0
$\begingroup$

I'm looking for a reference of the following statement (which can easily be proved by Laplace's formula and induction):

Let $R$ be a commutative ring with identity and let $A$ be an invertible matrix over $R$. Then there is a permutation matrix $P$ such that the diagonal of $PA$ has no zero.

Edit: Of course, $R$ has to be a domain to make the arguments (formula of Leibniz or Laplace) work. Futhermore, it's sufficient to require $det(A) \neq 0$ (instead of $A$ being invertible).

Improve this question
$\endgroup$
11
  • $\begingroup$ I'm hoping you won't find a reference of that statement, since it is false. (Consider the 1x1 matrix over the zero ring.) $\endgroup$
    – user5810
    Commented May 5, 2011 at 21:13
  • $\begingroup$ Doesn't a ring with identity have at least two elements? $\endgroup$
    – Tony Huynh
    Commented May 5, 2011 at 21:19
  • $\begingroup$ No, Tony. Otherwise algebra would be swamped in statements like "if $I$ is an ideal of $R$, then $R/I$ is a ring or zero". $\endgroup$
    – darij grinberg
    Commented May 5, 2011 at 21:21
  • $\begingroup$ @Ricky: I assume that the identity isn't zero and still hope for a reference. $\endgroup$
    – Ralph
    Commented May 5, 2011 at 21:56
  • 3
    $\begingroup$ I don't know of a reference, but you can also appeal to the determinant formula as a sum of (permuted) products for a shorter proof. Then you may not need a reference. Gerhard "Ask Me About System Design" Paseman, 2011.05.05 $\endgroup$
    – Gerhard Paseman
    Commented May 5, 2011 at 22:08

1 Answer 1

Reset to default
3
$\begingroup$

There is a strictly stronger fact about matrices:

Given $A\in{\bf M}_n(R)$, the following statements are equivalent.

  • There exists a permutation matrix $P$ such that the diagonal of $PA$ has no zero.
  • If an $m\times p$-block of $A$ is mades of zeroes, then $m+p\le n$.
Improve this answer
$\endgroup$
2
  • $\begingroup$ And it's called Hall's marriage theorem. $\endgroup$
    – darij grinberg
    Commented May 6, 2011 at 19:38
  • 1
    $\begingroup$ @darij, I'd rather say it can be proved using Hall's Marriage Theorem, or any of a number of other theorems equivalent to Hall's, including one, by Konig, that goes back at least 20 years before Hall, and was stated specifically in the context of matrices. $\endgroup$
    – Gerry Myerson
    Commented May 6, 2011 at 22:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .