2
$\begingroup$

$n$ 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.

$\endgroup$
0
1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Thanks. Then, do you have any idea to find the subspace annihilated by $f_1, ..., f_n$? $\endgroup$ – Majid May 18 '18 at 11:52
  • $\begingroup$ 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? $\endgroup$ – Majid May 18 '18 at 11:58
  • $\begingroup$ @Majid yes, $AX = 0$ is actually two linear equations with $n$ variables $\endgroup$ – Anton Grudkin May 18 '18 at 12:01
4
$\begingroup$

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 $$ \det\begin{pmatrix} 0 & -1 & -2 \\ 1 & 0 & -1 \\ 2 & 1 & 0 \end{pmatrix} = 0. $$

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.