4
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Not sure what the determinant is, but it's called a Hankel matrix. $\endgroup$
    – greg
    Feb 19 '19 at 21:49
  • $\begingroup$ 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.. $\endgroup$
    – Berci
    Feb 19 '19 at 22:43
  • 1
    $\begingroup$ 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. $\endgroup$
    – user1551
    Feb 19 '19 at 23:11
4
$\begingroup$

If my understanding of the problem is correct, the statement is actually wrong. So you won't be able to prove it.

Here is an counter example

$$ \left|\begin{aligned} 1 && 1 && 2 \\ 1 && 2 && 2 \\ 2 && 2 && 4 \end{aligned}\right| = 0 $$

but here $\{a_i\}$ is $\{1,1,2,2,4\}$, which does not follow the pattern $a_k = a_0 \cdot q^k$

$\endgroup$
1
  • 3
    $\begingroup$ actually $a_k = a_0\cdot q^k$ means the rank(A) = 1. But rank(A) < k is enough to get det(A) = 0. $\endgroup$
    – MoonKnight
    Feb 19 '19 at 23:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.