Exercise 9.23 of Advanced Linear Algebra by Steven Roman: Let $f$ be a nonzero linear functional on a subspace $S$ of a finite-dimensional inner product space $V$ and let $K=\text{ker}(f)$. Let $f(v)= \langle v, R_f \rangle$. Show that if $g \in V^*$ is an extension of $f$, that is, $g|_S=f$, then $R_g \in K^{\perp} - S^{\perp}$, where $R_g$ is the Riesz vector of $g$.
I do not know how to show that. Here are my thoughts:
$g|_S=f$ implies $\langle s, R_g \rangle =\langle s, R_f \rangle$ for all $s \in S$. So, $R_g-R_f \perp S$. Since $R_f \in S$, $R_f$ is a Fourier expansion of $R_g$ on $S$ and $R_g=R_f+w$ for some $w \in S^{\perp}$. In addition, $R_f \in K^{\perp} \cap S$. Hence, $R_g \in (K^{\perp} \cap S) \oplus S^{\perp}$. But this is not consistent with the aim of the question.
Would you tell me how I need to proceed?