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I have been reading chapter 1 of Shilov's Linear Algebra and get stuck at the part where he mentions determinant and permutation:

enter image description here

I just want to ask a couple questions regarding the part that is circled in red.I understand the inversion part, but I don't know how can one determine the formula for n order determinant by using the formula that is circled in red. How does he derive this formula to begin with?

Take the determinant of a 3x3 matrix:

$ \left( \begin{array}{cc} a_{11} & a_{21} & a_{31}\\ a_{12} & a_{22} & a_{32}\\ a_{13} & a_{23} & a_{33}\\ & \end{array} \right) %$

The third order determinant is $a_{11}a_{22}a_{13}+a_{21}a_{32}a_{13}+a_{31}a_{12}a_{23}-a_{31}a_{22}a_{13}-a_{11}a_{32}a_{23}-a_{21}a_{12}a_{33}$.

So the sign is negative for $a_{31}a_{22}a_{13}; a_{11}a_{32}a_{23}; a_{21}a_{12}a_{33}$ is because the inversion for each is only 1?

In another Russian book, the determinant is also defined as permutation as well, this image is taken from "Fundamentals of Linear Algebra and Analytical Geometry" by Bugrov and Nikolsky:

enter image description here

Russian books seem to define determinant early on, while more modern books seem to treat it much later.

James Warthington
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1 Answers1

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Are you confused about the meaning of the $\sum$ sign, as in, what the addends are, or are you confused about the signs?

  • The sum in Shilov's book is a sum over all sequences $\left(\alpha_1, \alpha_2, \ldots, \alpha_n\right)$ that are permutations (i.e., rearrangements) of the sequence $\left(1,2,\ldots,n\right)$. The sum in Bugrov and Nikolsky's book is a similar sum, except that they denote the sequences by $\left(j_1, j_2, \ldots, j_n\right)$ rather than by $\left(\alpha_1, \alpha_2, \ldots, \alpha_n\right)$.

  • The sign in the addend corresponding to a permutation $j$ is a $+$ if $j$ has an even number of inversions, and a $-$ if $j$ has an odd number of inversions.

See Appendix B in Neil Strickland's Linear Mathematics for Applications for a more modern exposition of this subject. Modern texts usually differ from older texts (like the one you are reading) in that

  • they define permutations as bijective maps from $\left\{1,2,\ldots,n\right\}$ to $\left\{1,2,\ldots,n\right\}$ (as opposed to defining them as sequences of numbers), and

  • they usually put stuff under $\sum$ signs that explains what the sum is ranging over (something that older texts don't do for reasons of typographical parsimony).

darij grinberg
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  • Yeah, in general, I am kind of confused about the sum sign and the rearrangement. How can the determinant be defined as rearrangement. Should I compute determinant of order 2 and 3 to see the pattern? Doesn't the product of the determinant mean the order of each element does not matter? – James Warthington Oct 27 '19 at 23:13
  • The determinant is not a rearrangement; it is a sum over all rearrangements. I suggest you read Strickland's Appendix B, which is more modern and more detailed than most Russian texts. Examples B.8 and B.9 will perhaps answer your question. – darij grinberg Oct 27 '19 at 23:13
  • So what do you think of my question about the 3rd order determinant? – James Warthington Oct 27 '19 at 23:17
  • Yes: The sign before $a_{11} a_{32} a_{23}$ is negative because the permutation $\left(1,3,2\right)$ has an odd number of inversions (namely, only $1$ inversion). – darij grinberg Oct 27 '19 at 23:18
  • So it is positive if the number of inversion is 0? – James Warthington Oct 27 '19 at 23:20
  • Yes, but it is also positive if the number of inversions is $2$ or $4$ or $6$ or.... (The number is $0$ only if the permutation is $\left(1,2,3\right)$.) – darij grinberg Oct 27 '19 at 23:21
  • Ok, I will read the text that you suggested, thank you. Hopefully I will come back with a clearer insight. – James Warthington Oct 27 '19 at 23:23
  • @ darij grinberg: I have read some sections of the note you sent me but I don't understand the diagram. Could you explain this a bit more? – James Warthington Oct 27 '19 at 23:56
  • Strickland defines a permutation of $N$ as a bijective map $f$ from $N$ to $N$. He draws such a map as a diagram of arrows, where an arrow from $i$ to $j$ stands for $f\left(i\right) = j$. – darij grinberg Oct 27 '19 at 23:59
  • Is this an efficient way to compute the n-order determinant? – James Warthington Oct 28 '19 at 00:03
  • See https://math.stackexchange.com/questions/2269267/determinant-of-large-matrices-theres-gotta-be-a-faster-way/2269279#2269279 . – darij grinberg Oct 28 '19 at 00:09