For positive semidefinite matrices $A,B,C \in \mathbb{R}^{n\times n}$, the following inequalities are well known:

$$(\det(A+B))^{1/n} \geq (\det A)^{1/n} + (\det B)^{1/n} $$


$$\det(A+B+C) + \det(C) \geq \det(A+C) + \det(B+C).$$

The first is, of course, Minkowski's determinant inequality. I'm not sure whether the second has a name, but it can be found, e.g., here and here.

I would like to know whether anyone knows of any reverse forms of these inequalities. To be more specific, I note that the Minkowski determinant inequality can be viewed as a consequence of the Entropy Power Inequality applied to Gaussian random vectors with covariance matrices $A$ and $B$. Applying similar reasoning to the entropy power inequality appearing in Theorem 4 of this paper, we may conclude that the following reverse counterparts to the above inequalities hold:

$$(\det(A+B))^{1/n} \leq (\det A)^{1/n}\left(\frac{\frac{1}{n}\operatorname{Tr}(B^{-1})}{(\det B^{-1})^{1/n}} \right) + (\det B)^{1/n}\left(\frac{\frac{1}{n}\operatorname{Tr}(A^{-1})}{(\det A^{-1})^{1/n}} \right),~~~~~(\star) $$


$$(\det(A+B+C) \det(C) )^{1/n} + (\det(A) \det(B) )^{1/n} \leq (\det(A+C) \det(B+C) )^{1/n} .~~~(\star\star)$$

The first may be obtained from the second by taking $C = \varepsilon I$ and letting $\varepsilon \downarrow 0$. I view $(\star)$ as a reverse counterpart to Minkowski's determinant inequality, since the ratio

$$\frac{\frac{1}{n}\operatorname{Tr}(B^{-1})}{(\det B^{-1})^{1/n}}\geq 1$$

by the AM-GM inequality applied to the eigenvalues of $B^{-1}$ (and similarly for the ratio involving $A^{-1}$), with near equality if the matrix $B$ has eigenvalues all roughly the same.

Has anyone seen these inequalities before, or does anyone have a direct proof of $(\star\star)$?


1 Answer 1


Inequality ($\star\star$) essentially follows from the original Minkowski plus an implication of Lidkskii's inequality (Fiedler's inequality, noted below). $\newcommand{\da}{\downarrow} \newcommand{\ua}{\uparrow}$

Assume $C$ is strictly positive definite (otherwise, $\det C=0$ rendering ($\star\star$) trivial), and let $C^{1/2}$ denote the positive square root of $C$. Renaming the matrices $A \leftarrow C^{-1/2}A C^{-1/2}$ and $B \leftarrow C^{-1/2}B C^{-1/2}$, the original inequality reduces to showing that \begin{equation*} \det(A+B+I)^{1/n} + [\det A\det B]^{1/n} \le [\det(I+A)\det(I+B)]^{1/n}. \end{equation*}

Let $a^\da$ denote the eigenvalues of $A$ in decreasing order; similarly, define $b^\ua$ for $B$. Then, we have \begin{equation*} \det(I+A)\det(I+B) = \prod_i(1+a_i^\da)\prod_i(1+b_i^\ua) = \prod_i(1+a_i^\da + b_i^\ua + a_i^\da b_i^\ua), \end{equation*} whence by the Minkowski inequality we get \begin{equation*} [\det(I+A)\det(I+B)]^{1/n} \ge \prod_i(1+a_i^\da+b_i^\ua)^{1/n} + \prod_i(a_i^\da b_i^\ua)^{1/n}. \end{equation*} Now from Fiedler's (1971) determinantal inequality we know that \begin{equation*} \prod_i(1+a_i^\da+b_i^\ua) \ge \det(I+A+B), \end{equation*} while clearly $\prod_i(a_i^\da b_i^\ua) = \det(A)\det(B)$. Thus, the inequality ($\star\star$) follows.

  • 1
    $\begingroup$ If $Q^{-1}CQ=I$ then $C=I$. What do you mean?? You have shown (**) for $C=I$, but I don't see how you can get the general case from that, given that the product of two positive definite matrices is not necessarily positive definite. $\endgroup$
    – Wolfgang
    Oct 9, 2016 at 19:11
  • $\begingroup$ I think it works by just considering positive square roots: $C = C^{1/2}C^{1/2}$. Then, the argument goes through as Suvrit suggested by considering $\tilde{A} = C^{-1/2}AC^{-1/2}$, $\tilde{B} = C^{-1/2}BC^{-1/2}$, and $\tilde{C}=I$. $\endgroup$
    – Tom
    Oct 9, 2016 at 19:59
  • $\begingroup$ @Suvrit, am curious as to why you say ($*$) is not a reverse Minkowski inequality (does such a thing have a precise definition?). That ($*$) follows from Minkowski (plus Fiedler's, plus linearizing ($**$)) should not preclude such a thing. E.g., 'reverse Holder' follows from Holder's inequality. In fact, ($*$) actually implies Minkowski's inequality, so could be regarded as an improvement. Indeed, Minkowski is equivalent to concavity of $A\mapsto (\det A)^{1/n}$, or $f(t)\leq f(0) + t f'(0)$ for $t$ small, and $f(t) := (\det(A+t I))^{1/n}$. But, this is precisely ($*$) with $B=tI$. $\endgroup$
    – Tom
    Oct 9, 2016 at 23:10
  • $\begingroup$ @Tom sorry, I meant to say $(**)$ is not a reverse Minkowski inequality (because to prove it I had to use Minkowski), but you never claimed that anyways. I'll edit out that part. $\endgroup$
    – Suvrit
    Oct 10, 2016 at 0:06
  • 1
    $\begingroup$ @Tom (1st comment) yes that completes Suvrit's argument, noting that $\tilde{A} $ and $\tilde{B} $ are psd. $\endgroup$
    – Wolfgang
    Oct 10, 2016 at 11:47

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