Answer: A Determinant of Sums

Key Idea / Intuition

The matrix looks complicated, but a clever sequence of row operations strips it down to something triangular almost immediately. The key observation is: subtract each row from the one below it. Since adjacent rows of $A$ differ by at most one entry, this telescoping turns the dense matrix into a bidiagonal upper-triangular matrix, whose determinant reads off instantly from the diagonal.

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Formal Proof / Solution

Step 1: Row reduction.

Perform the row operations $R_k \leftarrow R_k - R_{k-1}$ for $k = n, n-1, \ldots, 2$ (in that order, working from the bottom up).

For the original matrix, row $k$ has entries: $A_{k,j} = \min(k,j).$

After subtracting row $k-1$ from row $k$, the new row $k$ (for $k \geq 2$) has entries: $\min(k,j) - \min(k-1,j).$

Now observe:

• If $j < k$: $\min(k,j) = j = \min(k-1,j)$, so the difference is $0$.
• If $j = k$: $\min(k,k) - \min(k-1,k) = k - (k-1) = 1$.
• If $j > k$: $\min(k,j) = k$ and $\min(k-1,j) = k-1$, so the difference is $1$.

So the new row $k$ (for $k \geq 2$) is: $(\underbrace{0, \ldots, 0}_{k-1 \text{ zeros}}, 1, 1, \ldots, 1).$

Row $1$ is unchanged: $(1, 1, 1, \ldots, 1)$.

Step 2: The resulting matrix.

After these row operations, the matrix becomes:

1 & 1 & 1 & 1 & \cdots & 1 \\ 0 & 1 & 1 & 1 & \cdots & 1 \\ 0 & 0 & 1 & 1 & \cdots & 1 \\ \vdots & & & \ddots & & \vdots \\ 0 & 0 & \cdots & 0 & 1 & 1 \\ 0 & 0 & \cdots & 0 & 0 & 1 \end{pmatrix}.$$ This is upper triangular with all diagonal entries equal to $1$. **Step 3: Read off the determinant.** Since row operations of the form $R_k \leftarrow R_k - R_{k-1}$ do not change the determinant, and the resulting matrix is upper triangular with all $1$s on the diagonal: $$\det(A) = 1.$$ **Sanity check for small $n$:** For $n = 2$: $$\det\begin{pmatrix}1 & 1 \\ 1 & 2\end{pmatrix} = 2 - 1 = 1. \checkmark$$ For $n = 3$: $$\det\begin{pmatrix}1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3\end{pmatrix} = 1(6-4) - 1(3-2) + 1(2-2) = 2 - 1 + 0 = 1. \checkmark$$ **Conclusion:** $$\boxed{\det(A) = 1}$$ for all $n \geq 1$. **Bonus remark:** The matrix $A_{ij} = \min(i,j)$ is the covariance matrix of a standard Brownian motion $(B_1, B_2, \ldots, B_n)$ at integer times. The fact that its determinant is $1$ reflects a deep property of the increments: the Brownian increments $B_1, B_2 - B_1, \ldots, B_n - B_{n-1}$ are i.i.d. $N(0,1)$, so the transformation from $(B_k)$ to its increments is volume-preserving — exactly what determinant $1$ says!