Let $n \in \mathbb{N}$, and $U,V$ two linear subspaces of $\mathbb{R}^n$ of the same dimension. Could one always make a matrix $A \in \mathbb{M}^{n \times n}(\mathbb{R})$ such that $spanA = U$ and $spanA^T = V$?
I find it hard to find a counterexample. I know that for any $n \in \mathbb{N}$, there is such a matrix if $\dim U = \dim V \in \{0,n\}$, we could just take $0,I$ as matrices. Moreover, if $U,V$ are lines, that means $U = \mathbb{R}v$ and $V = \mathbb{R}w$ for some nontrivial vectors $v,w$, then the matrix below works.
$$ \left( \ w_1v \ | \ w_2v \ | \ \cdots \ | \ w_nv \ \right) $$ Here $w_i$ are coordinates of the vector $w$, and each entry represents a column this way.
Then I took two subspaces of dimension two in $\mathbb{R}^3$, namely $$ U \quad = \quad span \left\{ \left( \begin{array}{ccc} 1 \\ 0 \\ 0 \end{array} \right) , \left( \begin{array}{ccc} 0 \\ 1\\ 0 \end{array} \right) \right\} \qquad V \quad = \quad \left( \begin{array}{ccc} 0 \\ 1\\ 0 \end{array} \right)^\perp $$ The matrix we need here is $$ \left( \begin{array}{ccc} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 0 \end{array} \right) $$
I find it hard to deal with the general case. Could you give me some help to establish the statement, or give a counterexemple?