2
$\begingroup$

Det$ \begin{bmatrix} 1 & 2 & 3 &\ldots &n\\ 1& 2^3& 3^3& \ldots & n^3\\ 1 &2^5& 3^5& \ldots & n^5\\ \vdots & \vdots& & \vdots \\ 1&2^{2n-1}& 3^{2n-1}& \ldots &n^{2n-1} \end{bmatrix} $

If the powers were consecutively increasing down the rows than we could use the Vandermonde’s identity. However I am quite uncertain about how to tackle this. I am convinced that elementary row operations will be to no avail. Any hints?

$\endgroup$
2
$\begingroup$

If you multiply the second column by $2$, the third by $3$ and so on, you get $$ \det \begin{bmatrix} 1 & 2 & 3 &\ldots &n\\ 1& 2^3& 3^3& \ldots & n^3\\ 1 &2^5& 3^5& \ldots & n^5\\ \vdots & \vdots& & \vdots \\ 1&2^{2n-1}& 3^{2n-1}& \ldots &n^{2n-1} \end{bmatrix} = \frac{1}{n!}\det \begin{bmatrix} 1 & 2^2 & 3^2 &\ldots &n^2\\ 1& 2^4& 3^4& \ldots & n^4\\ 1 &2^6& 3^6& \ldots & n^6\\ \vdots & \vdots& & \vdots \\ 1&2^{2n}& 3^{2n}& \ldots &n^{2n} \end{bmatrix} $$ and the matrix is now Vandermonde.

$\endgroup$
1
$\begingroup$

Call $\Delta$ your determinant. It seems to me that $$\Delta = n! \det \left( \begin{array}{ccccc} 1 & 1 & 1 & \dots & 1 \\ 1 & 4 & 9 & \cdots & n^2 \\ 1 & 4^2 & 9^2 & \cdots & (n^2)^2 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 1 & 4^{n-1} & 9^{n-1} & \cdots & {(n^2)}^{(n-1)} \end{array} \right) = n! \left( \begin{array}{ccccc} 1 & 1 & 1 & \dots & 1 \\ x_1 & x_2 & x_3 & \cdots & x_n \\ x_1^2 & x_2^2 & x_3^2 & \cdots & x_n^2 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ x_1^{n-1} & x_2^{n-1} & x_3^{n-1} & \cdots & x_n^{n-1} \end{array} \right)$$ were $x_1 = 1, x_2=4, \cdots, x_i = i^2 , \cdots, x_n = n^2$. From which you can use the usual Vandermonde.

$\endgroup$

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.