# Taking powers of a Vandermonde matrix elementwisely

Consider a Vandermonde matrix $$V (x_1, x_2, \ldots , x_n) =\begin{pmatrix} 1 & x_1 & x_1^2 & \cdots & x_1^{n-1} \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-1} \\ 1 & x_3 & x_3^2 & \cdots & x_3^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-1}\end{pmatrix}$$ with positive and distinct $x_1, \dots, x_n$. Then it is well-known that $V(x_1, x_2, \ldots, x_n)$ is invertible. It is not hard to see that if we furthermore consider $$V_d(x_1, x_2, \ldots, x_n) = \begin{pmatrix} 1 & x_1^d & x_1^{2d} & \cdots & x_1^{(n-1)d} \\ 1 & x_2^d & x_2^{2d} & \cdots & x_2^{(n-1)d} \\ 1 & x_3^d & x_3^{2d} & \cdots & x_3^{(n-1)d} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_n^d & x_n^{2d} & \cdots & x_n^{(n-1)d}\end{pmatrix},$$ this is also invertible. This is because $$V_d(x_1, x_2, \ldots, x_n) = V(x_1^d, x_2^d, \ldots, x_n^d)$$ and by positivity and distinctness of $x_1, x_2, \ldots, x_n$, the values $x_1^d, x_2^d, \dots, x_n^d$ are also distinct.

I am asking myself the following. If we are now given a matrix $V(x_1,x_2, \dots, x_n)B$ with an invertible matrix $B$, and are taking its power element-wise, is the element-wise power of $V(x_1,x_2, \dots, x_n)B$ then also invertible? I think so, but I havent found a way to prove it.

• Have you worked out any simple examples. If there are counter-examples they should quickly show up. – Marc Bogaerts Sep 30 '14 at 11:18
• yes i have looked at simploe examples but could not find a counter-example – samsa44 Sep 30 '14 at 12:59
• I took the element-wise power of any invertible matrix and it campe up invertible too. Maybe it's valid for invertible matrices? – Marc Bogaerts Sep 30 '14 at 13:30
• No, this is not the case, consider for example $\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ – samsa44 Sep 30 '14 at 13:55
• And couldn't the same thing happen with a combination with a Vandermonde matrix? – Marc Bogaerts Sep 30 '14 at 14:08