Can anybody please help me to solve?
Let $F$ be a field. We define $n$ linear functionals on $F^n$, for $n \geq 2 $, by: $f_k(x_1,\ldots,x_n) = \sum \limits_{j = 1}^n (k-j)x_j $. What is the dimension of the subspace annihilated by $f_1,f_2,\ldots,f_n$?
I have tried to prove that the $n$ functionals are linear independent.
For $n=3$ the functionals are $$f_1=-x_2-2x_3,\\ f_2=x_1-x_3, \\ f_3=2x_1-x_2.$$ The space annihilated is then obtained by setting all these to zero, and one gets a matrix to row reduce which has rank $2$ making the subspace annihilated have dimension one. So in this case the functioals are not independent.
In general note that assuming only $f_1=f_2=0,$ we have $0=f_2-f_1=\sum x_j,$ and then since $f_1=\sum x_j-\sum jx_j$ we have also $\sum jx_j=0.$ So from these first two $f_j$ being zero the others are also zero as a consequence, because $f_k=k\sum x_j-\sum jx_j$ for each $k.$ It seems clear the dimension of the space annihilated by using just the first two functionals $f_1,f_2$ is $n-2.$ By the above, use of more of the functionals $f_3,f_4,$ etc. does not change the space annihilated, since their being zero is a consequence of the fist two being zero. If my analysis is right, this gives the dimension of the annihilated space as $n-2$ for any $n.$
