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Given $a,m,n,t\in\Bbb Z$, with $n=m^t$ and $a$ arbitrary, and given $\mathbb{Z}$-linearly independent vectors $v_1,\dots,v_n\in\Bbb Z^n$, and an arbitrary vector $w\in\Bbb Z^n$, such that $$\langle v_i-v_j,w\rangle=0$$ for all $1\leq i<j\leq n$, are there technical terms for the set of vectors

  1. $$\mathcal T_a=\{v\in\Bbb Q^n:\langle(\langle v_1,v\rangle,\langle v_2,v\rangle,\dots,\langle v_n,v\rangle),w\rangle=a\}?$$

  2. $$\mathcal U_t=\{v\in\Bbb Q^n:\forall\ 1\leq i\leq t \exists u_i\in\Bbb Z^{m}\mbox{ with }(\langle v_1,v\rangle,\dots,\langle v_n,v\rangle)=u_1\otimes \dots\otimes u_t\}?$$

Moreover: is there a reasonable way to check if $\mathcal T_a\cap\mathcal U_t\neq\emptyset$?

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    $\begingroup$ The last line is puzzling: the condition ' $v=(\langle v_1,v\rangle,\langle v_2,v\rangle,\dots,\langle v_n,v\rangle)\in\Bbb Z^n$ ' is equivalent to $Mv=v$, where $M$ is the (as per your hypothesis: full-rank) matrix which has the $v_i$ for it rows; but row-rank equals column-rank, also for linear algebra over $\mathbb{Z}$, so this 'matrix equation' is non-contradictory only if $M$ is the identity matrix, i.e., $v_i=e_i$=the $i$-th standard basis vector, whereupon your last line becomes tautologous. Would you please clarify what you mean? $\endgroup$
    – Peter Heinig
    Commented Sep 30, 2017 at 20:30
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    $\begingroup$ @PeterHeinig clarified but is this trivial? $\endgroup$
    – Turbo
    Commented Oct 1, 2017 at 16:22
  • $\begingroup$ Re "is this trivial": it isn't 'trivial' in any sense of trivial I know, I think, but for the time being, please let me continue critizsing the formulation of the OP: it seems to me that the 'parameter' " $a$ " is totally redundant and (to me) confusing. The given data are $v_1,\dotsc, v_n,w\in\mathbb{Z}^n$, and $b\in\mathbb{Z}$, subject to the 'axiom' that the $v_i$ be linearly independent. The equation containing '$a$' is a tautology/does-not-impose-any-condition. It seems to me that the '$a$' can be eliminated from the context for good, right? $\endgroup$
    – Peter Heinig
    Commented Oct 1, 2017 at 16:33
  • $\begingroup$ @PeterHeinig inner product all $v_i$ with $w$ is $a$. $\endgroup$
    – Turbo
    Commented Oct 1, 2017 at 16:40
  • $\begingroup$ One more attempt (starting with this comment MO is exhorting me not to): what I mean is that $a$ can be eliminated in the sense that your condition ' $\forall i\in \{1,..,n\}\quad\langle v_i,w\rangle = a$ ' is equivalent to ' $\color{blue}{\forall i,j\in\{1,..,n\}\quad \langle v_i-v_j,w \rangle = 0}$ ' , the latter being the form I would strongly recommend to write your condition in, removing the '$a$' completely from the OP. $\endgroup$
    – Peter Heinig
    Commented Oct 1, 2017 at 17:05

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