# Prove a special form of Schur polynomial identity [closed]

Let $A_k$ be the $n\times n$ matrix defined by $$A_k=\left[ \begin{array}{} 1 & x_1 & x_1^2 & \cdots & x_1^{n-2} & x_1^k \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-2} & x_2^k \\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-2} & x_n^k \end{array}\right].$$ If $k=n-1$, the matrix $A_{n-1}$ is called the Vandermonde matrix and the determinant of $A_{n-1}$ is given by $$\det (A_{n-1})= \prod_{1 \mathop \le i \mathop < j \mathop \le n} \left({x_j - x_i}\right).$$

My question is how to prove $$\det (A_n)= (x_1+x_2+\cdots + x_n)\prod_{1\leq i < j \leq n} \left({x_j - x_i}\right).\tag{1}$$ I know that $\det (A_n)$ is a kind of Schur polynomial, but I don't know how to apply the Schur polynomial to prove Eq. $(1)$.

• You want $k=n$. The formula is then just a special case of the bialternant formula for Schur functions. It is also easy to prove directly. – Richard Stanley May 2 '18 at 19:17
• @RichardStanley We know that if $x_i=x_j$, for some $i\neq j$ then two row of $A_n$ is zero and hence $\prod \left({x_j - x_i}\right)$ is apart of $\det (A_{n-1})$. Moreover, one of the term of $\det (A_{n-1})$ is multiplying of elements of diagonal of $A_n$ which is $x_2x_3^2\cdots x_{n-1}^{n-2}x_{n}^n$, but the degree of $x_n$ in $\prod \left({x_j - x_i}\right)$ is $n-1$ and therefore we need $x_n$. I dont know this argument can be applied to other $x_i$;s. Also I want to ask you to make a reference for "bialternant formula for Schur functions" that you mentioned. Thanks – user0410 May 2 '18 at 20:09
• Let $P(x) = (x - x_1) \ldots (x - x_n)$, the monic polynomial with the $x_i$ as roots. Then $P(x)$ is of the form $x^n - (x_1 + \cdots + x_n) x^{n-1}$ plus lower-degree stuff. So you can express $x^n$ as $(x_1 + ... + x_n) x^{n-1}$ plus stuff that will vanish in the determinant. Done. – Ravi Boppana May 2 '18 at 20:34
• @RaviBoppana Thanks for your comment. I have a question. Why elementary symmetric polynomials $e_i$, $0\leq i \leq n-2$ are zero in $P(x)$? – user0410 May 2 '18 at 21:02
• The elementary symmetric polynomials are not zero. Still the low-degree terms will not contribute to the determinant. That's because if you have a matrix with two equal columns, then its determinant is zero. – Ravi Boppana May 2 '18 at 21:13

This is exactly Exercise 6.16 in Darij Grinberg, Notes on the combinatorial fundamentals of algebra, 10 January 2019, except that the order of columns in my exercise is the other way round (which is why I end up with $$\prod\limits_{1\leq i instead of $$\prod\limits_{1\leq i). Of course, the difference is insubstantial: Reversing the order of the columns of an $$n\times n$$-matrix multiplies its determinant by $$\left(-1\right)^{n\left(n-1\right)/2}$$.
This is also (again, after the same reversal of the order of the columns) a consequence of "Corollary (The Bi-Alternant Formula)" in John R. Stembridge, A Concise Proof of the Littlewood-Richardson Rule, applied to $$\mu = \left(1,0,0,\ldots,0\right)$$ (with $$n-1$$ zeroes), since what he calls $$a_\rho$$ is the usual Vandermonde determinant $$\prod\limits_{1\leq i. This is the argument that Richard Stanley has been alluding to.
• @darijgrinberg By article you mentioned, $a_{\lambda +p}=s_{\lambda}a_p$, where $\lambda=\{0,0,0,\cdots ,0,1\}$ which result in $s_{\lambda}=e_1$ and finish. Am I right? – user0410 May 2 '18 at 22:37
• @user0410: Correct, except that the $p$ should be a $\rho$. – darij grinberg May 2 '18 at 23:20