# A determinant inequality

Notation: Suppose $\mathbf{A}$ and $\mathbf{B}$ are positive definite matrices in $\mathbb{R}^{n\times n}$ such that $\mathbf{A} \succeq \mathbf{B}$ (Loewner order). Let $\mathcal{S}(n,k)$ be the set of all $k$-subsets of $\{1,2,\dots,n\}$. For any $\mathcal{Q} \subset [n] \triangleq \{1,2,...,n\}$, $\mathbf{M}_\mathcal{Q}$ is the matrix obtained by deleting the rows and columns of matrix $\mathbf{M}$ whose indices are in $\mathcal{Q}$. Similarly, $\mathbf{x}_\mathcal{Q}$ is the vector obtained by deleting the rows of vector $\mathbf{x}$ whose indices are in $\mathcal{Q}$.

Conjecture: For any such $\mathbf{A}$, $\mathbf{B}$ and $\mathbf{x} \in \mathbb{R}^n$ and for all $k \in [n-1]$:

$$\frac{\sum_{\mathcal{Q} \in \mathcal{S}(n,k)} \det(\mathbf{A}_\mathcal{Q} + \mathbf{x}_\mathcal{Q}\mathbf{x}_\mathcal{Q}^\top)}{\sum_{\mathcal{Q} \in \mathcal{S}(n,k)} \det(\mathbf{A}_\mathcal{Q})} \leq \frac{\sum_{\mathcal{Q} \in \mathcal{S}(n,k)} \det(\mathbf{B}_\mathcal{Q} + \mathbf{x}_\mathcal{Q}\mathbf{x}_\mathcal{Q}^\top)}{\sum_{\mathcal{Q} \in \mathcal{S}(n,k)} \det(\mathbf{B}_\mathcal{Q})} \tag{1}$$

Progress so far:

1. We have $\det(\mathbf{A}_\mathcal{Q}+\mathbf{x}_\mathcal{Q}\mathbf{x}_\mathcal{Q}^\top) = \det(\mathbf{A}_\mathcal{Q})\,(1+\mathbf{x}_\mathcal{Q}^\top \mathbf{A}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q})$. Therefore, (1) can be rewritten as (2): $$\tag{2} \sum_{\mathcal{Q} \in \mathcal{S}(n,k)}\frac{ \det(\mathbf{A}_\mathcal{Q})\,}{\sum_{\mathcal{Q} \in \mathcal{S}(n,k)} \det(\mathbf{A}_\mathcal{Q})} \, \mathbf{x}_\mathcal{Q}^\top \mathbf{A}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q} \leq \sum_{\mathcal{Q} \in \mathcal{S}(n,k)}\frac{ \det(\mathbf{B}_\mathcal{Q})\,}{\sum_{\mathcal{Q} \in \mathcal{S}(n,k)} \det(\mathbf{B}_\mathcal{Q})} \, \mathbf{x}_\mathcal{Q}^\top \mathbf{B}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q}$$

2. Lemmas: It is easy to show that for any $\mathcal{Q} \subset [n]$:

• (2.1) $\quad \mathbf{A} \succeq \mathbf{B} \Rightarrow \mathbf{A}_\mathcal{Q} \succeq \mathbf{B}_\mathcal{Q}$
• (2.2) $\quad \det(\mathbf{A}_\mathcal{Q}) \geq \det(\mathbf{B}_\mathcal{Q})$
• (2.3) $\quad \mathbf{A}^{-1}_\mathcal{Q} \preceq \mathbf{B}_\mathcal{Q}^{-1} \Rightarrow \mathbf{x}_\mathcal{Q}^\top \mathbf{A}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q} \leq \mathbf{x}_\mathcal{Q}^\top \mathbf{B}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q}$
• (2.4) $\quad \det(\mathbf{A}_\mathcal{Q})\,\mathbf{x}_\mathcal{Q}^\top \mathbf{A}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q} \geq \det(\mathbf{B}_\mathcal{Q})\,\mathbf{x}_\mathcal{Q}^\top \mathbf{B}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q}$
3. Through recursion, it suffices to show that (1) or (2) hold for the special case of $\mathbf{A} = \mathbf{B} + \mathbf{p}\mathbf{p}^\top$ for some $\mathbf{p} \in \mathbb{R}^n$.
4. A very special case is posted here ($\mathbf{B} = \mathbf{I}$, $\mathbf{A} = \mathbf{I} + \mathbf{p}\mathbf{p}^\top$).
5. I haven't encountered any counterexample after running "many" simulations (I know this is not a strong argument).

Update: Let me explain the motivation as requested.

Motivation:

Suppose $\{\mathbf{a}_i\}_{i=1}^{m}$ are some vectors in $\mathbb{R}^n$ ($m \geq n$) and $\mathbf{A}_0 \succ \mathbf{0}$ is a given matrix in $\mathbb{R}^{n \times n}$. Now consider, \begin{align} f_k : 2^{[m]} &\to \mathbb{R}, \\ \mathcal{S} & \mapsto c_k(\mathbf{A}_0 + \sum_{i \in \mathcal{S}}\mathbf{a}_i\mathbf{a}_i^\top) \end{align} where $c_k(\mathbf{A})$ is the coefficient of $x^k$ in the characteristic polynomial of $\mathbf{A}$, i.e., $\det(x\mathbf{I} - \mathbf{A})$.

Conjecture: $f_k$ is monotone log-submodular (multiplicative submodular) for all $k \in \{0,1,\dots,n\}$.

I have proved the monotonicity (maybe for $|f_k|$).

Special Cases: This holds for $k=n-1$ (trace), $k=0$ (determinant is log-submodular) and $k=n$ (constant, $f_n(\mathcal{S}) = 1$).

Now (1) emerges from the proof of log-submodularity (multiplicative submodularity) of $f_k$ after expressing $c_k$ as the sum of determinants of principal minors.

Applications:

1. I came across this when working on graph Laplacian matrices. $f_k$ for Laplacian matrices (and $\mathbf{a}_i = \mathbf{e}_s - \mathbf{e}_r$ where $\{\mathbf{e}_s\}_{s=1}^n$ is the standard basis) is related to the weighted number of spanning trees. I recently showed that the weighted number of spanning trees is a monotone log-submodular function of the edge set (see a draft here). Other coefficients can be also be nicely related to the weighted number of spanning trees as shown by Alexander Kelmans (as a generalization of Kirchhoff's matrix tree theorem). For Laplacian matrices, (2) has a beautiful interpretation in terms of the expected value of the effective resistance ("distance") between two vertices after performing some random operations on the graph.

2. (1) and (2) also arise in $k$-DPPs (determinantal point process).

• quite an interesting looking inequality -- i'm curious to hear about the motivation behind this inequality! – Suvrit Dec 21 '16 at 4:54
• Thanks @Suvrit. Let me know if you need me to expand my answer. – Kasra Khosoussi Dec 22 '16 at 9:19
• @Kasra, look at my answer. It is now a completely different point of view, much more relevant. – Denis Serre Dec 22 '16 at 13:22
• @DenisSerre Thanks so much again, Denis. Give me some time to go through your answer carefully to make sure I understand everything. – Kasra Khosoussi Dec 22 '16 at 13:27

As suspected, the desired inequality actually holds for all hyperbolic polynomials; the inequality in the OP follows as corollary (Corollary 1) to Theorem 2 (which seems to be new).

We will need the following remarkable theorem to obtain our result.

Theorem 1 (Bauschke, Güler, Lewis, Sendov, 2001) Let $p$ be a homogenous hyperbolic polynomial; let $v$ be a vector in the strict interior of the hyperbolicity cone $\Lambda_{++}$ of $p$. Then, \begin{equation*} g(x) := \frac{p(x)}{Dp(x)[v]} \end{equation*} is concave on $\Lambda_{++}$.

This theorem helps prove the more general inequality (also conjectured by Denis Serre above).

Theorem 2. Let $p$ be a homogenous hyperbolic polynomial with hyperbolicity cone $\Lambda_{++}$. Let $a, b, c \in \Lambda_{++}$. Then, $p$ satisfies the (conic log-submodularity) inequality: \begin{equation*} \tag{0} p(a)p(a+b+c) \le p(a+b)p(a+c). \end{equation*}

Proof. Let $c \in \Lambda_{++}$ be arbitrary. Consider the function $f(a) := \frac{p(a+c)}{p(a)}$. Inequality (0) amounts to showing that $f(a)$ is monotonically decreasing on the cone $\Lambda_{++}$. Equivalently, we consider $\log f$ and show that its derivative is negative in the direction $v$. That is, for an arbitrary direction vector $v\in \Lambda_{++}$, we show that $$\tag{1} \frac{Dp(a+c)[v]}{p(a+c)} - \frac{Dp(a)[v]}{p(a)} \le 0\quad\Longleftrightarrow\quad \frac{p(a+c)}{Dp(a+c)[v]} \ge \frac{p(a)}{Dp(a)[v]}.$$ But from Theorem 1, we know that $\frac{p(x)}{Dp(x)[v]}$ is concave. Moreover, since $p$ is homogenous, from its concavity we obtain its superadditivity \begin{equation*} \frac{p(a+c)}{Dp(a+c)[v]} \ge \frac{p(a)}{Dp(a)[v]} + \frac{p(c)}{Dp(c)[v]}, \end{equation*} which is stronger than the desired monotonicity inequality (1) (since all terms are nonnegative).

Corollary 1. Let $E_k(A) = e_k \circ \lambda(A)$ denote the $k$-th elementary symmetric polynomial of a positive definite matrix $A$. Then for any positive definite $A, B, C$ we have \begin{equation*} E_k(A)E_k(A+B+C) \le E_k(A+B)E_k(A+C). \end{equation*} This log-submodularity, immediately implies the log-submodularity of $f_k(S) := E_k(A+\sum\nolimits_{i\in S}v_iv_i^T)$.

• @DenisSerre Thanks!! (also thanks for fixing the bad typo in the answer). – Suvrit Jan 3 '17 at 16:15

Edit. I think now that your question concerns Gårding's theory of hyperbolic polynomials.

A homogeneous polynomial of degree $d$ in $N$ real variables is hyperbolic in the direction $\bf e$ if for every vector $X$, the roots of the polynomial $t\mapsto p(X+t{\bf e})$ are real. We may suppose that $p({\bf e})>0$. The connected component of $\bf e$ in $\{p>0\}$ is the forward cone ; it is convex. Actually, $p$ is convex in the direction of any vector of the future cone. Let us denote $\Gamma$ the closure of the forward cone. Gårding proved a reverse Hölder inequality, in terms of the polar form associated with $p$ : $$p(x_1)^{1/d}\cdots p(x_d)^{1/d}\le\phi(x_1,\ldots,x_d)$$ for every $x_1,\ldots,x_d\in\Gamma$. He found also that $p^{1/d}$ is concave over $\Gamma$. Finally, the derivative of $p$ in a forward direction provides a hyperbolic polynomial, whose forward cone contains (strictly, in general) that of $p$.

How does this apply here ? The map $\sigma_d:S\mapsto \det S$ is a hyperbolic polynomial over the symmetric matrices, in the direction of $I_d$ ; we have $N=\frac{d(d+1)}2$. This is just saying that every symmetric matrix has real eigenvalues. Its forward cone is that of positive definite matrices. When differentiating in the direction of $I_d$, one obtains (up to a constant factor) $\sigma_{d-1}$, next $\sigma_{d-2}$, etc ... Their closed forward cones $\Gamma_d$, $\Gamma_{d-1}$ ... are larger and larger ; in particular, they all contain ${\bf SPD}_d$.

As mentionned below, your inequality amounts to $$(A+D=B+C,\, D\le B,C\le A)\Longrightarrow(\sigma_k(A)\sigma_k(D)\le\sigma_k(B)\sigma_k(C)).$$ I suspect that something stronger holds true, that is, if $p$ is hyperbolic, with closed forward cone $\Gamma$, then for every vectors $A,X,Y\in\Gamma$, the vectors $A,B=A+X,C=A+Y$ and $D=A+X+Y$ satisfy $p(A)p(D)\le p(B)p(C)$. I point out that if $X,Y$ are colinear (that is $A,B,C,D$ are colinear), then this is true because of the concavity of $p^{1/d}$.

I was able to prove the claim when $k=2$, in which case $p$ can be written in the Lorentz form $p(s,x)=s^2-|x|^2$, in some appropriate coordinates $X=(s,x)$. The proof is somewhat cumbersome, here are the main arguments. The quantity $$F(s,t,a,x,y)=p(B)p(C)-p(A)p(D),\qquad A=(1,a),\,X=(s,x)\,,Y=(t,y)$$ is a concave function of $x$ and $y$. Let us fix $s,t>0$. When minimizing over $|x|=s$ and $|y|=t$, the constraints must be equalities. There remains to minimize with respect to $a$ in the unit ball. If $a$ is on the unit sphere, $F$ is trivially $\ge0$. Otherwise, a minimum should be reached when $\nabla_aF=0$. An interesting calculation shows that this minimum is precisely zero. I can write out the details if you wish.

Let me begin with two observations. On the one hand, the quantity $$\sum_{Q\in S(n,k)}\det(A_Q)=:\sigma_k(A)$$ is nothing but the $k$-th symmetric polynomial in the eigenvalues of $A$, whence my notation.

On the other hand, if the required inequality is true, then a recursive use of it gives at well the inequality $$({\bf I}_k)\qquad \sigma_k(A)\sigma_k(D)\le\sigma_k(B)\sigma_k(C)$$ whenever $A,B,C,D$, symmetric positive definite, obey the constraints $$({\bf C})\qquad A+D=B+C,\qquad D\le B,C\le A.$$

I claim that this inequality is true at least for $k=1$ and $k=n$ (I guess that it remains true for every $k$). When $k=1$, this is because $\sigma_1$ is the trace, and the constraints imply $${\rm Tr}\,A+{\rm Tr}\,D={\rm Tr}\,B+{\rm Tr}\,C,\qquad0<{\rm Tr}\,D\le{\rm Tr}\,B,{\rm Tr}\,C\le{\rm Tr}\,A.$$ And we know that $a+d=b+c$ and $0<d\le b,c\le a$ imply $ad\le bc$.

For $k=n$, we must prove $\det A\det D\le \det B\det C$. To proceed, let us define $$X=\frac12(A+D)=\frac12(B+C),\qquad T=X-B,\qquad S=X-D.$$ The constraints are that $X>0$ and $\pm T\le S\le X$. We want to prove $$\det(X+S)\det(X-S)\le\det(X-T)\det(X+T).$$ Multiplying every matrix at left and right by $X^{-1/2}$, and using the multiplicativity of the determinant, we may restrict to the case where $X=I_n$. There remains to prove $$(|T|\le S\le I_n)\Longrightarrow(\det(I_n-S^2)\le\det(I_n-T^2)),$$ where $|T|$, the absolute value, is given by functional calculus. Remark that because the right-hand side involves only $T^2$, which equals $|T|^2$, we may also assume that $0_n\le T$. Therefore, there remains to check the monotonicity of $F:T\mapsto\det(I_n-T^2)$ over $0_n\le T\le I_n$. To this end, we differentiate $$DF(T)\cdot H={\rm Tr}(\widehat{I_n-T^2}(HT+TH)),$$ where $\hat M$ is the adjugate of $M$. Up to a density argument, we may assume that $T<I_n$ and therefore $I_n-T^2$ is invertible. Then $DF(T)\cdot H={\rm Tr}(HQ)$ where $$Q=\det(I_n-T^2)\,T^{1/2}(I_n-T^2)^{-1}T^{-1/2}.$$ Because $Q\ge0_n$, the monotonicity holds true and the proof is complete.

Edit. I find embarassing that the constraints (C) are invariant under congruence $M\mapsto P^TMP$, whereas the inequalities (I$_k$) to prove are not, except for $k=n$.

• Thanks Denis. Great observations. The original conjecture was to show that every coefficient of the characteristic polynomial is monotone log-submodular under the following construction: suppose $a_i$ ($i=1,\dots,m$) are real $n$-vectors ($m \geq n$). Then, $f_k : 2^{[m]} \to \mathbb{R}$ defined as the $k$th coefficient of the characteristic polynomial of $X + \sum_{i \in S} a_i a_i^\top$ is monotone log-submodular (where $X$ is any positive definite "offset"). As you mentioned, it is easy to show this holds for the trace. Determinant also follows from one of the inequalities in my post. – Kasra Khosoussi Dec 19 '16 at 14:41
• What do you mean by 'congruence'? – Fedor Petrov Dec 19 '16 at 16:43
• @Fedor. See my edited edits. – Denis Serre Dec 19 '16 at 16:50
• I'm afraid I don't understand all your notations. What is |T| for a matrix T? Just ±T? What is H? And is ${\rm Tr}(\widehat{I_n-T^2}(HT+TH))$ the same as ${\rm Tr}(({I_n-T^2})(HT+TH))$? – Wolfgang Dec 20 '16 at 8:52
• And $S\le X$ means $D\le 0_n$?? Shoudn't it be $T:=X-C$ and $S:=X-D$? – Wolfgang Dec 20 '16 at 8:53