Annihilator of subspace, linear functionals exercise I am trying to solve the following exercise:
Problem
Let $F$ be a subfield of the complex numbers. We define $n$ linear functionals in $F^n$ ($n \geq 2$) with $$f_k(x_1,...,x_n)=\sum_{j=1}^n(k-j)x_j$$
What is the dimension of the subspace annihilated by $f_1,...,f_n$?
I am not so sure what to do here. By the way these functionals are defined, it is clear that the $k-th$ coordinate of any vector $\alpha$ in $F^n$ is "annihilated" by the functional $f_k$. For example, if I consider the standard basis of the vector space, $\mathcal B=\{e_1,...,e_n\}$, then $f_i(e_j)=\delta_{ij}$. I don't realize how to deduce from these facts the dimension of the subspace. Any help would be appreciated.
 A: You can try thinking of the linear transformation $A: F^n \rightarrow F^n$ which is made up of the $n$ functionals, i.e., 
$Ax = (f_1(x),\ldots,f_n(x))$
and so the question becomes finding the dimension of the kernel. With respect to the standard basis, in the special case $n=4$, $A$ has matrix
$$
\left(\begin{array}{rrrr}
0 & -1 & -2 & -3 \\
1 & 0 & -1 & -2 \\
2 & 1 & 0 & -1 \\
3 & 2 & 1 & 0
\end{array}\right)
$$
Row reduce using the second row to see what the rank is. Your answer for the rank will hold for all $n$, not just 4.
A: Another solution is the following.
Notice that $f_s(x_1,\ldots,x_n)=sf(x_1,\ldots,x_n)-g(x_1,\ldots,x_n)$, 
where $f(x_1,\ldots,x_n)=\sum_{j=1}^nx_j$ and $g(x_1,\ldots,x_n)=\sum_{j=1}^njx_j$.
If $\displaystyle(x_1,\ldots,x_n)\in \bigcap_{s=1}^2\ker f_s$ then $0=f_1(x_1,\ldots,x_n)-f_2(x_1,\ldots,x_n)=(1-2)f(x_1,\ldots,x_n)$. Thus, $f_2(x_1,\ldots,x_n)-2f(x_1,\ldots,x_n)=-g(x_1,\ldots,x_n)=0$. So, $\displaystyle\bigcap_{s=1}^2\ker f_s \subset \ker f\cap\ker g $
Thus, if $k>1$ then $\displaystyle\bigcap_{s=1}^k\ker f_s\subset \bigcap_{s=1}^2\ker f_s \subset \ker f\cap\ker g $. Of course, $\displaystyle \ker f\cap\ker g \subset \bigcap_{s=1}^k\ker f_s$.
So, if $k>1$ then $\displaystyle\bigcap_{s=1}^k\ker f_s= \ker f\cap\ker g $.
Now, $\ker f\cap\ker g =\{(x_1,\ldots,x_n);\ x_1+\ldots+x_n=x_1+2x_2+\ldots+nx_n=0\}$.
It is easy to find a basis for this subspace.
