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?

 

[blue_raver22]

luk4,

The generalization of this relation to dimension 4 is given by the following formula:

det A = (1/12)((Tr A)^4 - 3(Tr A)^2Tr(A^2) + 2Tr(A^3) + 8 det(A^2))

This formula can be derived by considering the expansion of the determinant of a 4x4 matrix and using the properties of the trace and determinants of matrices. It is important to note that this formula only holds for square matrices of dimension 4. I hope this helps answer your question.