On sum of determinants

Question

$(1)$ Are there any special non-trivial classes of $n\times n$ square matrices where $$\det(A)=\sum_{i=1}^m\det(A_i)$$ at some (not necessarily any) $m$ satisfying $2\leq m\leq n$ where $A=\sum_{i=1}^mA_i$ holds?

$(2)$ Supposing if $A_i$ are symmetric and positive definite is it true that $$\det(A)\geq\sum_{i=1}^m\det(A_i)$$ holds at any $n\geq1$ if $A=\sum_{i=1}^mA_i$ holds (if true or not are there any other classes of matrices for which this holds)?

$(3)$ Are there any classes of non-trivial matrices for which we can have $$\det(A)>\sum_{i=1}^m\det(A_i)$$ holding true if $A=\sum_{i=1}^mA_i$ holds?

Answer 1

1. Not sure what is qualified as "non-trivial", but any singular matrix that can be split into the sum of a number of singular matrices will do. If you require nonsingular $A_i$s, then for any odd $n\ge3$, you may take a nilpotent Jordan form $A$ and set $A_1=pA+I$ and $A_2=(1-p)A-I$. There may be more elegant and more general examples, but I think the above ones are enough to illustrate the point.
2. Yes. By left- and right-multiplying $A^{-1/2}$ on both sides of $A=\sum_jA_j$, we may assume that $A=I$. Now, Hadamard's inequality states that the determinant of a positive semidefinite matrix $P$ is always bounded above by the determinant of its diagonal part, i.e. $\det P\le \det(P\circ I)$. Therefore $\sum_j\det(A_j)\le\sum_j\det(A_j\circ I)$. Denote the diagonal of $A_j$ by $u_j=(u_{1j},\ldots,u_{nj})^T>0$. Then $\sum_{j=1}^m\det(A_j\circ I) = \sum_{j=1}^m\prod_{i=1}^nu_{ij}\le \prod_{i=1}^n\sum_{j=1}^mu_{ij}=1=\det(A)$.
3. Yes. E.g. $\det\pmatrix{3&1\\ 1&3}=8>3+1=\det\pmatrix{2&1\\ 1&2}+\det I$. In fact, since tie occurs in Hadamard's inequality if and only if $P$ is diagonal, the argument in (2) shows that strict inequality must hold when $m>1$ and $A_1,\ldots,A_m$ are positive definite.