On $R^3$ let X,Y,Z be the vector fields
$X=z\frac{\partial}{\partial y}-y\frac{\partial}{\partial z}$
$Y=-z\frac{\partial}{\partial x}+x\frac{\partial}{\partial z}$
$Z=y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}$
There is a map $aX+bY+cZ \to (a,b,c) \in R^{3}$. I have shown that it is an isomorphism and that$ [U,V] \to $the cross-product of the images of U and V is also an isomorphism. But how to show the flow of aX+bY+cZ is a rotation of $R^{3}$ about some axis through 0?
So, $X$, $Y$, $Z$ are linear vector fields. If we rewrite the differential equation for each of vector field in form $\dot{\bf{\rm X}} = A \bf{\rm X}$, then we will find that $$ A_X = \left (\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{matrix} \right ),$$ $$ A_Y = \left (\begin{matrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{matrix} \right ),$$ $$ A_Z = \left (\begin{matrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right ).$$ It's not hard to see that all these matrices are skew-symmetric; moreover, they form the basis of space of $3 \times 3$ skew-symmetric matrices. Since the solution of $\dot{\bf{\rm X}} = A \bf{\rm X}$ is given by ${\bf\rm X}(t) = \exp(At) {\bf \rm X}(0)$ and the exponent of skew-symmetric matrice is orthogonal matrice, we get that for each $t$ we have a family of orthogonal transformations of $\mathbb{R}^3$. It's easy to show that ${\rm det} \exp(At) = 1$ and so we have that $\exp(At)$ is rotation.
Very late addition made for a sake of completeness
If we take a vector field $a\cdot A_X + b \cdot A_Y + z \cdot A_Z$, it's very easy to show (by direct check) that any point $(ta, tb, tc)$, $t\in \mathbb{R}$, is a steady point of the flow generated by this vector field. So, it's very clearly explains what these coefficients really define: the axis of rotation. Base vector fields $A_X$, $A_Y$ and $A_Z$ are just rotations along $X$, $Y$ and $Z$ axis respectively.