# A possible extension of a determinant inequality

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 $\det M\ge 0$.

• This seems to be true; is there further motivation for studying this question? – Suvrit Jul 2 '14 at 23:16
• @Suvrit: I don't have practical motivation for this. The problem occurred to me several months ago in a rather sudden occasion. At that time, I was thinking of block matrix extension some classical results. So far,I have one application in mind about this result. One way of tackling this problem may be the expression of determinant in terms of wedge product, but I am not quite confident about that. – M. Lin Jul 2 '14 at 23:31
• it was using the idea of exterior products that I concluded "it seems to be true" --- but the details seem a little murky... – Suvrit Jul 3 '14 at 0:01

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.

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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*}

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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 $X$.

• Very nice. That is the proof I am expecting. Thanks a lot. – M. Lin Jul 3 '14 at 6:21
• Just as a bit of a side note, Exercise 3.6.13 of "Positive definite matrices" by Rajendra Bhatia proves a similar statement using similar techniques. – Nathaniel Johnston Jul 22 '14 at 15:02
• @NathanielJohnston: nice find! Indeed that exercise of Bhatia proves $[\det(A_{ij})] \ge 0$, which seems to be an easier statement than the one shown above (more generally, the complete positivity of the immanant has been long known since at least the 60s) – Suvrit Jul 22 '14 at 16:33