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!

isa Lyapunov equation alright. $\endgroup$ – Felix Goldberg Sep 6 '12 at 19:43