Given $n$ linear functionals $f_k(x_1,\dotsc,x_n) = \sum_{j=1}^n (k-j)x_j$, what is the dimension of the subspace they annihilate? 
Let $F$ be a subfield of the complex numbers. We define $n$ linear functionals on $F^n$ ($n \geq 2$) by $f_k(x_1, \dotsc, x_n) = \sum_{j=1}^n (k-j) x_j$, $1 \leq k \leq n$. What is the dimension of the subspace annihilated by $f_1, \dotsc, f_n$?

Approaches I've tried so far:


*

*Construct a matrix $A$ whose $k$th row's entries are the coefficients of $f_k$, i.e., $A_{ij} = i - j$, and compute the rank of the matrix. Empirically, the resulting matrix has rank 2 for $n = 2$ to $n = 6$, but I don't see a convenient pattern to follow for a row reduction type proof for the general case.

*Observe that $f_k$, $k \geq 2$ annihilates $x = (x_1,\dotsc, x_n)$ iff $$\sum_{j=1}^{k-1} (k-j)x_j + \sum_{j = k+1}^n (j-k)x_j = 0$$ iff $$k\left(\sum_{j=1}^{k-1} x_j - \sum_{j=k+1}^n x_j\right) = \sum_{j=1}^{k-1} jx_j - \sum_{j=k+1}^n jx_j,$$ and go from there, but I do not see how to proceed.



Note: This is Exercise 10 in Section 3.5 ("Linear Functionals") in Linear Algebra by Hoffman and Kunze; eigenvalues and determinants have not yet been introduced and so I am looking for direction towards an elementary proof.

Here is the proof for rank of $A$ being 2. It is from @daw given as answer to another question with a little edition. 
It is clear for $n=2$. 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.
