# A determinantal inequality

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: $$\det \left( \left| a_1 \right| A_{1\ } + \left| a_2 \right| A_2 + \dots + \left| a_n \right| A_n \right) \geq | \det(a_1 A_1 + \dots + a_n A_n)|$$

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?

• Welcome to Mathematics SE. Take a tour. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. May 21, 2023 at 10:51

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.
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.
• okkk i got it, basically you said that the matrice $a_1 A_1 + ... + a_n A_n$ is also of dimension $m \times m$ with complex coefficients, so it is in fact diagonalizable in C right ? May 21, 2023 at 17:44
• @LéoFabrègues No, this has nothing to do with diagonalisation. E.g. if $A_1$ is of rank $3$ and $A_2$ is of rank $2$, we may write $A_1=x_1x_1^\ast+x_2x_2^\ast+x_3x_3^\ast$ and $A_2=x_4x_4^\ast+x_5x_5^\ast$ for some appropriate $x_j$s. Therefore $$a_1A_1+a_2A_2=\pmatrix{x_1&x_2&x_3&x_4&x_5}\pmatrix{a_1\\ &a_1\\ &&a_1\\ &&&a_2\\ &&&&a_2}\pmatrix{x_1&x_2&x_3&x_4&x_5}^\ast.$$ May 21, 2023 at 18:39
• If you don't like this perspective using rank-one matrices, you may also write $\sum_{k=1}^na_kA_k$ as $$\pmatrix{\sqrt{A_1}&\cdots&\sqrt{A_n}}\pmatrix{a_1I\\ &\ddots\\ &&a_nI}\pmatrix{\sqrt{A_1}\\ \vdots\\ \sqrt{A_n}}.$$ The current proof can still be applied with slight modifications (mostly about the sizes of the matrices). May 21, 2023 at 18:43