0
$\begingroup$

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 ?

$\endgroup$
4
  • 1
    $\begingroup$ What do you mean by "up to $n$"? $\endgroup$
    – Matemáticos Chibchas
    Jan 19, 2014 at 21:40
  • $\begingroup$ @MatemáticosChibchas mean if it is givin in terms of n. $\endgroup$
    – andre_lamothe
    Jan 19, 2014 at 21:54
  • $\begingroup$ I think the 3n+2 should be 4n+2. $\endgroup$
    – Edward ffitch
    Jan 19, 2014 at 22:01
  • $\begingroup$ @Edwardffitch I fixed it. $\endgroup$
    – andre_lamothe
    Jan 19, 2014 at 22:10

1 Answer 1

Reset to default
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .