Suppose that $A_1$ and $A_1$ are $\mathbb F$- representations of the matroids $M_1$ and $M_2$ respectively, show that
$$\begin{matrix}
A_1& 0\\
0 & A_2
\end{matrix}$$
is an $\mathbb F$- representation of the matroid $M_1 \oplus M_2.$
My idea is:
Construction a function that is not exactly the identity but the identity plus zero for each of the columns of the last matrix of $M_1 \oplus M_2.$ But I am confused of how to write this rigorously. Could someone help me in this please?
The independent sets of the direct sum are unions of independents from each of the original matroids (notice that the ground sets are disjoint, call them $X_1$ and $X_2$). Call $\mathcal{A}$ the matrix that you propose as the representation and consider $A\subseteq X_1\cup X_2$. The claim is that $$A\in \mathbb{I}_{M_1\oplus M_2} \text{ iff }c_A = \{c_x:x\in A\} \text{ are L.I},$$ where $c_x$ is a column indexed by $x$ in the matrix $\mathcal{A}$.
Let $A\in \mathbb{I}_{M_1\oplus M_2}$, then $A = A_1\cup A_2$ with $A_1\in \mathbb{I}_{M_1}$ and $A_2\in \mathbb{I}_{M_2}$, then $c_{A_1}=\{c_x:x\in A_1\}$ and $c_{A_1}=\{c_x:x\in A_2\}$ are independent. Now, to see that $c_{A}$ is L.I, suppose the opposite, then there are scalars $s_x$ not all zero such that $\sum _{x\in A}s_x\cdot c_x=0$, but clearly $\sum _{x\in A}s_x\cdot c_x=\sum _{x\in A_1}s_x\cdot c_x+\sum _{x\in A_2}s_x\cdot c_x$. Now, by construction of the matrix $\mathcal{A}$ one can see that $\sum _{x\in A_1}s_x\cdot c_x = (b_1,b_2,\cdots ,b_n,0,0,\cdots ,0)^T$ and $\sum _{x\in A_2}s_x\cdot c_x = (0,0,\cdots ,0,b'_1,b'_2,\cdots ,b'_m)^T$ with $n = |X_1|,m = |X_2|$, and so the only way they add up to the zero vector is if both of them are zero, but this is a contradiction in the independence of $A_1$ and $A_2$.
The other direction is immediate why?