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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.

Arnaud D.
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James
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2 Answers2

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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.

qualcuno
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  • How do I prove that this new matrix has the same determinant as the original one? – James Oct 02 '16 at 16:40
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    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. – qualcuno Oct 02 '16 at 21:05
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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 $

ImBatman
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