- #1

- 8

- 0

det A=((Tr A)

^{2}-Tr (A

^{2}))/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

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- Thread starter lukluk
- Start date

- #1

- 8

- 0

det A=((Tr A)

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

- #2

I like Serena

Homework Helper

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The page you refer to in turn refers to another page that contains the answer you're looking for:

http://en.wikipedia.org/wiki/Newton..._symmetric_polynomials_in_terms_of_power_sums

Recognize the formulas in this section?

- #3

- 8

- 0

so to see if I understand, the n=4 determinant can be written as

det A=(p

where

p

...right?

- #4

I like Serena

Homework Helper

- 6,577

- 176

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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