Consider a square real matrix $\mathbf{A}$ and a real symmetric matrix of the same size $\mathbf{S}$
If I can diagonalise $\mathbf{A}$, such that $\mathbf{A}=\mathbf{B}\mathbf{D}\mathbf{B}^{-1}$, then I can easily analyse t products of $\mathbf{A}$, that is, $\mathbf{A}^{t}$ as I have $\mathbf{A}^{t}=\mathbf{B}\mathbf{D}^{t}\mathbf{B}^{-1}$.
Is there a "similar" way to simplify $\mathbf{E}(t)=\big(\mathbf{A}^{t}\big)\mathbf{S}\big({\mathbf{A}^{t}}^{T}\big)$, which is symmetric for all $t\geq 0$ ?
PS: from this, my aim is to understand the "behavior" of the sum $\sum_{t\geq 0}\mathbf{E}(t)$
We can vectorize $E(t)$ as
$$ \operatorname{vec}(E(t)) = (A^t \otimes A^t) \operatorname{vec}(S) = (A \otimes A)^t \operatorname{vec}(S) $$
where $\otimes$ is the Kronecker product. If $A$ has eigenvalues in the unit circle, then the sum converges. Therefore,
$$ \operatorname{vec} \left(\sum_{t\geq0} E(t) \right) = (I - A \otimes A)^{-1} \operatorname{vec}(S) $$
Of course you can make calculations much easier by using decompositions of $A$. In particular, you can use Schur decomposition to make $A$ upper triangular and easily solve
$$ (I - A \otimes A) \operatorname{vec} \left(\sum_{t\geq0} E(t) \right) = \operatorname{vec}(S) $$
for the elements of the sum one by one.
Another alternative would be using SVD of $A$.
Edit. Another approach would be solving the following equivalent Lyapunov equation:
$$ E - A E A^T = S $$
where $E := \sum_{t\geq0} E(t)$.