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Suppose that $Q$ were not surjective. Its image would be a proper subspace of $F^n$, so there would be a nonzero functionalvanishing on $\mathrm{im}Q$. Thenfor every $x\in V$, contradicting the linear independence of the $q_j$. Hence the surjectivity of independent linear functionals gives

Let $K={\bigcap }_j\ker q_j=\ker Q$ and suppose $K\subseteq \ker f$. DefineThis 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 $a_1,\dots ,a_n$ such that $g(y)={\sum }_j{a}_j{y}_j$. Thereforeand hence