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Let $n \in\mathbb{N}$ and define $A \in M_{n}(\mathbb{R})$ by $A(i,j)= \min(i,j)$ for $i,j \in \{1, 2 ,3, 4,\cdots, n\}$.Compute $\det(A)$.

My try is with a example: 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.

Help, please.

Arnaud D.
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Jean Paul
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    See what happens when you replace row 4 with row 4 minus row 3... then replace R3 with R3 - R2. Then replace R2 with R2 - R1. What kind of matrix is that, and why is that noteworthy/good? Can you generalize that for arbitrary $n \in \Bbb N$? –  Feb 28 '18 at 00:50

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By looking at your example, it seems that by using Laplace expansion in the last column, you would get twice your induction assumption. Let me ellaborate: your determinant would be

$$-1\left|\begin{array}{ccc} 1 & 2 & 2 \\ 1 & 2 & 3 \\ 1 & 2 & 3\end{array}\right|+2\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & 3\end{array}\right|-3\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3\end{array}\right|+4\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3\end{array}\right|$$

The first two determinants are $0$, since there are repeated rows, and the last two are equal to $1$ by your induction assumption, so it is $4-3=1$. I am pretty sure that this approach Works in the general $n\times n$ matix, so try it out and see what happens.

Álvaro Serrano
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