Question-59
Consider the square matrix \(\displaystyle A_{n}\) of order \(\displaystyle n\) with entries from \(\displaystyle 1,\cdots ,n^{2}\) arranged in a row-wise manner, for \(\displaystyle n\geqslant 2\). For example:
\[ \begin{equation*} A_{2} =\begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix} ,\ A_{3} =\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} \end{equation*} \]
What is the rank of \(\displaystyle A_{n}\)?
Express every row of \(\displaystyle A_{n}\) as a linear combination of the first two rows.
Let us see the general form of \(\displaystyle A_{n}\) and study the pattern:
\[ \begin{equation*} A_{n} =\begin{bmatrix} 1 & 2 & \cdots & n\\ n+1 & n+2 & \cdots & n+n\\ 2n+1 & 2n+2 & \cdots & 2n+n\\ \vdots & \vdots & & \vdots \\ ( n-1) n+1 & ( n-1) n+2 & \cdots & ( n-1) n+n \end{bmatrix} \end{equation*} \]
We see that the element \(\displaystyle A_{rc}\) is \(\displaystyle ( r-1) n+c\). We also have \(\displaystyle A_{1c} =c\) and \(\displaystyle A_{2c} =n+c\). These are the two elements in column \(\displaystyle c\) in the first and second rows. We will now try to express \(\displaystyle A_{rc}\) as a combination of \(\displaystyle A_{1c}\) and \(\displaystyle A_{2c}\):
\[ \begin{equation*} \begin{aligned} A_{rc} =( r-1) n+c & =( r-1)( n+c-c) +c\\ & =( r-1)( n+c) +( 2-r) c\\ & =( 2-r) A_{1c} +( r-1) A_{2c} \end{aligned} \end{equation*} \]
In terms of row operations, the \(\displaystyle r^{th}\) row is:
\[ \begin{equation*} R_{r} =( 2-r) R_{1} +( r-1) R_{2} \end{equation*} \]
Rows 1 and 2 are linearly independent. Every other row can be expressed as a linear combination of the first two rows. Therefore, the rank of \(\displaystyle A_{n}\) is equal to \(\displaystyle 2\) for \(\displaystyle n\geqslant 2\)!