Let \(F\) be a subfield of the complex numbers. We define \(n\) linear functionals on \(F^n\) (\(n \geq 2\)) by \[ f_k(x_1, \ldots, x_n) = \sum_{j=1}^{n} (k - j)x_j, \quad 1 \leq k \leq n. \] What is the dimension of the subspace annihilated by \(f_1, \ldots, f_n\)?
Solution
A vector \(x=(x_1,\ldots,x_n)\) is annihilated by every \(f_k\) exactly when \[ 0=f_k(x)=k\sum_{j=1}^n x_j-\sum_{j=1}^n jx_j \qquad(1\le k\le n). \] Since \(n\ge2\) and \(F\) has characteristic zero, subtracting the equations for two distinct values of \(k\) gives \[ \sum_{j=1}^n x_j=0. \] Substituting back then gives \[ \sum_{j=1}^n jx_j=0. \] These are two independent linear equations: their coefficient vectors \[ (1,\ldots,1),\qquad(1,2,\ldots,n) \] are not scalar multiples. Hence the common annihilator has codimension \(2\) in \(F^n\), and therefore dimension \[ \boxed{n-2}. \]