in SO(3)$ find a vector $v\neq0$ fixed by $M$ using (1) and proceed to show that the set $\{\, A\in SO(3)\mid Av=v\,\}$ containing $M$ and $I$ is path connected, which will suffice. By (2) and$~$(3), restriction defines a homeomorphism of that set to the set of rotations of $v^\perp$, which is path connected, completing the proof. Proving each of (1), (2), (3) is easy.
A: I suggest using the fact that every matrix $X \in SO(3)$ is of the form $AX(\theta)A^{-1}$, where $A \in O(3)$ and $$X(\theta) = \begin{pmatrix} 1 & 0 & 0\\0 & \cos(\theta) & -\sin(\theta)\\0 &\sin(\theta) & \cos(\theta)\end{pmatrix}.$$ Then for every $X \in SO(3)$, the map $F_X : [0,1] \to SO(3)$ defined by the equation $F_X(t) = AX(t\theta)A^{-1}$ gives a path in $SO(3)$ from the identity to $X$.
