Editing Definite matrix (section)

=== Trace ===
The diagonal entries <math>m_{ii}</math> of a positive-semidefinite matrix are real and non-negative. As a consequence the [[trace (linear algebra)|trace]], <math>\operatorname{tr}(M) \ge 0.</math> Furthermore,<ref>{{harvtxt|Horn|Johnson|2013}}, p. 430</ref> since every principal sub-matrix (in particular, 2-by-2) is positive semidefinite,
<math display="block">\left|m_{ij}\right| \leq \sqrt{m_{ii}m_{jj}} \quad \forall i, j</math>
and thus, when <math>n \ge 1,</math>
<math display="block"> \max_{i,j} \left|m_{ij}\right| \leq \max_i m_{ii}</math>

An <math>n \times n</math> Hermitian matrix <math>M</math> is positive definite if it satisfies the following trace inequalities:<ref>{{cite journal | title=Bounds for Eigenvalues using Traces | last1=Wolkowicz | first1=Henry | last2 = Styan | first2 = George P.H. | journal=Linear Algebra and Its Applications | issue=29 | publisher=Elsevier | year=1980 | volume=29 | pages=471–506 | doi=10.1016/0024-3795(80)90258-X }}</ref>
<math display="block">\operatorname{tr}(M) > 0  \quad \mathrm{and} \quad \frac{(\operatorname{tr}(M))^2}{\operatorname{tr}(M^2)} > n-1 .</math>

Another important result is that for any <math>M</math> and <math>N</math> positive-semidefinite matrices, <math>\operatorname{tr}(MN) \ge 0 .</math> This follows by writing <math>\operatorname{tr}(MN) = \operatorname{tr}(M^\frac{1}{2}N M^\frac{1}{2}).</math> The matrix <math>M^\frac{1}{2}N M^\frac{1}{2}</math> is positive-semidefinite and thus has non-negative eigenvalues, whose sum, the trace, is therefore also non-negative.