ultrafilter said:
Since Josh_dePlume has not returned:
We have two vectors (a[sub]1c[/sub], …, a[sub]nc[/sub]) and (a[sub]c1[/sub], …, a[sub]cn[/sub]) and we can get all the entries of the matrix by multiplying entries in these two vectors, by the equation a[sub]ij[/sub] = a[sub]ic[/sub]a[sub]cj[/sub].
As Chronos wrote above, we can visualize this as a group multiplication table. The conclusion that the matrix is of rank 1 now follows from a standard theorem of linear algebra. Do you have a copy of Halmos’s Finite-Dimensional Vector Spaces? Section 51, Theorem 1 says
If a linear transformation A on a finite-dimensional vector space V is such that rank(A) [symbol]£[/symbol] 1, then the elements of the matrix of A have the form a[sub]ij[/sub] = b[sub]i[/sub]c[sub]j[/sub] in every coordinate system; conversely if the matrix of A has this form in some coordinate system, then rank(A) [symbol]£[/symbol] 1