A real matrix whith rows generating $U$ and columns generating $V$ 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?
 A: This is indeed always possible. Let $U$ be some $r$-dimensional subspace of $\mathbb{R}^n$. Let $R$ denote any matrix with rowspace $U$. Since $U$ is $r$-dimensional, it follows that the rank of $R$ is $r$ and so the image of $R$ is also $r$-dimensional. 
Let $\mathcal{B}'$ denote a basis for the image of $R$ and extend $\mathcal{B}'$ to a basis $\mathcal{B}$ of $\mathbb{R}^n$. Form the matrix $P$ with columns given by $\mathcal{B}$, with the first $r$ columns consisting of the vectors of $\mathcal{B}'$. It follows that $P$ maps the subspace $S$ spanned by $\left\{\mathbf{e}_1,\ \cdots, \mathbf{e}_r\right\}$ to $\mathrm{im}(R)$ and therefore $P^{-1}$ maps $\mathrm{im}(R)$ to $S$. 
Now let $\mathcal{C}'$ denote a basis for $V$ and extend $\mathcal{C}'$ to $\mathcal{C}$. Form the matrix $Q$ with the basis as the columns, with the first $r$ columns consisting of the vectors of $\mathcal{C}'$. It follows that $Q$ takes $S$ to $V$.
Then the matrix $QP^{-1}R$ is your desired matrix. Since $QP^{-1}$ is invertible, the rowspace of $R$ is unchanged, therefore $\mathrm{row}(QP^{-1}R) = U$. Also, $P^{-1}$ maps $\mathrm{im}(R)$ to $S$ and $Q$ maps $S$ to $V$. Therefore $\mathrm{im}(QP^{-1}R) = V$, as required.
A: Let $m$ be the dimension of $U$ and $V$. Let $M_U$ be an $n\times m$ matrix whose columns are a basis of $U$ and $M_V$ be an $n \times m$ matrix whose columns are a basis of $V$.
Now $U = \{M_U \vec{x} : \vec{x} \in \mathbb{R}^m \}$ and $V = \{M_V \vec{x} : \vec{x} \in \mathbb{R}^m \}$.
We wish to find a matrix $A$ such that 
$$
\begin{align*}
U &= \{M_U \vec{x} : \vec{x} \in \mathbb{R}^m \} = \{A \vec{x} : \vec{x} \in \mathbb{R}^n \}, \\
V &= \{M_V \vec{x} : \vec{x} \in \mathbb{R}^m \} = \{A^T \vec{x} : \vec{x} \in \mathbb{R}^n \}.
\end{align*}
$$
We note that $\{ M_V^T \vec{x} : \vec{x} \in \mathbb{R}^n\} = \mathbb{R}^m$, because the rank of $M_V^T$ is $m$ and it has $m$ rows. Thus
$$
U = \{M_U \vec{x} : \vec{x} \in \mathbb{R}^m \} = \{M_U M_V^T \vec{x} : \vec{x} \in \mathbb{R}^n \}.
$$
But this immediately suggests that $A=M_U M_V^T$, because $\mathrm{span}(M_U M_V^T) = U$. By symmetry, also $\mathrm{span}(M_V M_U^T) = V$, and because $A^T=M_V M_U^T$, we are done.
