Editing Determinant (section)

===Sum===

The determinant of the sum <math>A+B</math> of two square matrices of the same size is not in general expressible in terms of the determinants of ''A'' and of ''B''. 

However, for [[positive-definite matrix|positive semidefinite matrices]] <math>A</math>, <math>B</math> and <math>C</math> of equal size,
<math display=block>\det(A + B + C) + \det(C) \geq \det(A + C) + \det(B + C)\text{,}</math>
with the corollary<ref>{{cite arXiv| last1=Lin| first1=Minghua| last2=Sra|first2=Suvrit|title=Completely strong superadditivity of generalized matrix functions|eprint=1410.1958| class=math.FA| year=2014}}</ref><ref>{{cite journal|last1=Paksoy|last2=Turkmen|last3=Zhang|title=Inequalities of Generalized Matrix Functions via Tensor Products|journal=Electronic Journal of Linear Algebra|year=2014|volume=27|pages= 332–341| doi=10.13001/1081-3810.1622|url=https://nsuworks.nova.edu/cgi/viewcontent.cgi?article=1062&context=math_facarticles|doi-access=free}}</ref>
<math display=block>\det(A + B) \geq \det(A) + \det(B)\text{.}</math>  

[[Brunn–Minkowski theorem]] implies that the {{mvar|n}}th root of determinant is a [[concave function]], when restricted to [[Hermitian matrix|Hermitian]] positive-definite <math>n\times n</math> matrices.<ref>{{cite web|url=https://mathoverflow.net/questions/42594/concavity-of-det1-n-over-hpd-n|title=Concavity of det<sup><sup>1</sup>/<sub>''n''</sub></sup> over HPD<sub>''n''</sub>.|date=Oct 18, 2010|last1=Serre|first1=Denis|website=MathOverflow}}</ref> Therefore, if {{mvar|A}} and {{mvar|B}} are Hermitian positive-definite <math>n\times n</math> matrices, one has
<math display=block>\sqrt[n]{\det(A+B)}\geq\sqrt[n]{\det(A)}+\sqrt[n]{\det(B)},</math> since the {{mvar|n}}th root of the determinant is a [[homogeneous function]].

==== Sum identity for 2×2 matrices ====

For the special case of <math>2\times 2</math> matrices with complex entries, the determinant of the sum can be written in terms of determinants and traces in the following identity:
:<math>\det(A+B) = \det(A) + \det(B) + \text{tr}(A)\text{tr}(B) - \text{tr}(AB).</math>