Definition talk:Zero Vector

Subtle this: $\R^1$ is not a scalar ring, and so saying "where $0 \in \R^1$ is the zero scalar" is technically inaccurate.

This is despite the fact that $\R$ and $\R^1$ are the "same shape" - because $\R^1$ is a vector space. --prime mover 15:21, 14 March 2012 (EDT)
 * whoa, that's very subtle. Does $\R^1$ imply "$\R$ along with $+$ and $\times$"? --GFauxPas 15:27, 14 March 2012 (EDT)


 * $\R$ is a field and uses $+$ and $\times$ like for real numbers. $\R^1$ is a vector space which is a structure consisting of a group (in this case $\R$ with $+$) acted upon by a scalar field (in this case $\R$ with $+$ and $\times$). The two objects are isomorphic. But they are not the same. Study the abstract algebra leading up to the vector space definition and see if you can work out what's what. --prime mover 15:42, 14 March 2012 (EDT)