There is another way to view $vv^\dagger$, namely as an element in projective space. Consider a pair of real numbers $(b,c)$, and identify points on the same line through the origin. This can be though of as introducing an equivalence relation of the form \begin{equation} (b,c) \sim (b\chi,c\chi) \label{equiv2} \end{equation} where $0\ne\chi\in\RR$. The resulting space can be identified with the (unit) circle of all possible directions in $\RR^2$, with antipodal points identified. This is the real projective space $\RP^1$, also called the real projective line. But this space can also be identified with the squares of normalized column vectors $v$, that is, \begin{equation} \RP^1 = \{vv^\dagger: v\in\RR^2, v^\dagger v=1\} \end{equation} where we write $\dagger$ instead of $T$ for transpose, anticipating a generalization to the other division algebras.

The normalization condition can be written in terms of the trace of $vv^\dagger$, since \begin{equation} \tr(vv^\dagger) = v^\dagger v \end{equation} There is yet another way to write this condition, since \begin{equation} (vv^\dagger)(vv^\dagger) = v(v^\dagger v)v^\dagger = \left(\tr(vv^\dagger)\right) \, (vv^\dagger) \label{trID} \end{equation} Putting the pieces together, we obtain a matrix description of $\RP^1$ in terms of $2\times2$ real Hermitian matrices ($\bH_2(\RR)$), namely 1) \begin{equation} \RP^1 = \{\XX\in\bH_2(\RR): \XX^2 = \XX, \tr\XX=1\} \label{projective} \end{equation}

Not surprisingly, all of this works over the other division algebras as well; (\ref{trID}) holds even over $\OO$ since $v$ has only 2 components, so that the computation takes place in a quaternionic subalgebra. Thus, (\ref{projective}) can be used to define the projective spaces $\RP^1$, $\CP^1$, $\HP^1$, and $\OP^1$, which are again known as projective lines.

However, the traditional definition, in terms of (\ref{equiv2}), requires modification over the octonions, along the lines of the discussion in § 12.1, One possible choice would be \begin{equation} \OP^1 = \{(b,c)\in\OO^2:(b,c)\sim\left( (bc^{-1})\chi,\chi \right),0\ne\chi\in\OO\} \end{equation} with $c=0$ handled as a special case.