Maximal length vector under constraints - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T23:01:16Z http://mathoverflow.net/feeds/question/81269 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81269/maximal-length-vector-under-constraints Maximal length vector under constraints unknown (yahoo) 2011-11-18T17:40:43Z 2011-11-19T23:05:09Z <p>Consider a criculant symmetric $M$ an $n \times n$ matrix with $0$ and $1$ entries and $r$ entries of $1$ in each row with the diagonal values taken as $1$. I am looking for a $0-1$ vector $v$ with the largest hamming weight such that $\langle v , \frac{1}{r}Mv \rangle$ $=$ $\langle v , v \rangle$.</p> <p>I am also looking for the rate of growth in the hamming weight as $m \rightarrow \infty$, $\langle v , \frac{1}{r^{m}}M^{\otimes{m}}v \rangle$ $=$ $\langle v , v \rangle$. What are some good mathematical techniques/tools to study this kind of problems?</p> http://mathoverflow.net/questions/81269/maximal-length-vector-under-constraints/81383#81383 Answer by Gerry Myerson for Maximal length vector under constraints Gerry Myerson 2011-11-19T23:05:09Z 2011-11-19T23:05:09Z <p>Perhaps I misunderstand, but if $v$ is all ones (and you can't get a bigger Hamming weight than that) then $Mv$ is all $r$, so $(1/r)Mv$ is all ones again, so $(1/r)Mv=v$ and the inner product of $v$ with $(1/r)Mv$ is the inner product of $v$ with $v$. </p>