Frobenius norm and symmetric positive definite matrices: Is it true that $\|A_1\cdots A_n\|_F \leq \|B_1\cdots B_n\|_F$ where $0 < A_i^2< B_i^2$?

Question

Consider symmetric positive definite matrices $A_1, \cdots, A_n$ and $B_1,\cdots,B_n$ such that $$A_i^2 < B_i^2$$ for $i \in [n]$, where $A <B$ means that $B-A$ is positive definite.

Is it true that $$\|A_1\cdots A_n\|_F \leq \|B_1\cdots B_n\|_F$$ where $\|A\|_F$ denotes the Frobenius norm of matrix $A$?

According to this answer, the inequality is true for $n=2$.

If it is not true for $n>2$, is there a sufficient condition under which it can be true?

Answer 1

This is not true. When the constraint $A_i^2<B_i^2$ is relaxed to $A_i^2\le B_i^2$, there is a random counterexample: $$ \begin{aligned} &A_1=\pmatrix{0.2144&0\\ 0&0.8922}, \ A_2=\pmatrix{0.3455&-0.2883\\ -0.2883&0.8029}, \ A_3=\pmatrix{0.7817&-0.2633\\ -0.2633&0.5608},\\ &B_1=\pmatrix{0.273970&0.013878\\ 0.013878&0.897168}, \ B_2=\pmatrix{0.4402&-0.2299\\ -0.2299&0.8401}, \ B_3=\pmatrix{0.7818&-0.2635\\ -0.2635&0.5620}\\ \end{aligned} $$ where $$ \frac{\|A_1A_2A_3\|_F}{\|B_1B_2B_3\|_F}-1=0.020842>0. $$ So, if we replace each $B_i$ by $(B_i^2+\epsilon I)^{1/2}$ for a sufficiently small $\epsilon>0$, we obtain a counterexample with $A_i^2<B_i^2$.

Counterexamples like this (subject to the relaxed constraint) can be obtained easily by rejection sampling. Here is my Octave/Matlab script:

n=2;    % order of the matrices
t=.05;  % some small positive constant

for k=1:10000,
    [U,S,V]=svd(2*rand(n,n)-1);  % SVD of a random matrix
    AA1=diag(rand(n,1));         % A_1^2
    AA2=U*diag(rand(n,1))*U';
    AA3=V*diag(rand(n,1))*V';
    x=2*rand(n,1)-1;
    y=2*rand(n,1)-1;
    z=2*rand(n,1)-1;
    B1=sqrtm(AA1+t*x*x');  % thus B_1^2>=A_1^2
    B2=sqrtm(AA2+t*y*y');
    B3=sqrtm(AA3+t*z*z');
    A1=sqrtm(AA1);
    A2=sqrtm(AA2);
    A3=sqrtm(AA3);
    r=norm(A1*A2*A3,'fro')/norm(B1*B2*B3,'fro')-1;
    if r>0,
        A1,A2,A3,B1,B2,B3,r,break;
    end;
end

Answer 2

If $$||B_1\cdots B_n||_F=||B_1||_F\cdots||B_n||_F$$ holds, then the inequality is easy to prove:

For a $m\times m$ matrix, the Frobenius norm can be expressed as $$||A||_F=\sqrt{\sum \sigma_k^2(A)},$$ where $\sigma_k(A)$, $1\leq k\leq m$, are the eigenvalues of $AA^T$ (search Frobenius norm in this page of Wikipedia).

We can use Weyl's inequality to deduce that $$\sigma_k(A_i)<\sigma_k(B_i),$$ where $\sigma_k$ are ordered in increasing size. Therefore, \begin{align*}||A_1\cdots A_n||_F^2&\leq ||A_1||_F^2\cdots||A_n||_F^2\\&= \left(\sum \sigma_k^2(A_1)\right)\cdots\left(\sum \sigma_k^2(A_n)\right)\\&<\left(\sum \sigma_k^2(B_1)\right)\cdots\left(\sum \sigma_k^2(B_n)\right)\\ &=||B_1||^2_F\cdots||B_n||^2_F\\ &=||B_1\cdots B_n||^2_F.\end{align*}

PD: I think this restriction I've assumed is quite difficult to achieve, so this result might not be really helpful.