Find the determinant of this matrix

Question

$k_1,k_2,...,k_n$ are non-negative integers. Let $M$ be an $n\times n$ matrix with entries: $a_{i,1} = t^{k_i}, a_{i,j+1} = \frac{da_{i,j}}{dt}$ where $a_{i,j}$ is the element of $i$-th row and $j$-th column.

Prove that there exist $C$ and $r$ such that $\det(M) = Ct^r$

I was able to reduce it to:

$\det(M) = t^{(k_1+...+k_n) - {n(n-1)\over 2}}$ $\cdot \det\begin{pmatrix} 1& k_{1} & k_{1}(k_{1}-1) & ... & {k_1!\over (k_1-n+1)!}\\ 1& k_{2} & k_{2}(k_{2}-1) & \vdots & \vdots\\ \vdots& \vdots & \vdots & \ddots \\ 1& k_{n} & k_{3} & ... & {k_n!\over (k_n-n+1)!} \end{pmatrix}$

I have already determined $r$, the problem is to calculate determinant of this thing, any tips ?

Answer 1

$\displaystyle C=\pm \prod_{1\le i<j\le n} (k_j-k_i)$ .

I hope you know the Vandermonde Determinant and how to compute it. If $k_i=k_j$ for some $i\neq j$ or $max\{k_1,k_2,...,k_n\}\le n-2$ then $det(M)=0$.

Answer 2

It seems that taking $C$ to be the above determinant and $r$ to be exponent of $t$ in your expression solves your problem.