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$k_1,k_2,...,k_n$ 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 ?

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  • $\begingroup$ I'm not clear what the pattern ... means in your final matrix expression. You said $k_i$ is a "non negativ(e)", but unless it is an integer it's unclear what $k_i !$ should mean. $\endgroup$
    – hardmath
    Jan 13 '21 at 0:51
  • $\begingroup$ @hardmath Ops, they are integers yes. I forgot it, thank you :) $\endgroup$
    – Lac
    Jan 13 '21 at 0:52
  • $\begingroup$ @hardmath ... means... How to explain, it is to show that matrix go on without writing every terms. Actually the three points should be vertical in some parts, but i don't know how to write it $\endgroup$
    – Lac
    Jan 13 '21 at 1:00
  • $\begingroup$ Thank you @PM2Ring, now it is nice $\endgroup$
    – Lac
    Jan 13 '21 at 1:18
  • $\begingroup$ @hardmath The terms are permutation numbers, sometimes written as $$^nP_r=\frac{n!}{(n-r)!}$$ $\endgroup$
    – PM 2Ring
    Jan 13 '21 at 1:23
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$\displaystyle C=\pm \prod_{1\le i<j\le n} (k_j-k_i)$ .

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 $det(M)=0$.

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  • $\begingroup$ You might link to an account of the determinant of a Vandermonde matrix, and clarify how the matrix in the Question is related to a Vandermonde matrix, e.g. by elementary column operations. The plus-or-minus sign in your expression is not needed (it is + with subscripts as shown). $\endgroup$
    – hardmath
    Jan 14 '21 at 20:23
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It seems that taking $C$ to be the above determinant and $r$ to be exponent of $t$ in your expression solves your problem.

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