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: 
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.
Dimension of the subspace annihilated by $f\_i$'s is the dimension of the solution space of the system $$AX=0$$
where the $m^\text{th}$ row of $A$ is defined by $$A\_m=(m-1,m-2,\dots,m-n),1\le m\le n.$$
For $m\ge 3,$ perform the following elementary operations
\begin{align}
&(1)A\_m\to(1-m)A\_2+A\_m\\
&(2)A\_m\to \frac{1}{(m-2)} A\_m\\
&(3)A\_m\to A\_m+A\_1
\end{align}
On performing $(1)$ $$A\_m=\big(0,m-2,2(m-2),3(m-2),\dots,(n-1)(m-2)\big).$$
On $(2)$ $$A\_m=-A\_1.$$
On $(3)$ $$A\_m=0.$$
It is evident that $A\_1$ and $A\_2$ are linearly independent.

Thus, the dimension of the subspace annihilated = $n-2.$