I am taking a matroid theory class, and I am having trouble understanding an example we did in class:
Let $F$ be a field, $E$ a ground set, and $V$ a vector space over $F$. Let $\phi : E \longrightarrow V$ be a function. We define the support of a function $d:E \longrightarrow F$ to be $supp(d) = \lbrace e \in E : d(e) \neq 0 \rbrace$, and we define a linear dependence of $\phi$ to be a function $d : E \longrightarrow F$ with finite nonempty support such that $\displaystyle \sum_{e \in E} d(e)\phi(e) = 0$.
Lastly, we define $\mathcal{I} = \lbrace I \subseteq E : $there does not exist a linear dependence $d : E \longrightarrow F$ such that $supp(d) \subseteq I \rbrace$.
Then we say that the pair $(E, \mathcal{I})$ is a matroid, and we attempt to prove it.
It's obvious that $\emptyset \in \mathcal{I}$, and that if $I \in \mathcal{I}$ and $J \subseteq I$ then $J \in \mathcal{I}$.
I'm now trying to prove that if $I$ is maximal in $\mathcal{I}$ and $J \in \mathcal{I}$ is not maximal, then there exists some element $a \in I \setminus J$ such that $J \cup \lbrace a \rbrace \in \mathcal{I}$.
This is where I'm confused. The professor went over it in class, but very quickly and I did not follow much of it. Any help would be appreciated.
Note that you can essentially identify $E$ with its image under $\phi$; it's a collection of vectors, possibly with repetition. A subset of the vectors is in $\mathcal{I}$ if and only if it's a linearly independent set. Then the fact that it's a matroid follows immediately from the properties of linearly independent sets and bases in linear algebra.
In slightly more detail:
$I$ is maximal if and only if the vectors $\{\phi(e) : e \in I\}$ form a basis for the span of all the images of $E$, $\DeclareMathOperator{\span}{span}\span(\phi(E))$.
So if $J$ is not maximal, then the images $\{\phi(e) : e \in J\}$ span a proper subspace of $\span(\phi(E))$.
Hence, the vectors in $\phi(I)$ are not all contained in $\span(\phi(J))$ (otherwise, $\span(\phi(I))$ could not be larger than $\span(\phi(J))$); so at least one $a \in I$ has $\phi(a) \notin \span(\phi(J))$; then $J \cup \{a\}$ is independent.