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.

All the $f_k$ are linear combinations of the two linear functionals
$$
\sum_{j=1}^n x_j \quad\text{and}\quad \sum_{j=1}^n jx_j;
$$
therefore the dimension is at most $2$. Checking that the dimension is at least $2$ should be easy.
(For an exercise, you might want to use this observation to construct a solution along the lines of your idea #1.)
