TITLE: Solving Lyapunov-like equation
QUESTION [10 upvotes]: The following matrix equation might be a Lyapunov-like equation, but it seems hard for me to develop a simpler way to solve it. From the computation effort, I need some help for solving the special case of the following Lyapunov equation:
Let $X$ be an $n\times n$ symmetric matrix, and $I$ an identity matrix, and $A$ is a matrix whose entries are all between 0 and 1, and $A$ is invertible. I need to solve $X$ in the following equation:
$$AX+XA^T=I$$
Previously, I found some article discussing on using Krylov subspace to solve the following Lyapunov equation:
$$AX+XA^T=b \cdot b^T$$
where $b$ is a vector. Due to $b \cdot b^T$ being a rank-one matrix, Krylov subspace appraoch is highly efficient. Now in my case it is the identity matrix $I$, but $X$ in my case is symmetric. I found that in my equation $AX$ and $XA^T$ are symmetric. So by letting $Y=AX$, my equation can be reduced to :
$$Y+Y^T=I  \quad \textrm{with } \ \ Y=AX$$.
I don't know how to continue this. 
Another common way is to use tensor product to rewrite my equation as:
$$(I \otimes A + A \otimes I) vec(X) = vec(I)$$
but the LHS of the above equation is $n^2 \times n^2$ size, which is too large to solve.
Is there any other efficient way to solve this? Any advices are warmly welcome!

REPLY [5 votes]: I will summarize everything, for future reference.

All Lyapunov equations $AX+XA^T=B$ have a unique, symmetric solution $X=X^T$, unless there is a $\lambda\in\mathbb{C}$ such that $\lambda$ and $-\lambda$ are both eigenvalues of $A$. This holds, for instance, when $A$ is  Hurwitz stable, which is a common case.
As far as I know, reducing to $B=I$ (or to $Y+Y^T=I$ with a nontrivial constraint on $Y$) does not help in its numerical solution.
There are integral representations and reformulations as a system of $n^2$ linear equations in $n^2$ unknowns using Kronecker products, but they are also worthless from the point of view of numerical solution.
Up to $n\approx 1000$, the standard algorithm (also implemented in MATLAB's lyap) is the $O(n^3)$ Bartels-Stewart algorithm. The Hessenberg-Schur algorithm is a variation that helps for the more general Sylvester equation, but not in this case. For a general Sylvester equation $AX+XC=B$, the HS method requires one Schur decomposition of $A$ instead of one of $A$ and one of $C$, but if $C=A^T$ then the second decomposition comes for free.
For values of $n$ larger than $1000$, and in general for the case of $A$ large and sparse, the standard algorithms are either rational Krylov subspace methods (MATLAB code and papers on V. Simoncini's page, http://www.dm.unibo.it/~simoncin/software.html) or the ADI method (Matlab and C code on the page of Peter Benner's group, http://www.mpi-magdeburg.mpg.de/csc). Both can handle problems where $B$ has low rank. 
The case in which $A$ is large and sparse and $B$ has full rank (for instance $B=I$) is more difficult; solving it efficiently is still an active research problem. This preprint seems to go in the right direction http://www.math.cts.nthu.edu.tw/download.php?filename=683_a93bf34d.pdf&dir=publish&title=prep2012-05-005, but the methods is much newer and I have never tried it. EDIT: I should rather recommend this approach by a collaborator of mine https://arxiv.org/abs/1711.05493 ; it should solve exactly this problem for the case of large matrices.