I'm kind of embarassed for asking this question, but...
The problem: I'm having a hard time trying to establish a consistent notation for the spectral decomposition of the elements of a second-order cone $L_m\doteq \{(x_0,\bar{x})\in \mathbb{R}\times \mathbb{R}^{m-1} \colon x_0\geq \|\bar{x}\| \}$ when $m=1$. I could not find a standard treatment for this case anywhere.
Context: It is quite well-known that every $x\in L_m$ can be decomposed as $x=\lambda_1(x)v_1(x) + \lambda_2(x)v_2(x)$, where $\lambda_i(x)=x_0+(-1)^i\|\bar{x}\|$ and $v_i(x)=\frac{1}{2}(1,(-1)^iw)$, with $$w=\left\{\begin{array}{ll}\bar{x}/\|\bar{x}\| & \text{ if } \bar{x}\neq 0\\ \text{anything s.t. }\|w\|=1 & \text{ if } \bar{x}=0\end{array}\right.$$ However, I don't see how to denote this when $m=1$. In this case, it seems that the only reasonable way of defining these guys is to set $\lambda_1(x)=x$ and $v_1(x)=1$, since a defining property of $v_1$ and $v_2$ is their "orthonormality". And from the point of view of Euclidean Jordan Algebras, $L_1$ has rank 1. But what about $\lambda_2$ and $v_2$? Actually, I want to write that $v_2$ is not defined, in such a way that I can talk about $v_1$ and $v_2$ and it's clear for everyone that they should simply ignore $v_2$.
In other words, what I want to know is:
How to define $v_2(x)$ such that $\{v_1(x),v_2(x)\}=\{1\}$ when $m=1$, while keeping $v_1(x)v_2(x)=0$?
Even a clever notation would do the job, as long as it is not outrageous...
My (terrible) attempt to solve the problem: I am very inclined to make an abuse of notation and define $v_2(x)$ such that $\{v_2(x)\}\doteq\emptyset$ and $v_1(x)\doteq 1$, so I would have $\{v_1(x),v_2(x)\}=\{v_1(x)\}\cup\{v_2(x)\}=\{1\}$ and some sort of "vacuous orthogonality", but it makes me feel bad, and I have never seen this kind of abuse before. What do you think about it? Any suggestion is welcome!