Finding representation matrix of $M$*$(K_5)$ above $\mathbb F_2$.
$M$*$(K_5)$ is the dual matroid representing the graph $K_5$, that is, a complete graph with 5 vertices.
How do I solve this? We've never been shown this type of questions @class.
Thanks in advance for any assistance!
If a matroid of rank $m$ over a ground set of $n$ elements is representable over a field $F$, then it is representable by a matrix of the form $[I^m | B ]$, where $B$ is a $m \times (n-m)$ matrix over $F$. Also, any matrix representation can be transformed to one of this form by performing elementary row operations and (possibly) deleting rows. To see why, look at the observations after Definition 2 in this lecture. By Theorem 2 of the same lecture, if a matroid $M$ is representable in a field $F$ by a matrix $[I^m | B]$, then its dual matroid $M^\ast$ is also representable over $F$, by the matrix $[B^t | I^{n-m} ]$.
Moreover, by Lemma 1 of the lecture, the incidence matrix of a graph $G$ corresponds to a representation matrix of the graphic matroid of $G$ over $\mathbf{F}_2$.
So, to find a representation matrix of $M^\ast(K_5)$ over $\mathbf{F}_2$, we start with a representation matrix of $M(K_5)$ over $\mathbf{F}_2$, given by an incidence matrix of the graph:
$$ \left[ \begin{array}{cccccccccc} 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1 \\ \end{array} \right] $$
We perform row operations to obtain the form $[I^m | B]$. As the rank of $M(K_5)$ is $4$, we must delete one row, and it suffices to delete the first one to reach the desired form.
$$ \left[ \begin{array}{cccccccccc} 1 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1 \\ \end{array} \right] $$
Then, the representation matrix of $M^\ast(K_5)$, by the above remarks, is given by
$$ \left[ \begin{array}{cccccccccc} 1 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right] $$