Let $V$ be a vector space. Let $S=\{v_1,v_2,\cdots,v_n\}\subseteq V$. Let $I=\{X\subseteq S:X\text{ is linearly independent}\}$ ...

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I am learning matroids and I saw the following example:

Let $V$ be a vector space. Let $S=\{v_1,v_2,\cdots,v_n\}\subseteq V$. Let $I=\{X\subseteq S:X\text{ is linearly independent}\}$. Then $(S,I)$ is the vector matroid.

The problems is that I don't understand why this is true. If we have the $3\times 4$-Matrixenter image description here

and $A=\{e_1,e_2\}$, $B=\{e_2,e_3,e_4\}$, then the only vector that we can add to $A$ is $e_4$, but $A\times\{e_4\}$ isn't independent. I am sure that there is a mistake in my argument, but I can't find it.

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Here $B\setminus A = \{e_3,e_4\}$ so it is not the case that $e_4$ is the only vector that can be added to $A$. Clearly $A\cup\{e_3\}$ is independent.