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According to the direct sum theorem of matroid, (1) the union of 2 matroids whose ground sets are disjoint and (2)whose independent set is defined as the union of the independent sets of the two respective matroids, is a matroid.

when we say two matroid in this context, do we refer to the same types of matroids?

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What do you mean by the "type" of a matroid? – Calvin McPhail-Snyder Aug 14 '12 at 2:39
For example In linear algebra we can have a set of vectors as ground sets and the independent set correspond to the linearly independent vectors in the set. In graphs we can define the independent set as the set of edges which does not result to cycles. – dumbo Aug 14 '12 at 2:52
Then the answer is no. The type of the matroids doesn't matter. – Ido Aug 14 '12 at 6:55

Let's have a matroid built from a matrix and a matroid built from a graph and construct the new object and see if we encounter anything bothering us because of the difference of motivation of the two matroids or not.

Take $A$, a $2\times 2$ matrix as $\begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix}$ and Define the Graph $G$ be the complete graph on 3 vertices, meaning $K_3$ with edges called $a$, $b$ and $c$. Then if I call the matroids coming from them with $M_1$ and $M_2$ respectively, we have (index columns of $A$ with $1$ and $2$ from left to right);

$\begin{array}{l} E(M_1)=\{1,2\}\\ I(M_1)=\{\emptyset,\{1\}\}\\ E(M_2)=\{a,b,c\}\\ I(M_2)=\{\emptyset,\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\}\} \end{array}$

Conditions of your sum theorem are hold.

$\begin{array}{l} E(M):=\{1,2,a,b,c\}\\ I(M)=\{\emptyset,\{1\},\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\}\} \end{array}$

As you can check readably, this is indeed a matroid.

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