Minimum count of linear dependent columns in tensor product - MathOverflow most recent 30 from http://mathoverflow.net2013-05-25T15:50:27Zhttp://mathoverflow.net/feeds/question/61898http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/61898/minimum-count-of-linear-dependent-columns-in-tensor-productMinimum count of linear dependent columns in tensor productspk2011-04-16T08:08:56Z2011-04-19T15:02:44Z
<p>Good day!</p>
<p>I have a Vandermonde-like matrix $V = \left( \begin{array}{ccccc} 1 & \alpha & \alpha^2 & \ldots & \alpha^{n-1} \newline 1 & \alpha^2 & \alpha^4 & \ldots & \alpha^{2n - 2} \newline & & \vdots & & \newline 1 & \alpha^{n-1} & \alpha^{2n-2} & \ldots & \alpha^{(n-1)^2} \end{array} \right)$, where $\alpha$ is $n$-th root from unity in some extended field $GF(2^r)$. So sum of all columns of $V$ vanishes. And no other columns (of $V$) combinations vanishes over $GF(2)$.</p>
<p>After that I consider tensor product of $V$ with row $\mathbf{a} = (a_1, a_2, \ldots, a_{l-1})$ over the same field $GF(2^r)$. I know that elements of $\mathbf{a}$ are linear independent over $GF(2)$. </p>
<p>Now I consider tensor product $P = V \otimes \mathbf{a}$. Is it right, sum of no $p < n$ columns of $P$ vanishes over $GF(2)$?</p>
<p>Thank you!</p>