7
$\begingroup$

Determine all $n\times n$ matrices $A$ with the following property

For any square submatrix $B$ of $A$, $\det(B)\det(\bar B)=0$. Here, $\bar B$ is the complementary matrix of $B$ obtained by deleting from $A$ the rows and columns containing entries of $B$.

By Laplace expansion, this property automatically implies $\det(A)=0$. There are some easy-to-see sufficient conditions to guarantee this. For instance, $A$ admits a zero row(column); or $\operatorname{rank} A$ is less than half of $n$ (in this case, at least one of $B$ and $\bar B$ have size greater than $\operatorname{rank} A$).

Motivation

I have a collection of polynomials $\{p_i\}_{i=1}^m$ such that for each $p$ in the collection with monomial expansion $\sum a_I z_I$, the coefficients $a_I$ are expressed by entries of the aforementioned matrix $A$. My goal is to show that these polynomials are linearly dependent, while what I already have is an identity $\sum c_i p_i=0$ where each $c_i$ is a scalar of the form $\det(B) \det(\bar B)$. If there is a nonzero $c_i$, then I am done. Otherwise, all $\det(B) \det(\bar B)=0$ and I hope this is strong enough to force linear dependence as well.

Improve this question
$\endgroup$
11
  • 3
    $\begingroup$ Another sufficient condition is that $A$ has 3 identical/proportional lines (or columns), as then, for each submatrix $B$, either $B$ or $\bar B$ has 2 identical/proportional lines and thus is singular. In this case, the rank of $A$ can be as much as $n-2$. We may still consider this as a trivial case, so I guess the (most interesting part of the) question is whether such a matrix exists with rank $n-1$. $\endgroup$ Commented Jan 3 at 19:34
  • $\begingroup$ @Wolfgang Rank $n-1$ is possible with a zero row or column. I suppose you mean to ask if rank $n-1$ is possible without a zero row or column? $\endgroup$ Commented Jan 4 at 6:44
  • $\begingroup$ As a class of examples that contains both Lchencz's and Wolfgang's: If for some $k$ the matrix contains a set of $k$ rows or $k$ columns whose span has dimension strictly less than $k/2$, then the matrix has the desired property. $\endgroup$ Commented Jan 4 at 8:57
  • 1
    $\begingroup$ They can be both zero. I revise it into Det(B)Det(\bar B)=0. @Rodrigo de Azevedo $\endgroup$ Commented Jan 5 at 0:29
  • 3
    $\begingroup$ Here is a sufficient condition that includes Kevin P. Costello's condition and Wolfgang's second condition: If $A$ has an $a \times b$ submatrix of rank $r< \frac{ a+ b -n}{2}$ then $A$ has this property. Indeed, a $c \times d$ submatrix of an $m \times m$ invertible matrix has rank at least $c+d-m$. If $B$ and $\bar{B}$ are invertible than adding this formula applied to $B$ to this formula applied to $\bar{B}$ gives $2r \geq a+b-n$. $\endgroup$ Commented Jan 6 at 20:16

0

Reset to default

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.