Given two sets of dimension $n$ vectors
$\lbrace v_1 , v_2 , \ldots , v_m \rbrace$, $\lbrace u_1, u_2, \ldots , u_m \rbrace$,
where $m > n$, is there a computational method (in particular, using a program such as Mathematica, Maple, etc) to find an $n \times n$ matrix $A$ that gives a bijection between the two sets, so $A v_i = u_j$ for some $i,j$. In particular, the $n \times n$ matrix must have determinant $\pm 1$. The span of each set is the full $n$-dimensional space.
I have two sets of vectors that I think are the same up to a change in basis that preserves volume and permutes the vectors, but I can't think of an algorithmic way to approach the question.
Thanks!
Basically this is equivalent to solve overdetermined equations:
Let $$\mathbf{u}_i=[u_{i1},\cdots, u_{in}]^T,$$ $$\mathbf{v}_i=[v_{i1},\cdots, v_{in}]^T, i=1,\cdots, m$$ and $$\mathbf{A}=[\mathbf{a}_1,\cdots,\mathbf{a}_n]^T$$ where $\mathbf{a}_j=[a_{j1},\cdots, a_{jn}]$ ($j=1,\cdots, n$), so we have $$\mathbf{u}_i= \mathbf{A}\mathbf{v}_i$$.
To determine $\mathbf{A}$, we can solve the following equation \begin{equation} \begin{bmatrix} \mathbf{u}_1 \\ \vdots \\ \mathbf{u}_m \end{bmatrix} =\begin{bmatrix} \mathbf{V}_1 \\ \vdots \\ \mathbf{V}_m \end{bmatrix} \begin{bmatrix} \mathbf{a}_1^T \\ \vdots \\ \mathbf{a}_n^T \end{bmatrix} \end{equation} where \begin{equation} \mathbf{V}_i = \begin{bmatrix} \mathbf{v}_i^T & \mathbf{0} & \cdots & \mathbf{0}\\ \mathbf{0} & \mathbf{v}_i^T & \cdots & \mathbf{0}\\ \vdots & \vdots & \ddots & \vdots \\ \mathbf{0} & \mathbf{0}& \cdots &\mathbf{v}_i^T& \end{bmatrix}_{n \times n^2} \end{equation}