I am working through the book 'Quaternions and Rotation Sequences' by Jack B. Kuipers. In section 3.5, the fixed axis of rotation for a generic rotation matrix $A$ for $\mathbb{R}^3$ is derived as follows:
...to determine the rotation axis $v$...for any rotation operator ($A$), it remains that we must solve:
$[A-I]\mathbf{v} = \mathbf{0}$
which upon expansion gives the homogeneous equations
$(a_{11}-1)v_1 + a_{12}v_2 + a_{13}v_3 = 0$
$a_{21}v_1 + (a_{22}-1)v_2 + a_{23}v_3 = 0$
$a_{31}v_1 + a_{32}v_2 + (a_{33}-1)v_3 = 0$The standard technique is to set, say, $v_3=1$ and then use any of the two equations to solve for $v_1$ and $v_2$. We use:
$a_{21}v_1 + (a_{22}-1)v_2 = -a_{23}$
$a_{31}v_1 + a_{32}v_2 = -(a_{33}-1)$
The author then immediately arrives at the following possible expression for the composite rotation axis:
$v_1 = a_{12}a_{23} - (a_{22}-1)a_{13}$
$v_2 = a_{21}a_{13} - (a_{11}-1)a_{23}$
$v_3 = (a_{11}-1)(a_{22}-1) - a_{12}a_{21}$
I am most likely missing something fairly obvious here, but how is the solution for $v_1$, $v_2$, and $v_3$ obtained in this case?
Assuming it is possible to set $v_3=1$, it should be possible to solve for $v_1$ and $v_2$ using any two of the set of three homogeneous equations.
For some reason Kuipers suggested to use the last two equations, but the result seems to follow more naturally from the first two, which for $v_3=1$ give you \begin{align} (a_{11}-1)v_1 + a_{12}v_2 &= -a_{13}, \\ a_{21}v_1 + (a_{22}-1)v_2 &= -a_{23}. \end{align}
The usual ways for solving these equations should very quickly lead to \begin{align} v_1 &= \frac{a_{12}a_{23} - (a_{22} - 1)a_{13}} {(a_{11} - 1)(a_{22} - 1) - a_{12}a_{21}}, \\ v_2 &= \frac{a_{21}a_{13} - (a_{11} - 1)a_{23}} {(a_{11} - 1)(a_{22} - 1) - a_{12}a_{21}}, \\ v_3 &= 1. \end{align}
But you can multiply any solution of $[A-I]\mathbf{v} = \mathbf{0}$ by a scalar to obtain another solution. I think you can guess what scalar we choose to get the book's answer.