Here’s a solution to an interesting problem from an undergraduate linear algebra course. The problem statement is as follows:

Let \(V\) be a vector space, and \(f_1, \dots, f_n, g \in V^{*}\). Show that \(g \in \left< f_1, \dots, f_n \right>\) if and only if \(\bigcap_{i=1}^n \textrm{ker} f_i \subset \textrm{ker} g\)

One direction of this is of course trivial - if \(g\) is in the span of the \(f_i\), certainly their kernel intersection must lie within that of \(g\). The converse is much trickier (the finite dimensional case is a good toy version for grasping the point of the problem).

Here is an approach. Suppose \(\bigcap_{i=1}^n \textrm{ker} f_i \subset \textrm{ker} g\), and let the base field be \(k\). Fix some basis \((e_i)_i\) of \(k^n\), and define a map:

\begin{equation*} \begin{split} f : V &\longrightarrow k^n \\ v &\longmapsto \sum_i f_i(v) e_i \end{split} \end{equation*}

Evidently \(\textrm{ker} f = \bigcap_{i=1}^n \textrm{ker} f_i\). By our assumption that \(\textrm{ker} f \subset \textrm{ker} g\), we get that \(\bar{g}\) is well-defined, where \(\bar{\phi}\) represents the map induced by \(\phi\) under quotient by \(\textrm{ker} f\), and we see commutativity of the following diagram:

where the isomorphism theorem guarantees the invertibility of \(\pi \circ \bar{f}\). This diagram tells us the following:

\begin{equation*} g = \bar{g} \circ (\pi \circ \bar{f})^{-1} \circ \pi \circ f \end{equation*}

But as a linear map from \(k^n \rightarrow k\), we see that \(h := \bar{g} \circ (\pi \circ \bar{f})^{-1} \circ \pi\) is simply a \(k\)-linear combination of the dual basis elements (the component projections). But then since \(g\) factors as \(g = h \circ f\), this is just the statement that \(g\) is a linear combination of the \(f_i\) by our definition of \(f\), that is, \(g \in \left< f_1, \dots, f_n\right>\).