Prove an identity including determinant

Prove that:

$$$$\begin{vmatrix} x_0^{2n+1}&x_0^{2n}&\cdots&x_0&1\\ x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\ \vdots&\vdots&\cdots&\vdots\\ x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\ (2n+1)x_0^{2n}&2nx_0^{2n-1}&\cdots&1&0\\ \vdots&\vdots&\cdots&\vdots&\cdots\\ (2n+1)x_n^{2n}&2nx_n^{2n-1}&\cdots&1&0\\ \end{vmatrix}=(-1)^{n}\prod_{n\geq i>j\geq 0}(x_i-x_j)^4$$$$

where $(x_i^{2n+1})'=(2n+1)x_i^{2n}$,$\cdots$,$x_i'=1$,$1'=0$.

Each column contains the values and first derivatives of a power of $x$ evaluated at $x_0$ to $x_n$. Thus multiplying a column vector of coefficients by this matrix yields a column vector of values and first derivatives of the corresponding polynomial of degree up to $2n+1$ evaluated at $x_0$ to $x_n$. Since this linear evaluation map from the vector space of polynomials of degree up to $2n+1$ to the space of $(2n+2)$-dimensional vectors is an isomorphism if the $x_i$ are distinct (Hermite interpolation establishes a surjection in one direction, and this is a bijection because the dimensions coincide), the product of the matrix with a vector is non-zero unless the vector is zero or two of the $x_i$ coincide; thus the determinant is non-zero unless two of the $x_i$ coincide. Considering the determinant as a polynomial of degree $4n$ in, say, $x_0$, it must be a constant times linear factors $x_0-x_i$ for its roots. By symmetry, each of these factors must appear the same number of times, and thus $4$ times. Applying this argument for each $x_j$ shows that the determinant must be a constant times the product given in the question. To evaluate the constant, consider the term proportional to $x_n^{4n}x_{n-1}^{4(n-1)}\dotso x_{1}^4x_0^0$, which arises from $n+1$ determinants of $2\times2$ matrices containing only one $x_i$ and evaluating to $-x_i^{4i}$; then you just have to check that an odd permutation reorders the matrix such that these $2\times2$ matrices are on the diagonal.

• Extraordinary.... – Luqing Ye Dec 15 '12 at 11:12

Having seen the answer given by joriki,I think it is the time to share my own answer(I appreciate his answer,and need some time to digest...).

Let $$f(x)= \begin{vmatrix} x^{2n+1}&x^{2n}&\cdots&x&1\\ x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\ \vdots&\vdots&\cdots&\vdots\\ x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\ (2n+1)x^{2n}&2nx^{2n-1}&\cdots&1&0\\ \vdots&\vdots&\cdots&\vdots&\cdots\\ (2n+1)x_{n}^{2n}&2nx_n^{2n-1}&\cdots&1&0\\ \end{vmatrix}$$

It is easy to verify that $f(x)$ is a polynomial of degree $4n$.And $$f(x_1)=\cdots =f(x_n)=0$$

So $(x-x_1)(x-x_2)\cdots (x-x_n)|f(x)$.And

\begin{align*} f'(x)= \begin{vmatrix} x^{2n+1}&x^{2n}&\cdots&x&1\\ x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\ \vdots&\vdots&\cdots&\vdots\\ x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\ 2n(2n+1)x^{2n-1}&(2n-1)2nx^{2n-2}&\cdots&0&0\\ \vdots&\vdots&\cdots&\vdots&\cdots\\ (2n+1)x_{n}^{2n}&2nx_n^{2n-1}&\cdots&1&0\\ \end{vmatrix} \end{align*} So $f'(x_1)=\cdots f'(x_n)=0$.So $(x-x_1)^2(x-x_2)^2\cdots (x-x_n)^2|f(x)$.And

\begin{align*} f''(x)= \begin{vmatrix} (2n+1)x^{2n}&2nx^{2n-1}&\cdots&1&0\\ x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\ \vdots&\vdots&\cdots&\vdots\\ x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\ 2n(2n+1)x^{2n-1}&(2n-1)2nx^{2n-2}&\cdots&0&0\\ \vdots&\vdots&\cdots&\vdots&\cdots\\ (2n+1)x_{n}^{2n}&2nx_n^{2n-1}&\cdots&1&0\\ \end{vmatrix}+ \begin{vmatrix} x^{2n+1}&x^{2n}&\cdots&x&1\\ x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\ \vdots&\vdots&\cdots&\vdots\\ x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\ (2n-1)2n(2n+1)x^{2n-2}&(2n-2)(2n-1)2nx^{2n-3}&\cdots&0&0\\ \vdots&\vdots&\cdots&\vdots&\cdots\\ (2n+1)x_{n}^{2n}&2nx_n^{2n-1}&\cdots&1&0\\ \end{vmatrix} \end{align*}

It is easy to verify that $f''(x_1)=\cdots f''(x_n)=0$.So

$$(x-x_1)^3(x-x_2)^3\cdots (x-x_n)^3|f(x)$$ And it is also easy to figure out $f'''(x)$,so it is easy to verify that

$$f'''(x_1)=\cdots =f'''(x_n)=0$$ So $$(x-x_1)^4(x-x_2)^4\cdots (x-x_n)^4|f(x)$$

Because $f(x)$ is a polynomial of degree $4n$,so $f(x)=a(x-x_1)^4(x-x_2)^4\cdots (x-x_n)^4$.According to symmetry ammong $x_0,x_1,\cdots,x_n$ in this determinant,it is easy to verify that

$$f(x_0)=c\prod_{n\geq i>j\geq 0}(x_i-x_j)^4$$ Now we determine the constant $c$.We do it by induction.It is easy to verify that When $n=1$,

$$\det \begin{pmatrix} x_0^3&x_0^2&x_0&1\\ x_1^3&x_1^2&x_1&1\\ 3x_0^2&2x_0&1&0\\ 3x_1^2&2x_1&1&0\\ \end{pmatrix}=-(x_0-x_1)^4$$Then when $n=2$,Let's see the determinant

$$\det\begin{pmatrix} x_0^5&x_0^4&x_0^3&x_0^2&x_0&1\\ x_1^5&x_1^4&*x_1^3&*x_1^2&*x_1&*1\\ x_2^5&x_2^4&*x_2^3&*x_2^2&*x_2&*1\\ 5x_0^4&4x_0^3&3x_0^2&2x_0&1&0\\ 5x_1^4&4x_1^3&*3x_1^2&*2x_1&*1&*0\\ 5x_2^4&4x_2^3&*3x_2^2&*2x_2&*1&*0\\ \end{pmatrix}$$

The element marked * also form a determinant,we know that the constant of this determinant is -1,so the constant term of the determinant $$\det\begin{pmatrix} x_0^5&x_0^4&x_0^3&x_0^2&x_0&1\\ x_1^5&x_1^4&*x_1^3&*x_1^2&*x_1&*1\\ x_2^5&x_2^4&*x_2^3&*x_2^2&*x_2&*1\\ 5x_0^4&4x_0^3&3x_0^2&2x_0&1&0\\ 5x_1^4&4x_1^3&*3x_1^2&*2x_1&*1&*0\\ 5x_2^4&4x_2^3&*3x_2^2&*2x_2&*1&*0\\ \end{pmatrix}$$ is $-1\times (4-5)=1$ ……

So by induction,we know that $c=(-1)^n$.

In my answer I make use of the following result

If $$f(x)=\begin{vmatrix} f_{11}(x)&f_{12}(x)&\cdots&f_{1n}(x)\\ f_{21}(x)&f_{22}(x)&\cdots&f_{2n}(x)\\ \vdots&\vdots& &\vdots\\ f_{n1}(x)&f_{n2}(x)&\cdots&f_{nn}(x)\\ \end{vmatrix}$$ Then $$f'(x)=\sum_{i=1}^n \begin{vmatrix} f_{11}(x)&f_{12}(x)&\cdots&f_{1n}(x)\\ \vdots&\vdots & &\vdots\\ f'_{i1}(x)&f'_{i2}(x)&\cdots&f'_{in}(x)\\ \vdots&\vdots&&\vdots\\ f_{n1}(x)&f_{n2}(x)&\cdots&f_{nn}(x)\\ \end{vmatrix}$$

• This is a very nice answer, too. I think it would gain from having fewer formulas and more words. It essentially boils down to the fact that you have to differentiate at least four times to make the rows containing $x$ both different from all the other rows and different from each other. The $\TeX$ command for $\gg$ is \gg. Note that the $x_0\gg\dotso\gg x_n$ argument no longer appears in my answer because I realized that the constant can be determined from that term independent of whether it's the leading term or not. – joriki Dec 15 '12 at 11:52
• Yes,maybe I have to differentiate once more but I didn't do it because I thinked that $(x-x_1)^4(x-x_2)^4\cdots (x-x_n)^4$ is enough ,because $f(x)$ is degree $4n$.As for that constant,I also find that using the method of $\gg$ may not be rigourous,so I would change it according to your advice.@joriki – Luqing Ye Dec 15 '12 at 12:12
• I wasn't saying that you should have differentiated once more; I was just trying to summarize your rather formula-focussed proof in words. I could just as well have said that two rows always coincide as long as you differentiate less than $4$ times. – joriki Dec 15 '12 at 12:14
• @joriki I use a lot of formula because I am now using Emacs+$\LaTeX$ and quite enjoy its live preview feature,so I use a lot formula :-)I am not good at expressing formula into words because I am not good at expressing ideas in English... – Luqing Ye Dec 15 '12 at 12:19
• I see; sorry, I didn't mean to be critical; as I said, it's a very nice answer. I hope my summary can contribute to its understanding. – joriki Dec 15 '12 at 12:24

There is a sign error in the right-hand side; the term $(-1)^n$ should be $(-1)^{(n+1)(n+2)/2}$.

By the usual formula for the determinant of the Vandermonde matrix,

$$\begin{vmatrix} x_0^{2n+1}&x_0^{2n}&\cdots&x_0&1\\ x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\ \vdots&\vdots&\cdots&\vdots\\ x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\ (x_0+h)^{2n+1}&(x_0+h)^{2n}&\cdots&x_0+h&1\\ \vdots&\vdots&\cdots&\vdots&\cdots\\ (x_n+h)^{2n+1}&(x_n+h)^{2n}&\cdots&x_n+h&1\\ \end{vmatrix}=\prod_{i<j} (x_i-x_j)^2 \prod_{i\ne j} (x_i-x_j-h).$$ For each of the first $n+1$ rows of the matrix, subtract it from the row $n+1$ rows lower down. This does not change the determinant of the matrix. Then divide the right-hand side by $h^{n+1}$, and divide the left-hand side by $h^{n+1}$ by dividing each of the last $n+1$ rows of the matrix by $h$. Setting $h:=0$ then gives the identity.