First, I apologize if this is a duplicate question. I also must apologize if this has a trivial solution.
This question has two parts:
Let $R$ be a commutative ring with $1$, and let $F = R^n$ be a free module of finite rank.
- Is it possible to construct a set of vectors $Z = \{v_1,\ldots,v_{n+1}\}$ such that every subset of cardinality $n$ is a basis of $F$?
- Does there exist $A\in Mat_{n+1}(R)$ such that $\det(A) = 0$, and $$\text{adj}(A) = \left(\begin{array}{cccc} \alpha_1 & \cdots & \cdots & \alpha_{n+1}\\ 0 & \cdots & \cdots & 0\\ \vdots & \ & \ & \vdots\\ 0 & \cdots & \cdots & 0 \end{array}\right)$$ where $\alpha_i \neq 0$, $\forall \ 1\leq i \leq n+1$?
- I'm fairly certain this question is entirely trivial, as we could simply pick the standard basis $\{e_1,\ldots,e_n\}$ and then let $v = e_1 + \cdots + e_n$, and we claim that $Z = \{e_1,\ldots,e_n,v\}$ has the desired property. In fact, this should work for any basis $X = \{x_1,\ldots,x_n\}$, and setting $Z = \{x_1,\ldots,x_n,v\}$, where $v = x_1 + \cdots + x_n$, we claim $Z$ has the desired property. (Does this work?)
- This is a bit less clear. My initial instinct is that the answer is no, but I have not yet been able to show it.
Full disclosure: This is not for homework, but is related to an issue that came up in thinking through a problem for a class.
It is strange that you can answer (1) but not (2). With your example in (1), consider $$A= \begin{bmatrix} 0&v_1^T\\ 0&v_2^T\\ \vdots&\vdots\\ 0&v_{n+1}^T \end{bmatrix} =\left[\begin{array}{c|c} \begin{array}{c}0\\ \vdots\\ 0\end{array} & \Large I_n\\ \hline 0 & \begin{array}{ccc}1&\cdots&1\end{array}\\ \end{array} \right]. $$