Let $A \in M_n(\mathbb R)$ be fixed with spectral radius $\rho(A) < 1$. Then $T_1, T_2$ are two well-defined linear operators on $M_n(\mathbb R)$ given by \begin{align*} T_1(X) = \sum_{k=0}^{\infty} (A^T)^k X A^k, \\ T_2(X) = \sum_{k=0}^{\infty} A^k X (A^T)^k. \end{align*} Let us equip $M_n(\mathbb R)$ with trace product, i.e., $\langle A, B\rangle = \text{tr}(A^TB)$. I am wondering on the subspace of symmetric matrices $\mathbb S_n(\mathbb R)$, do $T_1, T_2$ have the same induced norm by Frobenius norm? For example, the induced norm of $T_1$ is \begin{align*} \| T_1\| = \sup_{\|X\|_F=1} \left(\sum_{k=0}^{\infty} (A^T)^k X A^k\right). \end{align*}
p.s. I have a bounty question here Operator norm (induced $2$-norm) of a Kronecker tensor (bounty (on elaborating an edit of an answer) is still open). If the answer there is positive, then the answer to this question should be true. Because that question will essentially affirm that norm of $T_i$ would be achieved on the subspace of symmetric matrices.
I get some thoughts: We note if $X \in \mathbb S_n(\mathbb R)$, $T_i(X)$ is symmetric. Let $S_i = T_i|_{\mathbb S_n(\mathbb R)}$. $S_i$ is well defined linear operator on $\mathbb S_n(\mathbb R)$. Note $S_1^T = S_2$ by looking at their Kronecker product, i.e., \begin{align*} \text{vec} (T_1(X) ) = (I \otimes I - A^T \otimes A^T)^{-1} \text{vec}(X), \\ \text{vec} (T_2(X) ) = (I \otimes I - A \otimes A)^{-1} \text{vec}(X). \end{align*} Then the answer of the question should follow directly. Is this correct?
