# trace inequality with symmetric Kronecker product

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 $$A=B=I$$.