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

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?

• Certainly not in terms of $\det A$ and $\det B$: let $A=\begin{pmatrix}a&0\\0&0\end{pmatrix}$ and $B=\begin{pmatrix}0&0\\0&b\end{pmatrix}$, for example. Jun 10, 2020 at 4:27
• Good point. Any thought on if in terms of $\det (A)$ and $B$ or in terms of $A$ and $\det (B)$? I edited my question. Jun 10, 2020 at 4:30
• @user1551 The example you provide is for A positive semi definite. What if either A or B are positive definite? Jun 11, 2020 at 13:39
• It doesn't matter. If $A=diag(a,1/a)$ and $B=I$, then $\det(A)$ and $B$ are constant, but $\det(A+B)$ is unbounded above. Jun 11, 2020 at 13:46
• No. Note that the $A$ in my previous comment has a constant determinant and $B$ is also constant. Jun 11, 2020 at 14:13

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.

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 $$B = I$$.

• Thanks for your answer. This doesn’t answer my question because the bound is a function of eigenvalues instead of A and det(B) (or B and det(A)). However, your answer is indeed insightful and interesting! Jun 13, 2020 at 0:28