Determinant as a function of trace

  • Thread starter lukluk
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  • #1
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Main Question or Discussion Point

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
 

Answers and Replies

  • #3
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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?
 
  • #4
I like Serena
Homework Helper
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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?
 

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