Question: > 
> 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 $f\_1, \dotsc, f\_n$?
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Approaches I've tried so far:

1. 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.
2. 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.

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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.

Answer: 
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.)
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.