# Rank of a matrix up to $n$

Suppose I have a matrix, but its elements, rows, and columns, are given in terms of $n$.

For example: $$A = \left[\begin{matrix} 2&6&10&\cdots &4n-2\\ 6&10&14&\cdots &4n+2\\ \vdots&\vdots&\vdots&\ddots &\vdots\\ 4n-2&4n+2&4n+6&\cdots&8n-6 \end{matrix}\right]$$ How would I get its rank ?

• What do you mean by "up to $n$"? – Matemáticos Chibchas Jan 19 '14 at 21:40
• @MatemáticosChibchas mean if it is givin in terms of n. – Ahmed Saleh Jan 19 '14 at 21:54
• I think the 3n+2 should be 4n+2. – Edward ffitch Jan 19 '14 at 22:01
• @Edwardffitch I fixed it. – Ahmed Saleh Jan 19 '14 at 22:10

Its rank is equal to $2$.
Let $a_1,\ldots,a_n$, be its rows, then $$a_2-a_1=a_3-a_2=\cdots=a_n-a_{n-1}.$$ This means that all the rows are linear combinations of the first two ones.