# Determinant of matrix with $A_{ij} = \min (i, j)$

Given a $n\times n$ matrix whose $(i, j)$-th entry is the lower of $i,j$, eg. $$\begin{pmatrix}1 & 1 & 1 & 1\\ 1 & 2 & 2 & 2 \\ 1 & 2 & 3 & 3\\ 1 & 2 & 3 & 4 \end{pmatrix}.$$ The determinant of any such matrix is $1$. How do I prove this? Tried induction but the assumption would only help me to compute the term for $A_{nn}^*$ mirror.

You can substract the $$j$$-th column to the $$(j+1)$$-th one. This will leave you with a lower-triangular matrix of all ones.

• How do I prove that this new matrix has the same determinant as the original one? Commented Oct 2, 2016 at 16:40
• Adding to a row/column a linear combination of other rows/colums doesn't change the determinant. This is derived from the properties of multilinearity and antisymmetry that the determinant has. Commented Oct 2, 2016 at 21:05

Rotate the matrix A by $$180$$ degree. Do you think the determinant would change? Have a look at this question if you have any doubt about this.
Does rotating a matrix change its determinant?
Let R denote the rotated matrix Let's have a look at the inverse of this matrix $$R$$
$$D_{n} = (R_{n})^{-1}= \begin{pmatrix} 1 & P^{T} \\ P & B_{n-1} \end{pmatrix}$$ where, $$B = \begin{pmatrix} 2 & -1 & 0 & \cdots &0 \\ -1 & 2 & -1 & \cdots & 0 \\ 0&-1&2&\cdots &0\\ \vdots & \vdots &\vdots &\ddots&\vdots\\ 0 & 0 & 0&\cdots& 2 \\ \end{pmatrix}$$ and $$P = (-1,0,0,0...,0)^{T}$$

You may find it a good exercise to verify this.

By expanding the determinant along row $$1$$ we have $$\forall n\geq 2$$ $$det(D_{n}) = det(B_{n-1})-det(B_{n-2})$$ where, B follows the recursion formulla $$det(B_{n}) = 2det(B_{n-1})-det(B_{n-2})$$ Base case $$det(B_{1}) = 2$$, which gives $$det(B_{n}) = n+1.$$

Hence, For such matrices D we have $$det(D_{n}) = 1 \implies det(R_{n}) = det(A_{n}) = 1$$