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I saw the following formula for the determinant of an $n \times n$ square matrix $A$: $$\det(A) = \sum (\pm) a_{1j_1}a_{2j_2}\cdots a_{nj_n}.$$

The sum being taken over all permutations of the second subscripts. The algebraic sign preceding each term is determined by the arrangement of the second subscripts. The formula is taken from Matrices and Linear transformations, 2nd edition, by Charles G. Cullen.

My question is, in the above formula, what is the purpose of the pair of parentheses that enclosed the symbol $\pm$? The same formula is also found in other texts, for example Elementary Linear algebra with Applications, 9th edition, by Bernard Kolman and David R. Hill.

dfeuer
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shuxue
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  • Here is a more understandable way of writing this formula. – Julien Oct 22 '13 at 15:19
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    So, at least two books on linear algebra uses this $(\pm)$ notation without any explanation? That's sad. – user1551 Oct 22 '13 at 15:35
  • Please click edit to see how I have changed the $\LaTeX$. – dfeuer Oct 22 '13 at 15:38
  • There is no any explanation on the notation $(\pm)$ in both books. – shuxue Oct 22 '13 at 15:40
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2 Answers2

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The same formula without the parentheses can also be found in places. So it's not necessary. All it means is $sgn(\sigma)$.

Daniel Donnelly
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In expressions where the $\pm$ is without parentheses, there is usually an understanding that "the" equation is actually two equations, one for each choice of sign, which is the same for each term.

This is just a guess since I don't have those books in front of me, but I think the parentheses are to emphasize that the sign depends on the term.

Usually I see this written as $\text{sgn}(\sigma)$, where $\sigma$ is the corresponding permutation.

Reply to comment: To calculate $\text{sgn}(\sigma)$, i.e., whether it is plus or minus, you have to look at the corresponding term $a_{1,j_1}a_{2,j_2}\cdots a_{n,j_n}$.

First, count the number of inversions: pairs $i,k$ such that $i<k$ but $j_i>j_k$. Then $$\text{sgn}(\sigma) = \begin{cases} +1 & \text{ if there are an even number of inversions,}\\ -1 & \text{ if there are an odd number of inversions.}\end{cases}$$

For example, if $n=4$, and you are looking at the term $a_{1,2}a_{2,1}a_{3,4}a_{4,3}$, then 1,2 forms an inversion since $j_1=2>j_2=1$. On the other hand, $1,3$ do not form an inversion since $j_1=1<j_3=4$. In fact there is only one other inversion (which I'll leave you to find) making a total of $2$, and so the coefficient of this term will be $+1$. There are alternate ways of calculating this if you know something about permutations.

Casteels
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