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 f1,…,fn,g∈V∗. Show that g∈⟨f1,…,fn⟩ if and only if ⋂ni=1kerfi⊂kerg

One direction of this is of course trivial - if g is in the span of the fi, 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 ⋂ni=1kerfi⊂kerg, and let the base field be k. Fix some basis (ei)i of kn, and define a map:

f:V⟶knv⟼∑ifi(v)ei

Evidently kerf=⋂ni=1kerfi. By our assumption that kerf⊂kerg, we get that ˉg is well-defined, where ˉϕ represents the map induced by ϕ under quotient by kerf, and we see commutativity of the following diagram:

where the isomorphism theorem guarantees the invertibility of π∘ˉf. This diagram tells us the following:

g=ˉg∘(π∘ˉf)−1∘π∘f

But as a linear map from kn→k, we see that h:=ˉg∘(π∘ˉf)−1∘π is simply a k-linear combination of the dual basis elements (the component projections). But then since g factors as g=h∘f, this is just the statement that g is a linear combination of the fi by our definition of f, that is, g∈⟨f1,…,fn⟩.