IMC 2005 · Day 1 · P1

For $n = 1$ the rank is 1. Now assume $n \ge 2$. Since $A = (i)_{i,j=1}^{n} + (j)_{i,j=1}^{n}$, matrix $A$ is the sum of two matrixes of rank 1. Therefore, the rank of $A$ is at most 2. The determinant of the top-left $2 \times 2$ minor is $-1$, so the rank is exactly 2.

Therefore, the rank of $A$ is 1 for $n = 1$ and 2 for $n \ge 2$.

Consider the case $n \ge 2$. For $i = n, n-1, \dots, 2$, subtract the $(i-1)$th row from the $n$th row. Then subtract the second row from all lower rows. $$\operatorname{rank} \begin{pmatrix} 2 & 3 & \dots & n+1 \\ 3 & 4 & \dots & n+2 \\ \vdots & & \ddots & \vdots \\ n+1 & n+2 & \dots & 2n \end{pmatrix} = \operatorname{rank} \begin{pmatrix} 2 & 3 & \dots & n+1 \\ 1 & 1 & \dots & 1 \\ \vdots & & \ddots & \vdots \\ 1 & 1 & \dots & 1 \end{pmatrix} = \operatorname{rank} \begin{pmatrix} 1 & 2 & \dots & n \\ 1 & 1 & \dots & 1 \\ 0 & 0 & \dots & 0 \\ \vdots & & \ddots & \vdots \\ 0 & 0 & \dots & 0 \end{pmatrix} = 2.$$