= Solution
Suppose that $Q$ were not surjective. Its image would be a proper subspace of $F^n$, so there would be a nonzero functional
$$
\ell(y_1,\ldots,y_n)=\sum_{j=1}^nc_jy_j
$$
vanishing on $\operatorname{im}Q$. Then
$$
0=\ell(Qx)=\sum_{j=1}^nc_jq_j(x)
$$
for every $x\in V$, contradicting the linear independence of the $q_j$. Hence the <surjectivity of independent linear functionals> gives
$$
\boxed{Q:V\to F^n\text{ is surjective}}.
$$
Let $K=\bigcap_j\ker q_j=\ker Q$ and suppose $K\subseteq\ker f$. Define
$$
g:F^n\to F,
\qquad
g(Qx)=f(x).
$$
This is well-defined: if $Qx=Qy$, then $x-y\in K$, so $f(x)=f(y)$. It is linear, and surjectivity of $Q$ means it is defined on all of $F^n$. Thus there are <scalars> $a_1,\ldots,a_n$ such that $g(y)=\sum_ja_jy_j$. Therefore
$$
f(x)=g(Qx)=\sum_{j=1}^na_jq_j(x),
$$
and hence
$$
\boxed{f\in\operatorname{span}\{q_1,\ldots,q_n\}}.
$$
Back to article page
