1
$\begingroup$

Consider following partial Vandermonde type, circulant matrix $\begin{bmatrix} x_1 & x_2 & 0 & \dots & 0 & x_n\\ x_1^2 & x_2^2 & x_3^2 & \dots & 0 & 0\\ \vdots &\vdots &\vdots &\ddots &\vdots &\vdots\\ x_1^n & 0 & 0 & \dots & x_{n-1}^n & x_n^n\\ \end{bmatrix}$.

Is there a closed form expression for this determinant like Vandermonde determinant expression?

$\endgroup$
5
  • 2
    $\begingroup$ I think your typesetting of the matrix (which seems not to be circulant) is off--it appears you have two more columns than rows. And can you be more clear about which entries are zero? $\endgroup$
    – Kimball
    Jun 7 '15 at 9:40
  • $\begingroup$ no $x_1$ to $x_n$ variables $1$ to $n$ exponents. $\endgroup$
    – Mr.
    Jun 7 '15 at 9:41
  • $\begingroup$ Every row/column has $3$ non-zero entries. $\endgroup$
    – Mr.
    Jun 7 '15 at 9:43
  • 6
    $\begingroup$ so it's not circulant, but a termwise product of the vandermonde matrix $V(x_1,\ldots,x_n)$ with the circulant matrix $circ(1,1,0,\ldots,0,1)$. $\endgroup$
    – kodlu
    Jun 7 '15 at 10:09
  • $\begingroup$ yes thats what I am meaning here. $\endgroup$
    – Mr.
    Jun 7 '15 at 18:29
1
$\begingroup$

The "closest" form you can expect is the well known determinant formula for tridiagonal matrices, which in your case can be written as $$ \Delta = \det \left[ \begin{pmatrix} -x_n^{-n} & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 0 & -x_1^{n} \end{pmatrix} \mathbf T_n \mathbf T_{n-1} \cdots \mathbf T_2 \mathbf T_1 \right] \,\,\prod_{k=1}^n (-x_k^{k+1})\,, $$ with transfer matrix $$ \mathbf T_k = \begin{pmatrix} -x_k^{-1} & -x_k^{-2} \\ 1 & 0 \end{pmatrix}. $$

$\endgroup$

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.