Determinant + indicial notation proof

Click For Summary

Homework Help Overview

The discussion revolves around proving a relationship involving the determinant of a second order tensor, specifically in the context of indicial notation. The original poster seeks assistance in demonstrating that the determinant can be expressed using the Levi-Civita symbol.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the 3x3 case of the determinant and its definition, with one suggesting to express the determinant as a sum. There is also a focus on how to incorporate the Levi-Civita symbol into the expression.

Discussion Status

The discussion is active, with participants providing insights and suggestions on how to approach the proof. Some guidance has been offered regarding the notation and potential starting points for expressing the determinant in the desired form.

Contextual Notes

There is an assumption that participants are familiar with the notation and concepts involved in tensor analysis and determinants. The specific definition of the determinant being used is also under consideration.

Mugged
Messages
103
Reaction score
0
Hello, I am supposed to prove that the determinant of a second order tensor (a matrix) is equal to the following:

det[A] = \frac{1}{6} \epsilon_{ijk} \epsilon_{pqr} A_{pi} A_{qj} A_{rk}

anyone have any idea how i would go about this? any method is welcome
 
Physics news on Phys.org
One thing which I would do is look at the 3X3 case amd see if you can see where the formula comes from. You also need to tell us the definition of the determinant that you're using.
 
well by det[A] i mean the following:

det[A_{ij}] = the determinant of a 3 by 3 matrix with the first row being A_{11}, A_{12}, A_{13} and the second row would be A_{21}, A_{22}, A_{23} and lastly the third row is A_{31}, A_{32}, A_{33}
 
How would you write that as a sum?
 
det[A] =
A_{11}(A_{23}A_{32}-A_{22}A_{33}) + A_{12}(A_{21}A_{33}-A_{23}A_{31}) + A_{13}(A_{22}A_{31}-A_{21}A_{32})
 
I am assuming that you're aware of the notation, how can you write the above sum as a sum which includes \varepsilon_{ijk}. A good start would be to write out one of the sums, say for example:
<br /> \varepsilon_{ijk}A_{1i}A_{1j}A_{3k}<br />
and see how this compares to your sum.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
10K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
12K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
14K
  • · Replies 15 ·
Replies
15
Views
5K