It is well known that if $A, B$ are positive semidefinite matrices, then $$\det (A+B)\ge \det A+\det B.$$ 


I am considering a possible extension of this result. Let $\mathbb{M}\_m(\mathbb{M}\_n)$ denote the set of $m\times m$ block matrices with each block the usual $n\times n$ matrix. 


Let $\mathbf{A}=[A\_{i,j}]\_{i,j=1}^m, \mathbf{B}=[B\_{i,j}]\_{i,j=1}^m \in \mathbb{M}\_m(\mathbb{M}\_n)$ be positive semidefinite. Define a new matrix $M=[m\_{ij}]\_{i,j=1}^m$ with $m\_{ij}=\det (A\_{i,j}+B\_{i,j})-(\det A\_{i,j}+\det B\_{i,j})$. Is it true that $M$ is positive semidefinite?


I ran some numerical simulations, yet no counterexamples showed up (of course, numerical simulations can never be thorough). I did not find a proof even when $m=2$.


Clearly, it suffices to show $\det M\ge 0$.



A possible extension of a determinant inequality
The claim is true. We prove it using a few block matrix manipulations. Note, in the proofs below $A \ge 0$ means $A$ is (symmetric) positive semidefinite.


$\newcommand{\trace}{\text{trace}}$



> 
> **Lemma** Let $X, Y \ge 0$. Then,
>  \begin{equation\*}
>  \otimes^k (X+Y) \ge \otimes^k X + \otimes^k Y.
> \end{equation\*}
>  *Proof.*
>  By induction on $k$. The case $k=1$ is trivial; $k=2$ shows us the crux. Indeed,
>  \begin{equation\*}
>  (X+Y)\otimes (X+Y) - X\otimes X - Y\otimes Y = X\otimes Y + Y \otimes X \ge 0,
>  \end{equation\*}
>  since $X, Y \ge 0$. The general case follows similarly. 
> 
> 
> 


-



> 
> **Corollary.** 
>  Let $X, Y \ge 0$. Then, by restricting to the suitable symmetry class of tensors we get
>  \begin{equation\*}
>  \wedge^k(X+Y) \ge \wedge^k X + \wedge^k Y
>  \end{equation\*}
> 
> 
> 


.



> 
> **Lemma.** Let $A=[A\_{ij}]$ be $mn\times mn$ with $n\times n$ blocks. Suppose $A$ is symmetric, positive semidefinite. Then, for $1\le k \le n$, the $m\binom{n}{k} \times m\binom{n}{k}$ matrix $C\_k := [\wedge^k A\_{ij}]$ is semidefinite.
> 
> 
> 


*Proof.* Some reflection shows that $C\_k$ is a principal submatrix of $\wedge^k A$, thus, $C\_k = P\_k^\*(\wedge^k A)P\_k \ge 0$ since $A\ge 0$ and wedge products preserve positivity.



> 
> **Theorem.** Let $A=[A\_{ij}] \ge 0$ and $B=[B\_{ij}] \ge 0$ be $mn\times mn$ block matrices composed of $n\times n$ blocks. Define $$M\_k = [\trace(\wedge^k(A\_{ij}+B\_{ij}))] - [\trace(\wedge^k A\_{ij})] - [\trace(\wedge^k B\_{ij})],$$ for any $1\le k \le n$. Then, $M\_k \ge 0$.
> 
> 
> 


*Proof*
 The Corollary above shows that $\wedge^k(A+B) \ge \wedge^k A + \wedge^k B$. Let $P\_k$ be as in the second Lemma; then $$H\_k = P\_k^\*(\wedge^k(A+B))P\_k- P\_k^\*(\wedge^k A)P\_k -P\_k^\*(\wedge^k B)P\_k \ge 0.$$ The matrix $M\_k$ is nothing but a (blockwise) partial trace of $H\_k$, so that $H\_k \ge 0 \implies M\_k \ge 0$.



> 
> **Corollary**
>  Let $A$ and $B$ be as above. Then,
>  \begin{equation\*}
>  M = [\det(A\_{ij}+B\_{ij})] - [\det A\_{ij}] - [\det B\_{ij}] \ge 0.
>  \end{equation\*}
> 
> 
> 


*Proof.*
 Observe that $\trace(\wedge^n X) = \det(X)$ for an $n\times n$ matrix $X$.


