Determinant as a function of trace

Click For Summary

Discussion Overview

The discussion revolves around the relationship between the determinant and the trace of square matrices, specifically exploring identities for dimensions 2, 3, and a proposed generalization for dimension 4. The focus includes theoretical aspects and mathematical reasoning related to these identities.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant states that for dimension 2, the determinant can be expressed in terms of the trace as det A = ((Tr A)² - Tr(A²)) / 2.
  • Another participant references a Wikipedia page that purportedly contains a similar identity for dimension 3.
  • A participant proposes a formula for the determinant in dimension 4 as det A = (p₁₄ - 6p₁₂p₂ + 3p₂² + 8p₁p₃ - 6p₄) / 24, where pᵢ = Tr(Aⁱ).
  • One participant confirms the proposed formula for dimension 4 and suggests checking it using knowledge of eigenvalues.
  • Additional information is provided about the relationship between eigenvalues, the determinant, and the trace, noting that the determinant is the product of eigenvalues and the trace is their sum.

Areas of Agreement / Disagreement

Participants generally agree on the identities for dimensions 2 and 3, and there is a proposed formula for dimension 4, but the discussion does not resolve whether this formula is universally accepted or verified.

Contextual Notes

The discussion does not clarify the assumptions or limitations regarding the generalization to higher dimensions, nor does it address potential dependencies on definitions or mathematical steps that may remain unresolved.

lukluk
Messages
8
Reaction score
0
for dimension 2, the following relation between determinant and trace of a square matrix A is true:

det A=((Tr A)2-Tr (A2))/2

for dimension 3 a similar identity can be found in http://en.wikipedia.org/wiki/Determinant

Does anyone know the generalization to dimension 4 ?

lukluk
 
Physics news on Phys.org
Thanks very much!
so to see if I understand, the n=4 determinant can be written as

det A=(p14-6p12p2+3p22+8p1p3-6p4)/24

where
pi=Tr (Ai)

...right?
 
Right!

Btw, you can check this yourself if you know a little bit about eigenvalues.

Did you know that each nxn matrix has n eigenvalues?
And that the determinant is the product of the eigenvalues?
And that the trace is the sum of the eigenvalues?
And the the trace of A^k is the sum of each eigenvalue^k?
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 15 ·
Replies
15
Views
5K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K