0
$\begingroup$

Consider the determinant $$ \Delta_n(k_1, x_1; ...; k_m, x_m) = \begin{pmatrix} M_{k_1}^n(x_1) \\ M_{k_2}^n(x_2)\\ \vdots \\ M_{k_m}^n(x_n) \end{pmatrix}. $$ where $ x_1 \cdots x_m$ are any variables; $ k_1 \cdots k_m$ are natural numbers such that $ k_1 + k_2 + \cdots + k_m = n$; $M_k^n(x)$ is a $k \times n $ matrix of the following form $$ M_k^n(x) = \begin{pmatrix} 1 & x & x^2 & \cdots & x^{n-1}\\ 0 & 1 & \binom{2}{1}x & \cdots & \binom{n-1}{1}x^{n-2}\\ 0 & 0 & 1 & \cdots & \binom{n-1}{2}x^{n-3}\\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & \binom{n-1}{k - 1}x^{n-k}& \end{pmatrix}. $$ The problem is to prove that the determinant can be computed as $$ \Delta_n(k_1, x_1; ...; k_m, x_m) = \prod_{1 \le j < i \le m}(x_i-x_j)^{k_ik_j} $$ (Hint: when $k_1 = \cdots = k_m = 1, $ i.e. $m = n$, it's a Vandermonde determinant )

I tried to reduce this to Vandermonde determinant as was menthioned in hint, but with no success. Also it looks like each row of the $M_k^n(x)$ matrix is a linear combination of all the succesive rows, but I do not know how to use it.

$\endgroup$
3
  • $\begingroup$ @Bernard, given matrix is a square matrix, since there are $\sum\limits_{i=1}^m k_i=n$ rows, which is equal to the number of columns in this matrix. $\endgroup$ – Martund Jan 12 at 13:36
  • $\begingroup$ @Martund: You're right. Apparently, I skimmed through the question. $\endgroup$ – Bernard Jan 12 at 13:54
  • $\begingroup$ I think I'd start with the usual Vandermonde determinant in variables $z_1, z_2,\dots, z_{k_1}, \dots$, diff $1\times$ wrt $z_2$, $2\times$ wrt $z_3$, \dots, $k_1 -1 \times$ wrt $z_{k_1}$ then put all these $z_i=x_1$; repeat for the other blocks of variables. You'll need to divide by various factorials to get it exactly right. $\endgroup$ – ancientmathematician Jan 12 at 16:25
0
$\begingroup$

[Revised and completed.]

Let me slightly change your notation, and replace your $x_i$ by $X_i$.

Now start with the usual $n\times n$ Vandermonde determinant in the variables $x_1,x_2,\dots, x_n$.

Your determinant is obtained from this determinant by operating on it in the following way:

(step 1) differentiate w.r.t $x_2$ once, $x_3$ twice, ..., $x_{k_1}$ $(k_1-1)$ times; then differentiate w.r.t $x_{k_1+2} $ once, $x_{k_1+3}$ twice, ..., $x_{k_1+k_2}$ $(k_2-1)$ times, and so on;

(step 2) substitute $x_1=x_2=\dots=x_{k_1}=X_1$, $x_{k_1+1}=x_{k_1+2}=\dots=x_{k_1+k_2}=X_2$ , and so on;

(step 3) divide by $\prod_{i=1}^{m}\prod_{r=1}^{k_i} r!$.

It remains to see what happens when we apply these steps to $\prod_{i<j}(x_i-x_j)$.

Firstly note that if we apply $t$ differentiations to this product we get a sum/difference of subproducts each of which "omits" $t$ of the factors. Secondly note that when we substitute for the variables in a block, the only subproducts which don't evaluate to $0$ are those where the variables come from different blocks.

So consider the first block: we perform $k_1 \choose 2$ differentiations, and the only summand which will survive the substitution is the unique one which omits the factor $\prod_{i<j\leqslant k_1} (x_i -x_j)$.

Arguing in the same way about each block of variables we see that the only summand of the derivative (after we have carried out all the differentiations) which can survive the substitutions is the subproduct of $\prod_{i<j}(x_i-x_j)$ which omits $\prod_{r=1}^{m}\prod_{k_{r-1}<i<j\leqslant k_r}(x_i -x_j)$.

When we substitute we will clearly get a product of the form $\prod_{r<s}(X_r-X_s)^{h(r,s)}$. To determine $h(r,s)$ note that these factors arise from the factors $(x_i-x_j)$, where $x_i$ is one of the $k_r$ variables in the $r$-th block and $x_j$ is one of the $k_s$ variables in the $s$-th block. So $h(r,s)=k_r k_s$ as required.

We have therefore proved that the determinant in question is of the form $C\prod_{r<s}(X_r-X_s)^{k_r k_s}$, where the constant $C$ depends on the $k_i$. It remains to determine this constant.

I would prefer to give a combinatorial explanation of why $C=1$, but for present such an explanation eludes me. Instead let us determine on both sides the coefficient of the "highest" monomial $X_1^{n_1}X_2^{n_2}\dots X_m^{n_m}$, where highest means maximising $n_m$, then $n_{m-1}$, and so on.

As far as $\prod_{r<s}(X_r-X_s)^{k_r k_s}$ is concerned, the coefficient is clearly some $\epsilon$ where $\epsilon=\pm1$.

As far as the determinant is concerned, we maximise the power of $X_m$ by choosing the final entry of the $(k_1+k_2+\dots+k_{m-1}+1)$-th row, the next to last entry of the $(k_1+k_2+\dots+k_{m-1}+2)$-th row and so on: the antidiagonal of the bottom right-hand $k_m\times k_m$ block of the determinant. So far the coefficient is $\pm 1$: it is not hard to check that the parity is the same as the parity of our previous $\epsilon$.

Now this power of $X_m$ is must be multiplied by the highest monomial in the determinant of the complementary block: inductively this is $1$ and we are done.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.