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let $W\in R^{d\times k},d>k$. Suppose that the associated gramian has the following structure: $$ W^TW=(P_{1}t,\cdots,P_{k}t) $$ with the set $\{P_{i},i=1,\cdots,k\}$ forming a group of permutations (on $k$ objects) and $t\in R^k$.

1) Is it true that $W$ must then be an orbit of a permutation group up to an isometry, i.e.

$$ W=UO,\;\;\; O=(\bar{P}_{1}s,\cdots,\bar{P}_{k}s),\;\;\;\;\;\;\;\;U^TU=Id $$ with the set $\{\bar{P}_{i},i=1,\cdots,k\}$ forming a group of permutations (on $d$ objects) and $s\in R^d$.?

2) Which other condition(s) (besides the particular structure of the gramian) should I had to have $U=Id$, i.e. $W=O$?

Thanks a lot!

Fabio

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  • $\begingroup$ think of the case $k=1$. Surely $W$ need not have any symmetry in it at all. $\endgroup$
    – Dima Pasechnik
    Commented Apr 12, 2016 at 20:31
  • $\begingroup$ Hi, I guess in that case P=e the identity,which is a permutation $\endgroup$
    – Fabio
    Commented Apr 13, 2016 at 7:32

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