Calculate the determinant of $a_{ij} = \frac{(1+x)^{i+j-1}-1}{i+j-1}$

Question

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 $c$.

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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!

Answer 1

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 $\det A=cx^{n^2}$.

Answer 2

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.