Expressing the determinant of a sum of two matrices? Can det(A+B) be expressed in terms of det(A),det(B),n where A,B are n×n matrices? I made the edit to allow n to be factored in.

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Expressing the determinant of a sum of two matrices?  Can det(A+B)  be expressed in terms of  det(A),det(B),n  where A,B are n×n matrices?  I made the edit to allow n to be factored in.

Pademagk71 · Accepted Answer

When n=2, and suppose A has inverse, you can easily show that  det(A+B)=detA+detB+detA⋅Tr(A−1B)  Let me give a general method to find the determinant of the sum of two matrices A,B with A invertible and symmetric (The following result might also apply to the non-symmetric case. I might verify that later...). I am a physicist, so I will use the index notation, Aij and Bij, with i,j=1,2,⋯,n. Let Aij donate the inverse of Aij such that AilAlj=δij=AjlAli. We can use Aij to lower the indices, and its inverse to raise. For example AilBlj=Bji. Here and in the following, the Einstein summation rule is assumed.  Let ϵi1…in be the totally antisymmetric tensor, with ϵ1⋯n=1. Define a new tensor ϵ−i1⋅in=ϵi1⋯in|detA|. We can use Aij to lower the indices of ϵ−i1⋯in, and define ϵi1⋯in=Ai1j1⋯Ainjnϵ−j1⋯jn. Then there is a useful property:  ϵ―i1⋯iklk+1⋯lnϵ−j1⋯jklk+1⋯ln=(−1)sl!(n−l)!δi1j1⋯δikjk,  where the square brackets [] imply the antisymmetrization of the indices enclosed by them. s is the number of negative elements of Aij after it has been diagonalized.   So now the determinant of A+B can be obtained in the following way  det(A+B)=1n!ϵi1⋯inϵj1⋯jn(A+B)i1j1⋯(A+B)injn

limacarp4 · Accepted Answer

When n≥2, the answer is no. To illustrate, consider A=In,B1=(1100)⊕0,B2=(1111)⊕0.  If det(A+B)=f(det(A),det(B),n) for some function f, you should get det(A+B1)=f(1,0,n)=det(A+B2). But in fact, det(A+B1)=2≠3=det(A+B2) over any field.

nick1337 · Accepted Answer

There is a proof in the article, but in general: det(A+B)=∑r∑α,β(−1)s(α)+s(β)det(A[α|β]) det(B(α|β)) where r runs over the integers from 0,…,n; then the inner sum runs over all strictly increasing sequences α and β of length r chosen from 1,…,n. A[α|β] is the r by r square submatrix of A lying in rows α and columns β. B(α|β) is the (n-r)-square submatrix of B lying in rows complementary to α and columns complementary to β. s(α) is the sum of the integers in α.

Expressing the determinant of a sum of two matrices? Can det(A+B) be

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