Let $A, B$ be two positive semi-definite $n \times n$ matrices and let $L$ be an $n \times n$ matrix that satisfies $\rho(L) < 1$, where $\rho(\cdot)$ denotes spectral radius.

Let $A \otimes B$ denote the regular Kronecker product and let $A \otimes_s B$ denote the symmetric Kronecker product as defined in Definition 3.3 here.

Define: $$ \begin{align*} T_1 &= \frac{1}{2} \mathrm{tr}( (I - L \otimes L)^{-1} (A \otimes B + B \otimes A) (I - L \otimes L)^{-T} ) \:, \\ T_2 &= \mathrm{tr}( (I - L \otimes_s L)^{-1} (A \otimes_s B) (I - L \otimes_s L)^{-T} ) \:. \end{align*} $$

I can prove that $T_2 \leq T_1$ always holds. I'm curious if it is possible to prove a reverse inequality, that $T_1 \leq C \cdot T_2$ for some absolute constant $C > 0$ that does not depend on $A, B, L, n$. In doing some simulations, I found that $C \geq 2$ was never violated. But I don't know how to prove this.

Edit: For instance, if $L = \rho \cdot I$ for some $|\rho| < 1$ and $A = B = I$, then it is not hard to see that the constant $C = n^2 / (n(n+1)/2) \leq 2$ works. More generally, if $L = \rho \cdot I$ and $A,B$ are simultaneously diagonalizable, this bound $C \leq 2$ also works. Another case where $C \leq 2$ works is if $L$ is symmetric and $A=B=I$.


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