TITLE: All the $k\times k$ minors determines the matrix?
QUESTION [5 upvotes]: Suppose two $n\times n$ matrices $A$ and $B$ with the same $k\times k$ minors, that is, for each $1\leq i_1<\cdots<i_k\leq n$, 
$$\det\left(\begin{matrix}
a_{i_1i_1}&\cdots&a_{i_1i_k}\\
\vdots& \ddots& \vdots\\
a_{i_ki_1}&\cdots&a_{i_ki_k}\end{matrix}\right)=\det\left(\begin{matrix}
b_{i_1i_1}&\cdots&b_{i_1i_k}\\
\vdots& \ddots& \vdots\\
b_{i_ki_1}&\cdots&b_{i_ki_k}\end{matrix}\right),$$
can we deduce that $A=B$?
If $k=1$, it is obvious right.
If $k=n\geq 2$, it is obvious wrong.
However for other $k=2,\cdots,n-1$, I could not be affirmative...
Another related question is that for any $$m=\left(n\atop k\right)$$ scalars $a_1,\cdots,a_m$, is there exists a $n\times n$ matrix $A$ such that the minors (in dictionary order) are $a_1,\cdots,a_m$.

REPLY [3 votes]: Let $A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$,
$B=0$. Then all the $2 \times 2$ minors are the same.