Let $X=[x_1,\dots,x_r]$ and $Y=[y_1,\dots,y_r]$ be two $n\times r$ matrices with rank $r$ (namely columns are linearly independent), where $x_i,y_j$ are all vectors in $\mathbb R^n$.
I wonder how to prove the following assertion I saw when I read a paper:
Suppose all the $r\times r$ minors of $X$ and $Y$ have the same determinants, then the columns of $X$ and $Y$ span the same subspace in $\mathbb R^n$.
If $X$ and $Y$ both have only one column, the this is obvious. But I found that the case even when $X$ and $Y$ have two columns is hard.
I believe this should be able to be found in some theoretical linear algebra book for math majors. Any solution or reference is appreciated!
Unfortunately I don't see how to associate those two notions.
Since $X$ has full rank, one of the $r \times r$ minors $A$ must be invertible (see determinental rank). By swapping rows of $X$ and $Y$ correspondingly as necessary, we may assume that $A$ is formed by taking the first $r$ rows of $X$. Let $B$ be the minor formed by the first $r$ rows of $Y$, and let $C = A^{-1} B$. I claim that $$XC = Y.$$ Because $C$ is invertible, $XC$ and $X$ have the same columnspace. Note also that $\det C = 1$, so the minors of $XC$ have the same determinants as the minors of $X$. Further, the first $r$ rows of $XC$ agree with the first $r$ rows of $Y$.
Now, we just need to establish the $(r + 1)$st row, if one exists, of $XC$ and $Y$ agree. We can then repeat this argument inductively to show that all the rows agree. In other words, we have reduced to the case where $X$ and $Y$ are $(r + 1) \times r$, and agree on all but their final row, with the expectation that the final rows must also agree.
We can now perform one final reduction. The first $r$ rows of $X$ and $Y$ are $B$. If we multiply both $X$ and $Y$ on the right by $B^{-1}$, then the determinants of $r \times r$ minors of $XB^{-1}$ and $YB^{-1}$ still agree. So, now $X$ and $Y$ are the identity matrix on the first $r$ rows, with another row adjoined.
From here, it's relatively simple. If $(x_1, \ldots, x_r)$ is the $(r + 1)$st row, then $-\operatorname{sgn}(i)x_i$ is the determinant of the minor excluding the $i$th row of $X$. Since this is the same for $Y$, the $(r + 1)$st rows must agree between $X$ and $Y$, so $X = Y$, as required.