Determinant as a function of trace

[lukluk]

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

 

[I like Serena]

Welcome to PF, lukluk!

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?

 

[lukluk]

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
p_i=Tr (Aⁱ)

...right?

 

[I like Serena]

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?