# Find the determinant of this matrix

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

• 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. Jan 13 '21 at 0:51
• @hardmath Ops, they are integers yes. I forgot it, thank you :)
– Lac
Jan 13 '21 at 0:52
• @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
– Lac
Jan 13 '21 at 1:00
• Thank you @PM2Ring, now it is nice
– Lac
Jan 13 '21 at 1:18
• @hardmath The terms are permutation numbers, sometimes written as $$^nP_r=\frac{n!}{(n-r)!}$$ Jan 13 '21 at 1:23

$$\displaystyle C=\pm \prod_{1\le i .
I hope you know the Vandermonde Determinant and how to compute it. If $$k_i=k_j$$ for some $$i\neq j$$ or $$max\{k_1,k_2,...,k_n\}\le n-2$$ then $$det(M)=0$$.
It seems that taking $$C$$ to be the above determinant and $$r$$ to be exponent of $$t$$ in your expression solves your problem.