# 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 $M=\mathrm{diag}(u,v)$. Then

$\mathrm{det}M$ | $=$ | $uv$ | ||

$\mathrm{tr}M$ | $=$ | $u+v$ | ||

$\mathrm{tr}{M}^{2}$ | $=$ | ${u}^{2}+{v}^{2}$ |

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

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

$$\mathrm{det}M=\frac{1}{6}{(\mathrm{tr}M)}^{3}-\frac{1}{2}(\mathrm{tr}{M}^{2})(\mathrm{tr}M)+\frac{1}{3}\mathrm{tr}{M}^{3}$$ |

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 |