# determinant in terms of traces of powers

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 DeterminantInTermsOfTracesOfPowers 2013-03-22 15:57:08 2013-03-22 15:57:08 Mathprof (13753) Mathprof (13753) 11 Mathprof (13753) Theorem msc 15A15