Evaluation of a specific determinant.

Question

Evaluate $\det{A}$, where $A$ is the $n \times n$ matrix defined by $a_{ij} = \min\{i, j\}$, for all $i,j\in \{1, \ldots, n\}$.

$$A_2 \begin{pmatrix} 1& 1\\ 1& 2 \end{pmatrix}; A_3 = \begin{pmatrix} 1& 1& 1\\ 1& 2& 2\\ 1& 2& 3 \end{pmatrix}; A_4 = \begin{pmatrix} 1& 1& 1& 1\\ 1& 2& 2& 2\\ 1& 2& 3& 3\\ 1& 2& 3& 4 \end{pmatrix}; A_5 = \begin{pmatrix} 1& 1& 1& 1& 1\\ 1& 2& 2& 2& 2\\ 1& 2& 3& 3& 3\\ 1& 2& 3& 4& 4\\ 1& 2& 3& 4& 5 \end{pmatrix}$$

$$A_6 = \begin{pmatrix} 1& 1& 1& 1& 1& 1\\ 1& 2& 2& 2& 2& 2\\ 1& 2& 3& 3& 3& 3\\ 1& 2& 3& 4& 4& 4\\ 1& 2& 3& 4& 5& 5\\ 1& 2& 3& 4& 5& 6 \end{pmatrix}; A_7 = \begin{pmatrix} 1& 1& 1& 1& 1& 1& 1\\ 1& 2& 2& 2& 2& 2& 2\\ 1& 2& 3& 3& 3& 3& 3\\ 1& 2& 3& 4& 4& 4& 4\\ 1& 2& 3& 4& 5& 5& 5\\ 1& 2& 3& 4& 5& 6& 6\\ 1& 2& 3& 4& 5& 6& 7 \end{pmatrix} $$

Answer 1

$$A_2 = \begin{bmatrix}1 & 1\\ 1 & 2 \end{bmatrix} = \begin{bmatrix}1 & 0\\ 1 & 1\end{bmatrix}\begin{bmatrix}1 & 1\\ 0 & 1\end{bmatrix}$$ $$A_3 = \begin{bmatrix} 1 & 1 & 1\\ 1 & 2 & 2\\1 & 2 & 3 \end{bmatrix} = \begin{bmatrix}1 & 0 & 0\\ 1 & 1 & 0\\ 1 & 1 & 1\end{bmatrix}\begin{bmatrix}1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1\end{bmatrix}$$ $$A_4 = \begin{bmatrix} 1 & 1 & 1 & 1\\ 1 & 2 & 2 & 2\\1 & 2 & 3 & 3\\ 1 & 2 & 3 & 4\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 & 0\\ 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 0\\ 1 & 1 & 1 & 1\end{bmatrix}\begin{bmatrix}1 & 1 & 1 & 1\\ 0 & 1 & 1 & 1\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 1\end{bmatrix}$$ Can you see the pattern here? Prove this is the case in general.

Use this, along with the fact that $\det(XY) = \det(X) \cdot \det(Y)$ and determinant of a triangular matrix is the product of the diagonal entries to conclude the answer.

EDIT

Note that $$A_{ij} = \min(i,j) = \sum_{k=1}^{\min(i,j)} 1 = \sum_{k=1}^{n} \mathbb{I}_{k \leq i}\mathbb{I}_{k \leq j} = \sum_{k=1}^n B_{ik} C_{kj}$$ where $B_{ik} = \begin{cases} 1 & k \leq i\\ 0 & \text{else}\end{cases}$ and $C_{kj} = \begin{cases} 1 & k \leq j\\ 0 & \text{else}\end{cases}$.

This should enable you to get the decomposition.

Answer 2

If you expand along the bottom row, then most of the $(n-1)\times(n-1)$ subdeterminants are $0$ because the $(n-1)\times(n-1)$ submatrices have their two right-most columns identical.

So expansion along the bottom row leaves only the last two terms: $$(-1)(n-1)\det(A_{n-1})+n\det(A_{n-1})$$ which is just $\det(A_{n-1})$. So inductively, $\det(A_{n})=\det(A_{n-1})=\cdots=\det(A_{1})=1$

Answer 3

If you consider the matrix $$ B_n=\begin{bmatrix} 1&0&\cdots&0\\ 1&1&0&\cdots\\ &&\cdots\\ 1&1&\cdots &1 \end{bmatrix} $$ (i.e. the $i,j$ entry of $B_n$ is $1$ is $i\geq j$ and $0$ otherwise), then the $k^{\rm t h}$ column of $A_n$ is obtained by adding the first $k$ columns of $B_n$. So $$ \det A_n=\det B_n=1. $$

Answer 4

Recall that adding a multiple or subtracting a multiple of one row does not change the value of the determinant, see, for example ProofWiki.

Using this fact and Laplace expansion you get

$$|A_4|= \begin{vmatrix} 1& 1& 1& 1\\ 1& 2& 2& 2\\ 1& 2& 3& 3\\ 1& 2& 3& 4 \end{vmatrix}= \begin{vmatrix} 1& 1& 1& 1\\ 0& 1& 1& 1\\ 0& 1& 2& 2\\ 0& 1& 2& 3 \end{vmatrix}= \begin{vmatrix} 1& 1& 1\\ 1& 2& 2\\ 1& 2& 3 \end{vmatrix}=|A_3|.$$ (We have subtracted the first row from the other rows.)

In the same way you get $|A_{n+1}|=|A_n|$.