# $A_{ij}=i-j$, for $i , j = 1, …, n$. What is the rank of $A$? ($n\geq 2$).

$n$ linear functionals are given as follows on $F^n$.

$$f_i(x_1, ..., x_n) = \sum_{j=1}^n(i-j)x_j,\ \ 1\leq i\leq n$$

And the dimension of the subspace annihilated by $f_1, ..., f_n$ is asked. I know this space is equal to the solution space of the system $Ax=0$, for the defined matrix in the question. So, if I find the rank of $A$, I have the solution.

I should mention that I just saw this question which makes my question somehow duplicate.

## 2 Answers

It is clear that $A$ is invertible for $n=2$. Now let $n\ge3$. And take $i\ge 3$. Then $$a_{ij} - a_{1j} = i-1, \ a_{2j}-a_{1j} = 1,$$ which implies $$a_{ij} - a_{1j}- (i-1)(a_{2j}-a_{1j}) =0.$$ So rank of $A$ is two, as the first two rows are clearly linearly independent.

• Thanks. Then, do you have any idea to find the subspace annihilated by $f_1, ..., f_n$? – Majid May 18 '18 at 11:52
• If rank of $A$ is two, then the dimension of the solution space of $AX=0$ is $n-2$. Is this the one I am looking for? – Majid May 18 '18 at 11:58
• @Majid yes, $AX = 0$ is actually two linear equations with $n$ variables – Anton Grudkin May 18 '18 at 12:01

For $n>2$ we have $$A_{i,j} + A_{i+2,j} - 2A_{i+1,j} = 0,$$ so "$i$-th row plus $(i+2)$-th row minus $2\times (i+1)$-th row" is zero, and thus $\det A = 0$ (if $n>2$). For example, if $n = 3$, we have $$\det\begin{pmatrix} 0 & -1 & -2 \\ 1 & 0 & -1 \\ 2 & 1 & 0 \end{pmatrix} = 0.$$