# Linear algebra problem **URL:** \ **Category:** Factual Questions\ **Created:** [October 22, 2003, 6:16am UTC](https://boards.straightdope.com/t/linear-algebra-problem/208853 "2003-10-22T06:16:34Z")\ **Posts on this page:** 2\ **Page:** 2
**Author:** ![Jabba](https://avatars.discourse-cdn.com/v4/letter/j/b5e925/32.png) [@Jabba](https://boards.straightdope.com/u/Jabba)\ **Post date:** [October 23, 2003, 11:20pm UTC](https://boards.straightdope.com/t/linear-algebra-problem/208853/21 "2003-10-23T23:20:53Z")
**ultrafilter** said: > [@](#): > > I need to see those bits spelled out in a little more detail. 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 ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [October 23, 2003, 11:27pm UTC](https://boards.straightdope.com/t/linear-algebra-problem/208853/22 "2003-10-23T23:27:54Z")
All right, I can buy that. And I see the error I made with the bit about determinants. [Previous page](https://boards.straightdope.com/t/linear-algebra-problem/208853.md?page=1)