Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $n,\ k$ be two positive integers such that $n \ge 2$ and $1 \le k \le n-1$. If the matrix $A\in \mathcal{M}_n(\mathbb{C})$ has exactly $k$ null minors of order $n-1$, then $\det A \neq 0$.

source: Romanian Mathematical Olympiad, final round , 2012

share|cite|improve this question
What do you mean by minor? – Davide Giraudo Jun 24 '12 at 15:12
And what is a "null minor"? Minor I know, and even principal or chief minor, but null minor I never heard of. – DonAntonio Jun 24 '12 at 15:15
@DonAntonio I understand the assumption as: among all $(n-1)\times (n-1)$ submatrices, exactly $k$ have zero determinant. – user31373 Jun 24 '12 at 15:30
@Don Antonio: Leonid Kovalev is right. Im sorry for my enghlish. Ive searched in the dictionary and "null" means equal to 0. – Claudiu Mindrila Jun 24 '12 at 15:32

I assume the olympiad is over and it's legitimate for us to discuss this problem. Anyway, I hide the answer as to not spoil the fun.

Let's use the adjugate matrix. We know that exactly $k$ entries of $\mathrm{adj}\,A$ are zero. Suppose that $\det A=0$; then the product of $A$ and $\mathrm{adj}\,A$ must be the zero matrix. Now, the rank of $A$ is exactly $n-1$, which implies that the kernel of $\mathrm{adj}\,A$ is $(n-1)$-dimensional. In other words, $\mathrm{adj}\,A$ has rank $1$. But any rank 1 matrix is of the form $u\otimes v$, making it impossible for it to have exactly $k$ zero entries.

share|cite|improve this answer
Yes, the olympiad is over (in fact I`ve recieved this problem in the contest). – Claudiu Mindrila Jun 24 '12 at 19:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.