Tuesday, 14 October 2014

linear algebra - Circulant determinants



Suppose that a1,a2,,an are n distinct real numbers; is the following statement true?



There is a permutation of a1,a2,,an, namely b1,b2,,bn, such that the determinant of the following matrix is nonzero:
[b1b2bnb2b3b1bnb1bn1]



(Such a matrix is called a circulant matrix.)


Answer



This statement is not true, without supplementary conditions on the ai's. Indeed,

suppose the nk=1ak=0, whatever your permutation is the vector [1,1,,1]T is in the kernel of the circulant matrix of the bi's, and consequently, its determinant is 0.


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