Let $n \in \mathbb{N}$ and $A = (a_{ij}) \in \mathbb{R}^{n \times n}$ where $a_{ij} := i +j \quad \forall 1 \leq i, j \leq n$. Find the rank of matrix $A$ and prove its value of every $n \in \mathbb{N}$.
$\begin{align} \operatorname{rank}( \begin{pmatrix}2 \end{pmatrix})&=1\\ \operatorname{rank}( \begin{pmatrix} 2 & 3 \\ 3 & 4 \\ \end{pmatrix})&=2\\ \operatorname{rank}( \begin{pmatrix} 2 & 3 & 4\\ 3 & 4 & 5\\ 4 & 5 & 6\\ \end{pmatrix})&=2\\ \operatorname{rank}( \begin{pmatrix} 2 & 3 & 4 & 5\\ 3 & 4 & 5 & 6\\ 4 & 5 & 6 & 7\\ 5 & 6 & 7 & 8\\ \end{pmatrix})&=2 \\ \operatorname{rank}( \begin{pmatrix} 2 & 3 & 4 & 5 & 6\\ 3 & 4 & 5 & 6 & 7\\ 4 & 5 & 6 & 7 & 8\\ 5 & 6 & 7 & 8 & 9\\ 6 & 7 & 8 & 9 & 10\\ \end{pmatrix})&=2 \\ \end{align}$
I have no idea how to start that task. Thank you in advance for your help.
Well, the rows $(2,3,\ldots,n+1)$ and $(3,4,\ldots,n+2)$ lie in the row span and so their difference $(1,1,\ldots,1)$.
Now, each row $(i,i+1,\ldots,n+i-1)$, $2\leq i\leq n+1$, is a linear combination of $(2,3,\ldots,n+1)$ and $(1,1,\ldots,1)$, namely
$$(i,i+1,\ldots,n+i-1) = 1 \cdot (2,3,\ldots,n+1) + (i-2)\cdot (1,1,\ldots,1).$$
This shows that the row space (rank) of the matrix is $2$.
For instance $(n=5)$:
$(3,4,5,6,7) = (2,3,4,5,6) + (1,1,1,1,1)$
and
$(4,5,6,7,8) = (2,3,4,5,6)+ 2\cdot (1,1,1,1,1)$.