A determinantal inequality

Question

The following is an exercise from the 2023 international selection at École Normale Supérieure (ENS):

Prove that for any complex numbers $a_1, \dots, a_n$ and positive semi-definite complex matrices $A_1, \dots ,A_n$ , the following inequality is satisfied: $$ \det \left( \left| a_1 \right| A_{1\ } + \left| a_2 \right| A_2 + \dots + \left| a_n \right| A_n \right) \geq | \det(a_1 A_1 + \dots + a_n A_n)| $$

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So far, I have tried using the Hadamard inequality on determinants but it doesn't seem to lead anywhere. Same for expanding the determinant using the permutations formula.

Since it's an exercise from ENS, it is supposed to be somehow hard and I don't know how to "start". The exercises from ENS are known to be very hard to approach since it's literally the best mathematics department of France. Any hints, please?

Answer 1

Let $A_k$ be $m\times m$. By splitting each $A_k$ into a sum of rank-one positive semidefinite matrices, you may assume that each $A_k$ is rank-one. In this case, if $m>n$, then both sides are zero and we have nothing to prove. Suppose $m\le n$. Let $A_k=x_kx_k^\ast$, $D=\operatorname{diag}(a_1,a_2,\ldots,a_n)$ and $D=\operatorname{diag}(|a_1|,|a_2|,\ldots,|a_n|)$. By Cauchy-Binet formula, $$ \begin{aligned} &|\det(a_1A_1+\cdots+a_nA_n)|\\ &=|\det(XDX^\ast)|\\ &=\left|\,\sum_{S\in\binom{[n]}{m}}\det\left(X_{[m],S}\,(DX^\ast)_{S,[m]}\right)\,\right|\\ &=\left|\,\sum_{S\in\binom{[n]}{m}}\det\left(X_{[m],S}\,D_{S,S}\,(X^\ast)_{S,[m]}\right)\,\right|\quad\text{(because $D$ is diagonal)}\\ &\le\sum_{S\in\binom{[n]}{m}}\left|\,\det\left(X_{[m],S}\,D_{S,S}\,(X^\ast)_{S,[m]}\right)\,\right|\quad\text{(triangle inequality)}\\ &=\sum_{S\in\binom{[n]}{m}}\det\left(X_{[m],S}\,|D|_{S,S}\,(X^\ast)_{S,[m]}\right)\quad\text{(because $|\det D_{S,S}|=\det|D|_{S,S}$)}\\ &=\sum_{S\in\binom{[n]}{m}}\det\left(X_{[m],S}\,(|D|X^\ast)_{S,[m]}\right)\quad\text{(because $|D|$ is diagonal)}\\ &=\det\left(X\,|D|\,X^\ast\right)\\ &=\det\left(|a_1|A_1+\cdots+|a_n|A_n\right). \end{aligned} $$ So, the inequality in question is basically the triangle inequality in disguise.

Answer 2

I assume the matrices $A_i$ are in $\Bbb C^{d\times d}$. Let $M:=a_1A_1+\cdots+a_nA_n$ and $N:=|a_1|A_1+\cdots+|a_n|A_n$.

• Suppose $N$ is invertible (so positive definite). The goal is to show $\det N\ge|{\det M}|$, likewise $$\lvert{\det(N^{-1}M)}\rvert\le 1.$$ It then suffices to show that any eigenvalue $\lambda$ of $N^{-1}M$ has $|\lambda|\le1$. Let $x\neq0$ be an associated eigenvector. Then $N^{-1}Mx=\lambda x$, so $Mx=\lambda Nx$ and $$|\lambda|\,x^*Nx=|x^*Mx|=\left\lvert\sum_{i=1}^na_i\,x^*A_ix\right\rvert\le\sum_{i=1}^n|a_i|\,|x^*A_ix|=\sum_{i=1}^n|a_i|\,x^*A_ix=x^*Nx,$$ which shows that $|\lambda|\le1$.
• If $N$ is not invertible, there still exists a sequence $\varepsilon_k\searrow0$ such that $N+\varepsilon_k\,I_d$ is positive definite. Then $$\det(|a_1|A_1+\cdots+|a_n|A_n+\varepsilon_k\, I_d)\ge\left\lvert{\det(a_1A_1+\cdots+a_nA_n+\varepsilon_k\,I_d)}\right\rvert$$ by the previous point. Letting $k\to\infty$ yields $$\det(|a_1|A_1+\cdots+|a_n|A_n)\ge\left\lvert{\det(a_1A_1+\cdots+a_nA_n)}\right\rvert$$ by continuity of the determinant.