Re: Dse2011 question

Posted by knowpraveen on Jun 17, 2016; 7:12am
URL: http://discussion-forum.276.s1.nabble.com/Dse2011-question-tp7601218p7601474.html

For n=2, let the relation be a1v1+a2v2=0 which goes on to mean that (1)v1+(a2/a1)v2=0 which means there are two scalars. That's the minimum number of nonzero scalars possible in such an equation.

Forgot to add that this is the same for any value n since all other scalar values could be taken as zero.