Calculation of the determinant

Question

Question: Let $x$ be indeterminate and $m$ be a positive integer,$m'=2m-1$,proof: $$\left( \begin{matrix}1&0&...&0&1&0&...&0\\ x^{{}2}&x^{2}&...&x^{2}&1&1&...&1\\ x^{4}&2x^{4}&...&2^{{}m-1}x^{4}&1&2&...&2^{{}m-1}\\ x^{6}&3x^{6}&...&3^{m-1}x^{6}&1&3&...&3^{m-1}\\ \vdots &\vdots &&\vdots &\vdots &\vdots &&\vdots \\ x^{2m^{\prime }}&m^{\prime }x^{2m^{\prime }}&...&(m^{\prime })^{m-1}x^{2m^{\prime }}&1&m^{\prime }&...&\left( m^{\prime }\right)^{m-1} \end{matrix} \right) =\left( \prod^{m-1}_{i=1} i!\right)^{2} x^{m^{2}-m}\left( 1-x^{2}\right)^{m^{2}} $$ My attempt: All the roots of a determinant(in fact,0 or 1) are determined by taking its derivative, the difficulty lies in determining the coefficient of the highest term($\left( \prod^{m-1}_{i=1} i!\right)^{2} $)

Answer 1

Here is my attempt(maybe it is not a strict proof) for this question.

1. The result of this determinant must be a polynomial, all roots are determined by assignment and derivation(for determinant), we may get $ax^{m^{2}-m}\left( 1-x^{2}\right)^{m^{2}} $, where $a$ is to be determined.

   For example, consider $$\left| \begin{matrix}1&0&1&0\\ x^{2}&x^{{}2}&1&1\\ x^{4}&2x^{4}&1&2\\ x^{6}&3x^{6}&1&3\end{matrix} \right| $$ Obviously,$0$ and $\pm 1$ are roots(assign values of $0$ and $\pm 1$ respectively),let's take the derivative to determine its multiplicity. $$\left| \begin{matrix}1&0&1&0\\ x^{2}&x^{2}&1&1\\ x^{4}&2x^{4}&1&2\\ x^{6}&3x^{6}&1&3\end{matrix} \right|^{\prime } =\left| \begin{matrix}0&0&1&0\\ 2x&x^{{}2}&1&1\\ 4x^{3}&2x^{4}&1&2\\ 6x^{5}&3x^{{}6}&1&3\end{matrix} \right| +\left| \begin{matrix}1&0&1&0\\ x^{2}&2x&1&1\\ x^{4}&8x^{3}&1&2\\ x^{6}&18x^{5}&1&3\end{matrix} \right| $$ Obviously,$0$ and $\pm 1$ are still roots, take derivative again. $$\left| \begin{matrix}1&0&1&0\\ x^{2}&x^{2}&1&1\\ x^{4}&2x^{4}&1&2\\ x^{6}&3x^{6}&1&3\end{matrix} \right|^{\prime \prime } =\left| \begin{matrix}0&0&1&0\\ 2&x^{{}2}&1&1\\ 12x^{2}&2x^{4}&1&2\\ 30x^{4}&3x^{{}6}&1&3\end{matrix} \right| +\left| \begin{matrix}1&0&1&0\\ 2x&2x&1&1\\ 4x^{3}&8x^{3}&1&2\\ 6x^{{}5}&18x^{5}&1&3\end{matrix} \right| $$ $$+\left| \begin{matrix}0&0&1&0\\ 2x&2x&1&1\\ 4x^{3}&8x^{3}&1&2\\ 6x^{5}&18x^{5}&1&3\end{matrix} \right| +\left| \begin{matrix}1&0&1&0\\ x^{2}&2&1&1\\ x^{4}&24x^{2}&1&2\\ x^{6}&90x^{4}&1&3\end{matrix} \right| $$ Obviously,$\pm 1$ are roots, but $0$ is not, take derivative again. $$\left| \begin{matrix}1&0&1&0\\ x^{2}&x^{2}&1&1\\ x^{4}&2x^{4}&1&2\\ x^{6}&3x^{6}&1&3\end{matrix} \right|^{\prime \prime \prime } =\left| \begin{matrix}0&0&1&0\\ 0&x^{2}&1&1\\ 24x&2x^{4}&1&2\\ 120x^{3}&3x^{6}&1&3\end{matrix} \right| +\left| \begin{matrix}0&0&1&0\\ 2&2x&1&1\\ 12x^{2}&8x^{3}&1&2\\ 30x^{4}&18x^{5}&1&3\end{matrix} \right| $$ $$+\left| \begin{matrix}0&0&1&0\\ 2&2x&1&1\\ 12x^{2}&8x^{3}&1&2\\ 30x^{4}&18x^{5}&1&3\end{matrix} \right| +\left| \begin{matrix}0&0&1&0\\ 2x&2&1&1\\ 4x^{3}&24x^{2}&1&2\\ 6x^{5}&90x^{4}&1&3\end{matrix} \right| $$ $$+\left| \begin{matrix}0&0&1&0\\ 2&2x&1&1\\ 12x^{2}&8x^{3}&1&2\\ 30x^{4}&18x^{5}&1&3\end{matrix} \right| +\left| \begin{matrix}0&0&1&0\\ 2x&2&1&1\\ 4x^{3}&24x^{2}&1&2\\ 6x^{5}&90x^{4}&1&3\end{matrix} \right| $$ $$+\left| \begin{matrix}0&0&1&0\\ 2x&2&1&1\\ 4x^{3}&24x^{2}&1&2\\ 6x^{5}&90x^{4}&1&3\end{matrix} \right| +\left| \begin{matrix}1&0&1&0\\ x^{2}&0&1&1\\ x^{4}&48x&1&2\\ x^{6}&360x^{3}&1&3\end{matrix} \right| $$ $\pm 1$ are still roots, but not obvious($\left| \begin{matrix}0&0&1&0\\ 2&2x&1&1\\ 12x^{2}&8x^{3}&1&2\\ 30x^{4}&18x^{5}&1&3\end{matrix} \right| $).So far,$0$(the multiplicity is $2$) and $\pm 1$(the multiplicity is $2$) are roots, the degree of this determinant is at most $10$, hence $0$ and $\pm 1$ are all roots: $$\left| \begin{matrix}1&0&1&0\\ x^{2}&x^{2}&1&1\\ x^{4}&2x^{4}&1&2\\ x^{6}&3x^{6}&1&3\end{matrix} \right| =ax^{2}\left( 1-x^{2}\right)^{4} $$

2. Determine $a$ of $ax^{m^{2}-m}\left( 1-x^{2}\right)^{m^{2}} $. $$\left| \begin{matrix}1&0&\cdots &0&1&0&\cdots &0\\ x^{2}&x^{2}&\cdots &x^{2}&1&1&\cdots &1\\ \vdots &\vdots &&\vdots &\vdots &\vdots &&\vdots \\ x^{2\left( m-1\right) }&\left( m-1\right) x^{2\left( m-1\right) }&\cdots &\left( m-1\right)^{m-1} x^{2\left( m-1\right) }&1&m-1&\cdots &\left( m-1\right)^{m-1} \\ x^{2m}&mx^{2m}&\cdots &m^{m-1}x^{2m}&1&m&\cdots &m^{m-1}\\ x^{2\left( m+1\right) }&\left( m+1\right) x^{2\left( m+1\right) }&\cdots &\left( m+1\right)^{m-1} x^{2\left( m+1\right) }&1&m+1&\cdots &\left( m+1\right)^{m-1} \\ \vdots &\vdots &&\vdots &\vdots &\vdots &&\vdots \\ x^{2\left( 2m-1\right) }&\left( 2m-1\right) x^{2\left( m-1\right) }&\cdots &\left( 2m-1\right)^{m-1} x^{2\left( m-1\right) }&1&2m-1&\cdots &\left( 2m-1\right)^{m-1} \end{matrix} \right| =\left| \begin{matrix}A&B\\ C&D\end{matrix} \right| $$ Notice that $$A=\left( \begin{matrix}1&&&&\\ &x^{2}&&&\\ &&x^{4}&&\\ &&&\ddots &\\ &&&&x^{2\left( m-1\right) }\end{matrix} \right) B:=EB$$ $$C=\left( \begin{matrix}x^{2m}&&&&\\ &x^{2\left( m+1\right) }&&&\\ &&x^{2\left( m+2\right) }&&\\ &&&\ddots &\\ &&&&x^{2\left( 2m-1\right) }\end{matrix} \right) D:=FD$$ So $$\begin{gathered}\left| \begin{matrix}A&B\\ C&D\end{matrix} \right| =\left| \begin{matrix}FB&B\\ FD&D\end{matrix} \right| =\left| B\right| \left| E\right| \left| D-FD\left( EB\right)^{-1} B\right| \\ =\left( \prod^{m-1}_{i=1} i!\right) x^{m^{2}-m}\left| D-FD\left( EB\right)^{-1} B\right| \end{gathered} $$ We have know that $$\left| \begin{matrix}A&B\\ C&D\end{matrix} \right| =ax^{m^{2}-m}\left( 1-x^{2}\right)^{m^{2}} $$ So $$\left| D-FD\left( EB\right)^{-1} B\right| =b\left( 1-x^{2}\right)^{m^{2}} $$ Take $x=0$, we get $$b=\left| D\right| =\prod^{m-1}_{i=1} i!$$ To sum up, $$a=(\prod^{m-1}_{i=1} i!)^{2}$$ I still crave a more "standard solution"!