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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?

melatonin15
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2 Answers2

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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(2rand(n,n)-1);  % SVD of a random matrix
    AA1=diag(rand(n,1));         % A_1^2
    AA2=Udiag(rand(n,1))U';
    AA3=Vdiag(rand(n,1))V';
    x=2rand(n,1)-1;
    y=2rand(n,1)-1;
    z=2rand(n,1)-1;
    B1=sqrtm(AA1+txx');  % thus B_1^2>=A_1^2
    B2=sqrtm(AA2+tyy');
    B3=sqrtm(AA3+tzz');
    A1=sqrtm(AA1);
    A2=sqrtm(AA2);
    A3=sqrtm(AA3);
    r=norm(A1A2A3,'fro')/norm(B1B2B3,'fro')-1;
    if r>0,
        A1,A2,A3,B1,B2,B3,r,break;
    end;
end
user1551
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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.

moqui
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    I'm afraid $|B_1B_2\cdots B_n|_F=|B_1|_F|B_2|_F\cdots|B_n|_F$ is an impossible condition unless the matrices are $1\times1$. Since $$ |B_1B_2\cdots B_n|_F \le|B_1B_2|_F\big(|B_3|_F\cdots|B_n|_F\big) \le|B_1|_F|B_2|_F\big(|B_3|_F\cdots|B_n|_F\big), $$ if the LHS is equal to the RHS, we must have $|B_1B_2|_F=|B_1|_F|B_2|_F$. ... – user1551 Nov 16 '22 at 11:38
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    ...Therefore $$ \begin{aligned} |B_1|_F^2|B_2|_F^2 &=|B_1B_2|_F^2\ &=\operatorname{tr}\big((B_1B_2)(B_2B_1)\big) =\operatorname{tr}(B_1^2B_2^2)\ &=\langle B_1^2,B_2^2\rangle_F\ &\le|B_1^2|_F|B_2^2|_F\ &\le|B_1|_F^2|B_2|_F^2. \end{aligned} $$ Hence $|B_1^2|_F=|B_1|_F^2$, i.e., $\sqrt{\sum_i\lambda_i^4}=\sum_i\lambda_i^2$ where the $\lambda_i$s are the eigenvalues of $B_1$. This is impossible if $B_1$ is at least $2\times2$, because $\lambda_i>0$. – user1551 Nov 16 '22 at 11:38