Is there an upper bound on the determinant of the sum of positive definite (or semidefinite) matrices?

Question

The Minkowski inequality, $$ (\det (A+B) )^{1/n} \geq (\det(A))^{1/n} + (\det(B))^{1/n},$$ implies $$ \det (A + B) \geq \det (A) + \det (B)$$ where $A$ and $B$ are $n \times n$ Hermitian matrices.

If the matrices $A$ and $B$ are positive definite (or positive semidefinite), is there an upper bound on the determinant of the sum of matrices $A$ and $B$ (possibly in terms of $A$ and $\det(B)$ or in terms of $\det(A)$ and $B$)?

While not possible in general (as pointed out in the comments), what if we have some constraint on the matrices or determinant (even if the constraint is on the entries of the matrices)? Is there any case/constraints we can add to compute an upper bound as a function of det(A) and B or det(B) and A?

Answer 1

As pointed out by @user1551, no upper bound exists as a function of $A$ and $\det B$, or $B$ and $\det A$. In other words, knowing $B$ and $\det A$ (or knowing $A$ and $\det B$) is not enough for $\det (A + B)$ to be bounded. We need more information about $A$ or/and $B$.

Let $A$ and $B$ be both $n\times n$ positive definite Hermitian matrices. We have the following upper bounds: $$\det (A + B) \le \det A \det \left(I + \frac{1}{\lambda_{\min}(A)}B\right) \tag{1}$$ and $$\det (A + B)\le \left(\frac{1}{\lambda_{\min}(A)} + \frac{1}{\lambda_{\min}(B)}\right)^n\det A \det B \tag{2}$$ where $\lambda_{\min}(A)$ is the smallest eigenvalue of $A$. Indeed, first, we have \begin{align} \det (A + B) &= \det A \det (I + A^{-1}B)\\ & = \det A \det (I + B^{1/2} A^{-1}B^{1/2}) \\ &\le \det A \det \left(I + B^{1/2} \frac{1}{\lambda_{\min}(A)}I B^{1/2}\right)\\ & = \det A \det \left(I + \frac{1}{\lambda_{\min}(A)}B\right) \end{align} where we have used $A^{-1} \le \frac{1}{\lambda_{\min}(A)}I$ (from $A\ge \lambda_{\min}(A) I$) and $\det X \ge \det Y$ if $X, Y, X-Y$ are all positive semidefinite. Second, we have $\det (A + B) = \det A \det (I + A^{-1}B) = \det A \det B \det (B^{-1} + A^{-1})$ and get (2) similarly.

Answer 2

The examples in the comments show a bound on terms of only $\det(A)$ and $\det(B)$ is not possible. However, a bound based on the eigenvalues of $A$ and $B$ is possible. This may or may not answer your question (it doesn't answer the bolded question), but it may be interesting (or already known to you).

Let $\lambda^\uparrow(M)$ and $\lambda^\downarrow(M)$ denote vectors of the eigenvalues of $M$ listed in ascending and descending order respectively. We have the well-known majorization relation:

$$ \lambda^\downarrow(A)+\lambda^\uparrow(B)\prec \lambda^\downarrow(A+B). $$

We know that the map $(x_1,\ldots,x_n)\mapsto x_1\cdots x_n$ is an elementary symmetric polynomial and thus Schur concave. Thus

$$ \det (A+B)=\prod_{j=1}^n \lambda_j^\downarrow(A+B)\le \prod_{j=1}^n (\lambda_j^\downarrow (A)+\lambda_j^\uparrow(B)). $$

The bound is tight for user1551's example from the comments $A = \operatorname{diag}(a,a^{-1})$ and $B = I$.