[energycode] - Max Determinant

Identity

The sequence A181983(n) gives the determinants of the square matrices: $M_{n}$, with elements $m_{i,j}= \max(i,j)$, where $\max$ the is the maximum function.

This identity appears as follows:

\[\begin{equation} det{\bigg[ max(i,j) \bigg] }_{1\leq i,j \leq n} = -n \cdot {(-1)}^{n} = A181983(n) \end{equation}\]

Proof

A new matrix with the same determinant can be created by subtracting row $i$ from row $i+1$ starting from the 2nd row. The determinant of this new matrix can then be computed easily using the expansion by minors technique at element $m_{1,n}$

This can be better illustrated with an example:

We can transform:

\[\begin{equation} M_{10} = \left(\begin{array}{cccccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 2 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 3 & 3 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 4 & 4 & 4 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 5 & 5 & 5 & 5 & 5 & 6 & 7 & 8 & 9 & 10 \\ 6 & 6 & 6 & 6 & 6 & 6 & 7 & 8 & 9 & 10 \\ 7 & 7 & 7 & 7 & 7 & 7 & 7 & 8 & 9 & 10 \\ 8 & 8 & 8 & 8 & 8 & 8 & 8 & 8 & 9 & 10 \\ 9 & 9 & 9 & 9 & 9 & 9 & 9 & 9 & 9 & 10 \\ 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 \end{array} \right) \end{equation}\]

Into:

\[\begin{equation} M^{*}_{10} =\left(\begin{array}{cccccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \end{array} \right) \end{equation}\]

This proof can be generalized to a very similar kind of matrices, resulting in:

\[\begin{equation} det{\bigg[ max(i,j)^k \bigg]}_{1\leq i,j \leq n} = {(-1)}^{n-1} \cdot n^{k} \cdot \prod_{s=1}^{n-1}{(s+1)^k-s^k} \end{equation}\]

\[\begin{equation} det{\bigg[ min(i,j)^k \bigg]}_{1\leq i,j \leq n} = \prod_{s=1}^{n-1}{(s+1)^k-s^k} \end{equation}\]

References

[1]

N. J. A. Sloane, “A051125: Table t(n,k) = maxn,k read by antidiagonals (n >= 1, k >= 1).” Available: https://oeis.org/A051125/

[2]

M. Somos, “A181983: A(n) = (-1)^(n+1) * n.” 2012. Available: https://oeis.org/A181983/

[3]

E. Deutsch, “A161124: Number of inversions in all fixed-point-free involutions of 1,2,...,2n.” Jun. 2009. Available: https://oeis.org/A161124/

[4]

N. J. A. Sloane, “A001147: Double factorial of odd numbers: A(n) = (2*n-1)!! = 1*3*5*...*(2*n-1).” Available: https://oeis.org/A001147/

--- title: Max Determinant description: Determinants of special square matrices with elements being the maximum or minimum between the indices of the elements. author: 'Enrique Pérez Herrero' date: '2022-03-05' image: plot3dmaxmatrix.png categories: - math - linear algebra - determinant editor: source bibliography: references.bib nocite: | @* --- ![Plot3D of max matrix elements](plot3dmaxmatrix.png) ## Identity The sequence [A181983](https://oeis.org/A181983)(n) gives the determinants of the square matrices: $M_{n}$, with elements $m_{i,j}= \max(i,j)$, where $\max$ the is the maximum function. This identity appears as follows: \begin{equation} det{\bigg[ max(i,j) \bigg] }_{1\leq i,j \leq n} = -n \cdot {(-1)}^{n} = A181983(n) \end{equation} ## Proof A new matrix with the same determinant can be created by subtracting row $i$ from row $i+1$ starting from the 2nd row. The determinant of this new matrix can then be computed easily using the expansion by minors technique at element $m_{1,n}$ This can be better illustrated with an example: We can transform: \begin{equation} M_{10} = \left(\begin{array}{cccccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 2 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 3 & 3 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 4 & 4 & 4 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 5 & 5 & 5 & 5 & 5 & 6 & 7 & 8 & 9 & 10 \\ 6 & 6 & 6 & 6 & 6 & 6 & 7 & 8 & 9 & 10 \\ 7 & 7 & 7 & 7 & 7 & 7 & 7 & 8 & 9 & 10 \\ 8 & 8 & 8 & 8 & 8 & 8 & 8 & 8 & 9 & 10 \\ 9 & 9 & 9 & 9 & 9 & 9 & 9 & 9 & 9 & 10 \\ 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 \end{array} \right) \end{equation} Into: \begin{equation} M^{*}_{10} =\left(\begin{array}{cccccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \end{array} \right) \end{equation} This proof can be generalized to a very similar kind of matrices, resulting in: \begin{equation} det{\bigg[ max(i,j)^k \bigg]}_{1\leq i,j \leq n} = {(-1)}^{n-1} \cdot n^{k} \cdot \prod_{s=1}^{n-1}{(s+1)^k-s^k} \end{equation} \begin{equation} det{\bigg[ min(i,j)^k \bigg]}_{1\leq i,j \leq n} = \prod_{s=1}^{n-1}{(s+1)^k-s^k} \end{equation} ## References ::: {#refs} :::