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?
Best Answer
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.