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There is a question asked by my classmate. Looking forward to some ideas, thanks.

Set $A=\{a_{ij}\}_{n\times n}$, where $$a_{ij}=\frac{(1+x)^{i+j-1}-1}{i+j-1}.$$ Prove that $\det A=cx^{n^2}$ for some $c$.


I have tried to calculate it, but failed. I computed $$\frac{(1+x)^{i+j-1}-1}{i+j-1}=\sum_{k=1}^{i+j-1}\frac{(i+j-2)!}{k!(i+j-1-k)!}x^k,$$ but I have no idea how to continue. I know when $a_{ij}=\frac{1}{i+j-1}$, it is the Hilbert matrix, and we can get its determinant, but I don’t know how to calculate the above determinant. Are there some hints? Looking forward to your answer. Thanks!

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    $\begingroup$ If you take entry-wise derivative, then $a'_{ij} = (1+x)^{i+j-2}$ and this determinant you can compute. Then, you could try using the Jacobi formula for determinants. $\endgroup$
    – dezdichado
    Jan 15, 2022 at 3:33
  • $\begingroup$ actually, did you check it for $n=2 ?$ If I am not mistaken, the determinant in that case is $$\dfrac{x^4}{12} + \dfrac{x^3}{3}+x^2+x$$ $\endgroup$
    – dezdichado
    Jan 15, 2022 at 3:40
  • $\begingroup$ @dezdichado I think there is something wrong in your calculation. When $n=2$, it is $\frac{x^4}{12}$. $\endgroup$
    – ling
    Jan 15, 2022 at 6:40

2 Answers 2

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Firstly, we know there is a nondegenerate matrix $J(n)$ such that $$(1,x,\dots,x^{n-1})=(1,(1+x),\dots,(1+x)^{n-1})J(n).$$ Then we have $$ \begin{aligned} &\begin{pmatrix} 1&x&\cdots&x^{n-1}\\ x&x^2&\cdots&x^{n}\\ \vdots&\vdots&\ddots&\vdots\\ x^{n-1}&x^n&\cdots&x^{2n-2} \end{pmatrix} =(1,x,\cdots,x^{n-1})^T(1,x,\cdots,x^{n-1})\\[7pt]&= J(n)^T\begin{pmatrix} 1&1+x&\cdots&(1+x)^{n-1}\\ 1+x&(1+x)^2&\cdots&(1+x)^{n}\\ \vdots&\vdots&\ddots&\vdots\\ (1+x)^{n-1}&(1+x)^n&\cdots&(1+x)^{2n-2} \end{pmatrix}J(n). \end{aligned} $$ Hence, integrating respect to $x$ yields (Note that $J(n)$ does not depend on $x$) $$ \begin{pmatrix} \frac{x}{1}&\frac{x^2}{2}&\cdots&\frac{x^{n}}{n}\\ \frac{x^2}{2}&\frac{x^3}{3}&\cdots&\frac{x^{n+1}}{n+1}\\ \vdots&\vdots&\ddots&\vdots\\ \frac{x^{n}}{n}&\frac{x^{n+1}}{n+1}&\cdots&\frac{x^{2n-1}}{2n-1} \end{pmatrix}= J(n)^T\begin{pmatrix} \frac{(1+x)-1}{1}&\frac{(1+x)^2-1}{2}&\cdots&\frac{(1+x)^{n}-1}{n}\\ \frac{(1+x)^2-1}{2}&\frac{(1+x)^3-1}{3}&\cdots&\frac{(1+x)^{n+1}-1}{n+1}\\ \vdots&\vdots&\ddots&\vdots\\ \frac{(1+x)^{n}-1}{n}&\frac{(1+x)^{n+1}-1}{n+1}&\cdots&\frac{(1+x)^{2n-1}-1}{2n-1} \end{pmatrix}J(n). $$ Then it is easy to see that $$\det \begin{pmatrix} \frac{x}{1}&\frac{x^2}{2}&\cdots&\frac{x^{n}}{n}\\ \frac{x^2}{2}&\frac{x^3}{3}&\cdots&\frac{x^{n+1}}{n+1}\\ \vdots&\vdots&\ddots&\vdots\\ \frac{x^{n}}{n}&\frac{x^{n+1}}{n+1}&\cdots&\frac{x^{2n-1}}{2n-1} \end{pmatrix}=ax^{n^2},$$ which implies $\det A=cx^{n^2}$.

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  • $\begingroup$ See my answer employing partitioned matrix with associated vectors, which actually gives a good information that there does, in fact, exist such 'some c' . $\endgroup$
    – AJKOER
    Jan 15, 2022 at 15:53
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Let assume the case of dimension k is correct and now we must demonstrate the validity for dimension k+1.

I proceed by quoting work from a prior thread by Siong Thye Soh for the case of dimension k+1 which can be presented now as a problem in determining the determinant of a special case of a partitioned matrix with associated vectors:

"We have if $\det(A) \ne 0$,

$$\det\begin{pmatrix} A & b \\ a' & \alpha\end{pmatrix}=\det(A) \det(\alpha-a'A^{-1}b)=(\alpha - a'A^{-1}b) |A|"$$

from which it is evident that that case k+1 follows.

[EDIT] Interestingly, my approach to problem also provides information that there is, fact, "some constant c". This follows as the associated constant term above is a function of the inverse of matrix $A$, which as my proof requires the $\det(A) \ne 0$, confirming that the inverse of matrix $A$ exist. Also somewhat obvious, the constant term $(\alpha - a'A^{-1}b)$ involves all constants being product of a 1 x k vector with k x k matrix and a k x 1 vector.

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  • $\begingroup$ Actually, I do not see any issue with my proof other than it is easy to understand and concise without making claims like "Firstly, we know there is a nondegenerate matrix" or for some complex matrix "Then it is easy to see that". Or, perhaps I choose to give credit to the vector partition matrix answer, which by the way, is very useful, in applications when working on large regression problems. Really a downgrade!? $\endgroup$
    – AJKOER
    Jan 15, 2022 at 15:43
  • $\begingroup$ I have perform an edit on my answer to do what no other 'answer' has, support for the claim there is 'some constant c'. $\endgroup$
    – AJKOER
    Jan 15, 2022 at 16:20
  • $\begingroup$ you got it backwards when you said "... as my proof requires $\det(A)\neq 0$, confirming that the inverse matrix $A$ exist..." The reduction formula you are citing is only valid when you know beforehand $A$ is invertible, not vice versa. Besides, you now need to prove that $\det(\alpha - a'A^{-1}b) = \text{constant}\times x^{n^2-(n-1)^2}$ in order for your induction step to be complete. $\endgroup$
    – dezdichado
    Jan 15, 2022 at 16:22
  • $\begingroup$ Thanks dezdichado: I should have been more clear. I have first demonstrated a simple matrix equation relationship between case k and k+1. If we assume the cited relationship for case k is not otherwise true (meaning the cited general equation is inaccurate) and then show it is true for case k+1, then this constitutes a proof by contradiction. Do you agree? In which case, I will edit to makes this proof's logic to be more evident. $\endgroup$
    – AJKOER
    Jan 15, 2022 at 17:39
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    $\begingroup$ no, you still have not proved the inductive step based on your inductive hypothesis. You need to prove that: $$\det(\alpha - a'A^{-1}b) = Cx^{2n-1}$$ and before that you need to prove that $A$ is invertible. $\endgroup$
    – dezdichado
    Jan 15, 2022 at 18:45

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