Suppose $t = [t_1,t_2,t_3]^T\in \mathbb R^3,t \neq 0$. Then define $$t^{\land} = \begin{bmatrix} 0 & -t_3 & t_2 \\ t_3 & 0 & -t_1\\ -t_2 & t_1 & 0\end{bmatrix}, Z= \begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}.$$ According to exponential map, there exist a corresponding rotation matrix $$R = \exp(t^{\land})$$ if $t^{\land}$ can be block diagonalized, that is $$t^{\land} = U\begin{bmatrix} aZ & 0 \\ 0 & 0\end{bmatrix}U^T,a>0$$ then we get $R = \exp(t^{\land})=U\begin{bmatrix}\cos(\|t\|_2)&\sin(\|t\|_2) & 0\\ -\sin(\|t\|_2)&\cos(\|t\|_2)&0 \\ 0 & 0 &1\end{bmatrix}U^T,a=\|t\|_2$.
(refer to R.Hartley & A.Zisserman Multiple View Geometry in Computer Vision page 583 to 584.) So how to do block diagonalization in 3*3 case? And what is $U$ like?
Update
let $w=t/\|t\|$, then we take three orthonormal basis which form a orthonormal matrix $U = [u,v,w]$.Therefore $$\begin{align} Udiag(aZ,0)U^T&=[u,v,w]\begin{bmatrix} 0 & a & 0 \\ -a & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix} u^T \\ v^T \\ w^T \end{bmatrix} \\ &=a(uv^T-vu^T) \\ &=a\begin{bmatrix} 0 & u_1v_2-u_2v_1 & u_1v_3-u_3v_1 \\ u_2v_1-u_1v_2 & 0 & u_2v_3-u_3v_2 \\ u_3v_1-u_1v_3 & u_3v_2-u_2v_3 & 0\end{bmatrix} \\ &=a(v \times u)^{\land}\end{align} $$ So above derivation leads to $$t^{\land} = a(v \times u)^{\land}$$that is $$t = a*(-w) = \|t\|(-w)$$ but it contradicts with $t = \|t\|w$, who can help?