Find the rank of $A_n$ in terms of $n$

Question

Let $A_n$ be the real $n × n$ matrix $(n ≥ 2)$ whose $(i, j)$ entry is $i − j$. What is the rank of $A_n$ as a function of $n$?

My attempt:- $A_2$ has rank $2$. It is obvious. $A_3$ has also rank $2$. Since, $A_3$ has determinant $0$ and has a submatrix of order $2$ with determinant non-zero. Similarly, I could conclude that $A_4$ has rank $2$. Using this method I can't go beyond. I also know that $A_n$ is a skew-symmetric matrix. $\det A_n=0,\forall n.$ I am not able to draw conclusion beyond this. Please help me.

Answer 1

Hint: Can you show that the matrix $B = \begin{bmatrix}1 & 1 & \cdots & 1 \\ 2 & 2 & \cdots & 2 \\ \vdots & \vdots & \ddots & \vdots \\ n & n & \cdots & n\end{bmatrix}$ (i.e. $B_{i,j} = i$) has rank $1$?

Similarly, can you show that the matrix $C = \begin{bmatrix}1 & 2 & \cdots & n \\ 1 & 2 & \cdots & n \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 2 & \cdots & n\end{bmatrix}$ (i.e. $C_{i,j} = j$) has rank $1$?

Once you do that, what can you say about the rank of $A = B-C$?

Answer 2

Hint: Show that column $k$ of the matrix is given by $A + k B$. (Find these column vectors)

Hence, the matrices have rank 2.

Answer 3

The upper left $2\times 2$ matrix is always the same and has rank $2$, so $\operatorname{rank}A_n\ge 2$ for all $n\ge 2$. Moreover, $$ a_{1j}-a_{ij} = 1-j-(i-j) = 1-i $$ means that subtracting row $i$ from row $1$ is always a multiple of the vector $(1,1,\ldots,1)$ and thus shows that $\operatorname{rank}A_n = 2$ for all $n$.