determinant in terms of traces of powers

It is possible to express the determinant of a matrix in of traces of powers of a matrix.

The easiest way to derive these expressions is to specialize to the case of diagonal matrices. For instance, suppose we have a $2\times 2$ matrix $M=\operatorname{diag}(u,v)$ . Then

$\displaystyle\operatorname{det}M$	$\displaystyle=$	$\displaystyle uv$
$\displaystyle\operatorname{tr}M$	$\displaystyle=$	$\displaystyle u+v$
$\displaystyle\operatorname{tr}M^{2}$	$\displaystyle=$	$\displaystyle u^{2}+v^{2}$

From the algebraic identity $(u+v)^{2}=u^{2}+v^{2}+2uv$ , it can be concluded that $\operatorname{det}M=\frac{1}{2}(\operatorname{tr}M)^{2}-\frac{1}{2}% \operatorname{tr}(M^{2})$ .

Likewise, one can derive expressions for the determinants of larger matrices from the identities for elementary symmetric polynomials in of power sums. For instance, from the identity

xyz=\frac{1}{6}(x+y+z)^{3}-\frac{1}{2}(x^{2}+y^{2}+z^{2})(x+y+z)+\frac{1}{3}(x% ^{3}+y^{3}+z^{3}),

it can be concluded that

\operatorname{det}M=\frac{1}{6}(\operatorname{tr}M)^{3}-\frac{1}{2}(% \operatorname{tr}M^{2})(\operatorname{tr}M)+\frac{1}{3}\operatorname{tr}M^{3}

for a $3\times 3$ matrix $M$ .

Title	determinant in terms of traces of powers
Canonical name	DeterminantInTermsOfTracesOfPowers
Date of creation	2013-03-22 15:57:08
Last modified on	2013-03-22 15:57:08
Owner	Mathprof (13753)
Last modified by	Mathprof (13753)
Numerical id	11
Author	Mathprof (13753)
Entry type	Theorem
Classification	msc 15A15