Let F be subfield of complex number We define n linear functional as $f_k(x_1,....x_n)=\sum^n_{i=1}(k-i)x_i$
I wanted to find dimension of subspace annihilated by $f_1,...f_k$
My attempt I had use 2 method which gives me 2 different answer. I convinced with both but I do not know which is right or wrong.
1 Way:
Rank of matrix formed by the above functional is $2$.
Implies Null space union of all functionals is $n-2$ dimensional
2 ways:
$f_1(1,0,0,0...,0)=f_1(e_1)=0 $ by putting in the above formula
Similary $f_k(e_k)=0$ for every $k=1,...,n$.
Also $f_k(e_j)\neq 0 ,j\neq k$ so $e_1$ is annihilated by $f_1$ but by no other functional. So if we take intersection of subspace annihilated by all then it will be $0$.
This implies only $0$ is annihilated by all functionals.
So dimension of subspace annihilated by that functionals is 0.
Where does my mistake occur?
Any help will be appreciated