Can we determine the determinant?

Question

Could someone prove that this determinant is not zero? $$\left| \begin{array}{cccc} 1^n & 2^n & \cdots & (n+1)^n \\ 2^n & 3^n & \cdots & (n+2)^n \\ \vdots & \vdots & & \vdots \\ (n+1)^n & (n+2)^n & \cdots & (2n+1)^n \\ \end{array} \right|\neq0$$ How can we compute it?

Answer 1

These determinants are equal to $(-1)^{\textrm{Floor}((n+1)/2)}(n!)^{n+1}$

For the proof see:

J. M. Monier, Algebre & geometrie, Dunod (1996), p.216.

Added: I cannot find the original proof (I used to have the original edtion but the only printings I can find are the 5th edition where he's taken this proof out--apparently) but I do have a heuristic method:

Let $R = \left(\begin{array}{cccc}1^n & 2^n &\cdots & (n+1)^n\\2^n & 3^n & \cdots & (n+2)^n\\ \vdots & \vdots & \ddots & \vdots \\ (n+1)^n & (n+2)^n & \cdots & (2n+1)^n\end{array}\right)$

If we row operate by subtracting the previous row from each row beginning with row 2.

i.e. $\tilde{R}_{i,j} \gets R_{i,j} - R_{i-1,j}$

Then repeat this same but starting this time with row 3. And continue repeating--each time beginning one row further down you will wind up with:

$$\tilde{R}_k =R_k + (-1)^1{k-1 \choose 1}R_{k-1} + (-1)^2{k -1\choose 2}R_{k-2} +\cdots+(-1)^{k-1}{k -1\choose k-1}R_{1}$$ where$R_j$ is an original row vector of $R$. Part of the magic of the proof is that this operation will transform the final row to be a constant value of $n!$. This is because $$(k+n)^n - {n \choose 1}(k+n-1)^n + {n\choose 2}(k+n-2)^n +\cdots+(-1)^n{n\choose n}k^n = n!$$for each $k$.

Since the original matrix is symmetric, you can do the same operation on the columns and get a similar result. In fact if you do this operation on the rows, then the columns you'll get A matrix with $n!$ down the anti-diagonal and zeros below the anti-diagonal. The values above the anti-diagonal appear to be OEIS: A142071 although I have no idea why. The important thing for the determinant is that you can rearrange the rows to get an upper diagonal matrix with $n!$ on the diagonal. This rearranging gives $\det R =(-1)^{\textrm{Floor}((n+1)/2)}(n!)^{n+1}$.

Answer 2

First we apply binomial theorem, e.g. write $2^n$ as $(1+1)^n$, then $2^n=1^n+C_n ^1 1^{n-1}1^1+ \cdots+C_n ^n 1^0 1^n$. And note that $\det(AB)=\det(A)\det(B)$ for square matrix $A,B$ with same order. $$\left| \begin{array}{cccc} 1^n & 2^n & \cdots & (n+1)^n \\ 2^n & 3^n & \cdots & (n+2)^n \\ \vdots & \vdots & & \vdots \\ (n+1)^n & (n+2)^n & \cdots & (2n+1)^n \\ \end{array} \right| = \left| \begin{array}{cccc} (0+1)^n & (0+2)^n & \cdots & (0+(n+1))^n \\ (1+1)^n & (1+2)^n & \cdots & (1+(n+1))^n \\ \vdots & \vdots & & \vdots \\ (n+1)^n & (n+2)^n & \cdots & (n+(n+1))^n \\ \end{array} \right| = \left| \begin{array}{cccc} 0 & 0 & \cdots & 1 \\ 1^n & 1^{n-1} & \cdots & 1^0 \\ \vdots & \vdots & & \vdots \\ n^n & n^{n-1} & \cdots & n^0 \\ \end{array} \right| \cdot \left| \begin{array}{cccc} 1 & 1 & \cdots & 1 \\ C_n^1 1^1 & C_n^1 2^{1} & \cdots & C_n ^1 (n+1)^1 \\ \vdots & \vdots & & \vdots \\ C_n^n 1^n & C_n^n 2^{n} & \cdots & C_n^n (n+1)^n \\ \end{array} \right| $$ Notice that this 2 matrix are nearly of the Vandermonde form. See here. Except that the first one should have all its columns reversed; the second one should extract the factors $C_n ^1, \dots,C_n ^n$. Hence, $$ \left| \begin{array}{cccc} 0 & 0 & \cdots & 1 \\ 1^n & 1^{n-1} & \cdots & 1^0 \\ \vdots & \vdots & & \vdots \\ n^n & n^{n-1} & \cdots & n^0 \\ \end{array} \right| \cdot \left| \begin{array}{cccc} 1 & 1 & \cdots & 1 \\ C_n^1 1^1 & C_n^1 2^{1} & \cdots & C_n ^1 (n+1)^1 \\ \vdots & \vdots & & \vdots \\ C_n^n 1^n & C_n^n 2^{n} & \cdots & C_n^n (n+1)^n \\ \end{array} \right|= \\ (-1)^{n(n+1)/2}C_n ^1 C_n ^2 \cdots C_n ^n \left| \begin{array}{cccc} 1 & 0 & \cdots & 0 \\ 1 & 1^{1} & \cdots & 1^n \\ \vdots & \vdots & & \vdots \\ 1 & n^{1} & \cdots & n^n \\ \end{array} \right| \cdot \left| \begin{array}{cccc} 1 & 1 & \cdots & 1 \\ 1^1 & 2^{1} & \cdots & (n+1)^1 \\ \vdots & \vdots & & \vdots \\ 1^n & 2^{n} & \cdots & (n+1)^n \\ \end{array} \right| \\ =(-1)^{n(n+1)/2}C_n ^1 C_n ^2 \cdots C_n ^n \prod_{0 \leq i <j \leq n}(j-i) \prod_{1 \leq i <j \leq n+1}(j-i) \neq 0. $$