Here is something I want to learn ( Problem #2.2.7 in Matroid Theory, first edition( page 88 ) , by James Oxley:
Finding an $\mathbb R$-representation for the dual matroid $M^*$ when $M$ is the vector matroid of the following matrix over $\mathbb R:$( sorry I was not able to type the matrix so I had to take a picture of it):
My questions are:
I know the following theorem:
Theorem 2.2.8
Let $M$ be the vector matroid of the matrix $[I_r | D]$ where the columns of this matrix are labelled, in order, $e_1, e_2, \dots , e_n$ and $1\leq r < n.$ Then $M^*$ is the vector matroid of $[ - D^T | I_{n-r} ]$ where its columns are also labelled $e_1, e_2, \dots , e_n$ in that order.
So I am guessing that the answer to this question is just to calculate $[ - D^T | I_{n-r} ],$ am I correct? But if so, the given matrix is not a square matrix( I do not know the number of rows of the matrix), so it is not very clear to me how to calculate $[ - D^T | I_{n-r} ],$ we can only find inverses of square matrices so I am confused now.
I tried to see the matrix for different $n$ values and here is what I came up with:
$n = 1,2$ the pattern cannot be satisfied. For $n=3,$ I do not know where exactly will be the $-1$ and what will be beside it.
Note that, in the second edition of "Matroid Theory" book by Oxley, the matrix was different and it sounds like it does not have a pattern like this in it, I suspect it is a typo.
What do you mean by inverse? This is still a square matrix by construction. Call the matrix $A$. Notice that if we consider the rows of $A$, say $\{r_i\}_{i\in [n]}$, then for $1< i<n$ we have that $r_i = e_i+e_{i-1}$, $r_1 = e_1+e_n$ and $r_n = e_{n-1}+(-1)^ne_n$, where $e_j = (0,\cdots ,1,\cdots 0)$ has a $1$ exactly at position $j$ and has zeroes everywhere else.
We can apply row operations to this matrix and maintain the matroid, so replace $r_2$ by $r_2-r_1=(e_2+e_{1})-(e_1+e_{n})=e_2-e_n$, and replace $r_3$ by $r_3-r_2=(e_3+e_{2})-(e_2-e_n)=e_3+e_n$ and notice that doing this for every row $r_i$ with $1\leq i<n$ you get $r_i = e_i+(-1)^{i-1}e_n$. Now row $r_n$ replace it with $r_n-r_{n-1}=(e_{n-1}+(-1)^ne_n)-(e_{n-1}+(-1)^{n}e_n)=0$. Erase the last row and you end up with the matrix $$A = [I_{n-1}|D]$$ with $D = (1,-1,\cdots ,(-1)^n)^T$ and so, using the theorem you copied here, the representation of the dual is given by $$[-1,1,\cdots ,(-1)^{n+1}|I_1].$$