Determining the rank of an $n \times n$ matrix

Question

If I have the following $n \times n$ matrix

$$\begin{bmatrix} 2 & 3 & \cdots & n+1\\ 3 & 4 & \cdots & n+2\\ \vdots & \vdots & \ddots & \vdots\\ n+1 & n+2 & \cdots & 2n\\ \end{bmatrix}$$

how can I determine its rank?

I know that I need to check for the maximum number of linearly independent columns. However, I’m having a hard time determining the linearly independent columns. Thanks for those who’ll help.

Answer 1

Note that elementary column operations do not change the column space of the matrix; in particular they do not alter its rank. Perform the following sequence of column operations: subtract the $(i-1)$th column from the $i$th, for each $i$ starting from $i=n$ down to $i=2$, to get

$$\begin{bmatrix} 2 & 1 & \cdots & 1\\ 3 & 1 & \cdots & 1\\ \vdots & \vdots & \ddots & \vdots\\ n+1 & 1 & \cdots & 1 \end{bmatrix},$$

which has rank $2$, assuming the matrix wasn't $1\times 1$ to begin with. So the original matrix has rank $2$ if $n>1$, and rank $1$ if $n=1$.

Answer 2

Let $v:=\pmatrix{2\\ 3\\ \dots \\ n+1}$ and $w:=\pmatrix{1\\ 1\\ \dots \\ 1}$. If we call $v_i$ the $i$-th column of the matrix, then $v_i=v+\left(i-1\right)w$ hence the vector space generated by the $v_i$'s is the same as the subspace generated by $v$ and $w$.