20
$\begingroup$

This question is most probably not research level, but I thought that the MO folks might like it... Feel free to close.

Here is the motivation: If you have ever taught a maths course for engineers which covered determinants and which included a written exam then you often ask the students to calculate a 4 by 4 determinant to check if they got the basic rules (e.g. using Laplace's formula to reduce to 3 by 3 if the structure is favorable or use Gauß elimination). If you did this you most probably have seen a student solving this problem by applying the "Sarrus rule for 4 by 4 matrices". Usually, the students memorize for 3 by 3 a pattern like

Sarrus' rule

(of course, the lines mean that you should multiply the numbers along the lines; green lines get a $+$, red lines get a $-$, finally add everything up). My colleagues told me, that in every exam there is at least one smart guy who happily generalizes this rule to 4 by 4 matrices with a scheme like this:

Sarrus' false rule

which I am going to refer to as False Sarrus Rule. Indeed, one could turn this into a working rule by assigning the right signs and repeating the procedure two times more in a different way. I wrote a small blog post here (and there is even paper on this (German description, Russian description)). Basically, I wrote this blog post to give the people who search the net for a generalized Rule of Sarrus some visual reminder that there is no easy "Sarrus Type Rule" available. Believe it or not: The post is found frequently via search terms like "sarrus rule", sarrus 4*4", "sarrus matrice 4 4" or the like. Discussing this with a colleague today, we asked ourselves the following question:

How does the set of 4 by 4 matrices for which this "False Sarrus Rule" gives the correct determinant looks like?

Basic thoughts: A matrix $A=(a_{ij})$ is in this set, if and only if the following equation is fulfilled $$\sum_{\text{eight special permutations}\ \pi_j} \pm a_{1\pi_j(1)}\cdots a_{4\pi_j(4)} = \det(A).$$ Four out of the eight summands on the left have the right sign, the other four have the wrong sign, and hence, one could simplify a bit. However, the bottom line is: There is just this one equation which has to be fulfilled for all the sixteen entries of a 4 by 4 matrix (and this equation is a homogeneous polynomial of degree four) and hence, the set of matrices for which the False Sarrus Rule gives the right result is a 15-dimensional variety, but I have no clue how it looks like. Probably some algebraic geometers could step in and provide some insight?

Final remark: I do not plan to include this discussion in a math course for engineers (although it may help to scare some people away from the thought that "there could be an easy $4\times 4$ Rule of Sarrus").

$\endgroup$
10
  • 4
    $\begingroup$ @Gerhard, if a hypersurface contains another, both of the same degree, one can say a lot. I do not think this happens here. $\endgroup$ Aug 28, 2012 at 0:07
  • 1
    $\begingroup$ @Gerhard Paseman: There exists a slow algebraic tests - the $k+1\times k+1$ minors. Given the existence of a slow algebraic test, the existence of a quick algebraic test seems unlikely. In particular, the space of matrices is dimension $n^2$, and rank $k$ matrices have dimension $2kn-k^2$, so you need at least $(n-k)^2$ polynomials to cut out the set of rank $k$ matrices. $\endgroup$
    – Will Sawin
    Aug 28, 2012 at 3:07
  • 1
    $\begingroup$ Is the set you mentioned closed under multiplication or inverse, provided the matrix is non singular? $\endgroup$ Apr 19, 2018 at 13:56
  • 2
    $\begingroup$ "the set of matrices for which the False Sarrus Rule gives the right result is a 15-dimensional variety, but I have no clue how it looks like." I'm not sure what "looks like" means, in the context of 15-dimensional varieties. I'm not even sure I know what a 15-dimensional hyperplane looks like, or what a 15-sphere looks like. $\endgroup$ Apr 19, 2018 at 21:38
  • 2
    $\begingroup$ @gerrymyerson Well, "looks like" may be bad wording - varieties are not my piece of cake, sorry. I hoped that algebraic geometry had some notions that help to understand the set. A hyperplane, for example, has a constant normal vector, constant tangent space, zero curvature and no singularities... $\endgroup$
    – Dirk
    Apr 20, 2018 at 4:26

1 Answer 1

10
$\begingroup$

Here is one result suggested by Gerhard Paseman's comments. The False Sarrus Rule is correct on all matrices of rank $1$ and $4\times 4$ and $5\times 5$ matrices of rank $2$. It does not hold in general on matrices of rank $3$ for $n\times n$ matrices with $n\gt 3$. It also fails for some matrices of rank $2$ and dimension $6$ or greater.

To see that it fails on matrices of rank $3$, consider block diagonal matrices with a nonsingular $2\times 2$ block $A = {a~~b \choose c~~d}$ and a $J_{n-2}$ block. This is singular for $n \gt 3$ but the False Sarrus Rule produces $\det(A) \ne 0.$

$$\begin{pmatrix} a & b & 0 & \cdots & 0 \\\ c & d & 0 & \cdots & 0 \\\ 0 & 0 & 1 & \cdots & 1 \\\ \vdots & \vdots & \vdots & \ddots & \vdots \\\ 0 & 0 & 1 &\cdots &1\end{pmatrix}$$

If $M$ has rank $\le 2$, let the columns be linear combinations of $\vec{v}$ and $\vec w$, $a_i \vec{v} + b_i\vec{w}.$ Then the monomials of the False Sarrus Rule are of the form $\prod_{i\in I} a_{\pi(i)} v_i \prod_{i \in I^c} b_{\pi(i)} w_i.$ If $|I| \le 2$ or $|I^c| \le 2$ then the coefficient of the monomial is $0$ by collecting terms ($n-n$ for ranks $0$ and $1$, and $1-1$ for rank $2$).

This fails for $|I|,|I^c| \ge 3$. For example, the False Sarrus Rule evaluates to $1$ on $P (J_3 \oplus J_{n-3}) P $, where $P$ is the permutation matrix for the $(3 ~4)$ transposition since only the main diagonal contributes. For $n=6$, this is

$$\begin{pmatrix}1 & 1 & 0 & 1 & 0 & 0 \\\ 1 & 1 & 0 & 1 & 0 & 0 \\\ 0 & 0 & 1 & 0 & 1 & 1 \\\ 1 & 1 & 0 & 1 & 0 & 0 \\\ 0 & 0 & 1 & 0 & 1 & 1 \\\ 0 & 0 & 1 & 0 & 1 & 1 \end{pmatrix}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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