Talk:Vector Times Magnitude Same Length As Magnitude Times Vector

The $\|$ need to be replaced with $\Vert$ for compatibility with the eqn template; also, strictly speaking $\mathbf u \cdot \left\Vert{v}\right\Vert$ is not defined. --Lord_Farin 05:45, 28 May 2012 (EDT)

Notation should now be consistent, and $\|$'s are dealt with. Also, I seem to have discovered that this only holds over a scalar field, not any scalar division ring. I can't come up with an easy example to show that this requirement is in fact necessary, but it seems like it would be. Anyone have a non-abelian vector space to check if it holds? --Alec (talk) 12:50, 28 May 2012 (EDT)

You could try a function vector space over the quaternions. --Lord_Farin 13:07, 28 May 2012 (EDT)

There seems to be a fundamental flaw in the approach, apparent therein that we presume $\left\Vert{\mathbf u}\right\Vert \circ \mathbf v$ is defined, but this only holds generally if $\R_{\ge0} \subseteq K$... This seems to imply that the scope of the theorem needs to be restricted further (to a field containing $(\R,+,\cdot)$ as a subfield would be enough). --Lord_Farin 13:13, 28 May 2012 (EDT)

Yet another edit: a division ring $K$ with subfield $\R$ such that $\R \subseteq Z(K)$ with $Z(K)$ the center of $K$. --Lord_Farin 13:15, 28 May 2012 (EDT)

Try that. --Alec (talk) 13:40, 28 May 2012 (EDT)