Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 teached 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: Obviously, 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 the 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 Rule of Sarrus).

share|improve this question
    
Of particular interest: does this set contain the set of singular matrices? Gerhard "Ask Me About System Design" Paseman, 2012.08.27 –  Gerhard Paseman Aug 27 '12 at 21:41
    
This might merit the more general question: is there a quick algebraic test to determine if the rank of a matrix exceeds k, for some small number k? It would be really cool, but unlikely, for a sarrus type term to yield that information. Gerhard "Ask Me About System Design" Paseman, 2012.08.27 –  Gerhard Paseman Aug 27 '12 at 21:47
1  
@Gerhard, if a hypersurface contains another, both of the same degree, one can say a lot. I do not think this happens here. –  Mariano Suárez-Alvarez Aug 28 '12 at 0:07
    
@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. –  Will Sawin Aug 28 '12 at 3:07
    
There are some exciting results (to me) regarding determinants of minors of binary matrices by Brent and Osborn on arxiv this month. They use a result announced by a Hungarian (whose name is like F. Szollozi) on unitary matrix determinants. I now wonder if there are Sarrus terms that work well for some unitary matrices. Gerhard "Ask Me About Binary Matrices" Paseman, 2012.08.28 –  Gerhard Paseman Aug 28 '12 at 16:35
add comment

1 Answer 1

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

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

share|improve this answer
add comment

Your Answer

 
discard

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.