Let $m_1,\dots,m_n$ be real numbers $\ge n-1$.

How can I find the determinant of the matrix $A$ defined by $(a_{i,j})=\binom{m_i}{j-1}$, for $1\le i\le n$ and $1 \le j \le n$ ?

This all looks pretty VanDerMonde-ish, but I haven't found a good way to solve it.

  • $\begingroup$ What are the usage of these $m_1, m_2, \cdots, m_n$ here? $\endgroup$
    – KON3
    Oct 7, 2014 at 5:44
  • 1
    $\begingroup$ @AnjanDebnath It was a typo, edited. $\endgroup$ Oct 7, 2014 at 10:28

1 Answer 1



$$\det \binom{m_i}{j-1} = \frac{1}{\prod_{k=1}^{n-1} k^{n-k}}\prod_{0<i_1<i_2\leq n}\left(m_{i_2}-m_{i_1}\right)$$


Let us simplify the proof by breaking it into several elementary facts.

Fact 1

$$\det \binom{m_i}{j-1} = Const_n\prod_{0<i_1<i_2\leq n}\left(m_{i_2}-m_{i_1}\right)$$

Proof 1

If we compute $\det \binom{m_i}{j-1}$ per definition, as a sum of the products of the elements taken with the appropriate sign, we will get a sum of products of type $m_1^{q_1}m_2^{q_2}\dots m_n^{q_n}$, where $\sum_{i=1}^n q_i \leq \frac{n(n-1)}{2}$ (each column will contribute $number\ of\ column-1$ to this sum). In other words, we will get a $n$-dimensional polynom, and the degree of each summand is at most $\frac{n(n-1)}{2}$.

On the other hand, if $\exists i_1,i_2 : m_{i_1}= m_{i_2}$, then 2 rows will be the same and $\det \binom{m_i}{j-1}$ is $0$. Thus, our determinant can be presented as $$\det\binom{m_i}{j-1} =SomePolynom\left(m_{1},\dots, m_n\right)\prod_{i_1< i_2}\left(m_{i_1}- m_{i_2}\right)$$ (it can be easily proved by absurd).

So we have two representations of the same polynom, the first one has "overall" degree of at most $\frac{n(n-1)}{2}$, the second one will have "overall" degree of at least $\frac{n(n-1)}{2}$. Thus, the "overall" degree of this polynom is exactly $\frac{n(n-1)}{2}$, and, equivalently, $$SomePolynom\left(m_{1},\dots, m_n\right)\equiv SomeConstant$$ This constant is the same for any given $n$, but can vary from $n$ to $n$, so we will add a subscript to recognise it: $Const_n$.

Fact 2

Recursive formula for $Const_n$ is $$Const_n = \frac{Const_{n-1}}{(n-1)!}$$

Proof 2

Let us compute the determinant of matrix $\binom{m_i}{j-1}$ by expanding it over the last column. We have $$Const_n\prod_{0<i_1<i_2\leq n}\left(m_{i_2}-m_{i_1}\right)= \sum_{I=1}^n (-1)^{n+I}\binom{m_I}{(n-1)!} Const_{n-1}\prod_{0<i_1<i_2\leq n,\; i_1\neq I, i_2\neq I}\left(m_{i_2}-m_{i_1}\right)$$

We consider this as a polynom wrt $m_n$ and check the coefficient of $m_n^{n-1}$. This coefficient from the left hand side expression is $Const_n \prod_{0<i_1<i_2< n} \left(m_{i_2}-m_{i_1}\right)$; the one taken from the right hand side is $(-1)^{n+n}\frac{1}{(n-1)!}Const_{n-1} \prod_{0<i_1<i_2< n} \left(m_{i_2}-m_{i_1}\right) = Const_{n-1} \prod_{0<i_1<i_2< n} \left(m_{i_2}-m_{i_1}\right)$.

Since these values are equal, we have $$Const_{n} = \frac{Const_{n-1}}{(n-1)!}$$

Fact 3


Fact 4

$$Const_{n} = \prod_{k=1}^{n-1} \frac{1}{k!} = \frac{1}{\prod_{k=1}^{n-1} k^{n-k}}$$

Proof 4

$\frac{Const_{n-1}}{Const_{n}} = \frac{\prod_{k=1}^{n-1} k^{n-k}}{\prod_{k=1}^{n-2} k^{n-k}}=(n-1)!$

  • $\begingroup$ Although it may seem basic to you, I'm not sure how you go from your first to your second line $\endgroup$ Oct 7, 2014 at 18:33
  • $\begingroup$ I am reasonably happy with the new version, let me know if anything is unclear! $\endgroup$
    – Yulia V
    Oct 8, 2014 at 8:47
  • $\begingroup$ In proof $1$, how do you go from $\det \left( m_i^{j-1} \right)$ to $\det \binom{m_i}{j-1}$ ? $\endgroup$ Oct 8, 2014 at 10:25
  • $\begingroup$ I don't, this is a residual of the previous version of the proof :) Let me fix it $\endgroup$
    – Yulia V
    Oct 8, 2014 at 10:26
  • 1
    $\begingroup$ Thanks for this answer, your formatting makes things really clear ! $\endgroup$ Oct 8, 2014 at 10:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .