# prove the relation for the matrix element

Let $$a_0, a_1, ... a_n$$ be non-zero numbers such that $$det \begin{pmatrix} a_0&a_1&a_2...&a_k\\ a_1&a_2&a_3...&a_{k+1}\\ a_2&a_3&a_4...&a_{k+2}\\ . & . &. & .\\ . & . &. & .\\ a_k&a_{k+1}&a_{k+2}...&a_{2k}\\ \end{pmatrix}=0$$ prove that $$a_k = a_0\cdot q^k, k= 1,....,n, q \neq 0$$

I thought it was circulant matrix.

I also tried to find a pattern by looking at the lower order determinants, but it didn't work

I have no idea now how to do it.

• Not sure what the determinant is, but it's called a Hankel matrix.
– greg
Feb 19 '19 at 21:49
• The determinant is zero iff the matrix has nontrivial nullspace (or, iff its columns are linearly dependent). Basically, we have to prove that the nullspace is actually of dimension $k$. This might get you started.. Feb 19 '19 at 22:43
• Are you sure the statement is correctly stated? What is $n$? It doesn't appear in your matrix. As an answer below shows, the statement as it stands is false. Feb 19 '19 at 23:11

\left|\begin{aligned} 1 && 1 && 2 \\ 1 && 2 && 2 \\ 2 && 2 && 4 \end{aligned}\right| = 0
but here $$\{a_i\}$$ is $$\{1,1,2,2,4\}$$, which does not follow the pattern $$a_k = a_0 \cdot q^k$$
• actually $a_k = a_0\cdot q^k$ means the rank(A) = 1. But rank(A) < k is enough to get det(A) = 0. Feb 19 '19 at 23:31