solving Lyapunov-like equation

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!

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How large is your $n$ in practice? (an order of magnitude will be enough) –  Federico Poloni Sep 5 '12 at 13:15
What exactly does "solve" mean for you in this contex? Do you want a numerical solution for a given $A$ or a closed-form formula that you can analyze? Or do you want to know when is the equation solvable? (That's an easy one, iff $A$ has "regular inertia", see the Ostrowski-Schneider theorem). Btw, it is a Lyapunov equation alright. –  Felix Goldberg Sep 6 '12 at 19:43

As Federico Poloni pointed out, the Hessenberg-Schur algorithm, used by MATLAB's lyap.m function is a much better choice. It is a refined version of the older Bartels-Stewart algorithm (which also works pretty well). Here's the original paper for Hessenberg-Schur, by Golub-Nash-van Loan:

https://www.cs.cornell.edu/cv/ResearchPDF/Hessenberg.Schur.Method.pdf

In your case, since $A=B^T$, things are even a bit simpler, in that only one matrix needs to be decomposed.

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If the $n\times n$ matrix $A$ is negatively stable (i.e. $\mbox{Re} \; \lambda_i <0$ for all $i=1,...,n$ where $\lambda_i$ are the eigenvalues of $A$), then for any $n\times n$ matrix $C$ there exists a unique $X$ such that

$$AX+XA^T = C.$$

See Theorem 6.4.2 of Ortega ("Matrix Theory: A Second Course" 1987).

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This follows from the integral representation in Suvrit's answer. –  Federico Poloni Jun 14 '14 at 9:39

Let me first summarize some basic facts.

It is known that the equation

\begin{equation*} AX + XA^T = B, \end{equation*}

has a unique solution if the matrix $A$ is positively stable (i.e., has spectrum in the right half plane). If $A$ is diagonal with entries $a_1,\ldots,a_n$, then the solution to the equation can be given in closed form

\begin{equation*} X = D \circ B, \end{equation*} where [EDIT:] $D$ is a matrix with entries $1/(\bar{a}_i+a_j)$.

In the more general case, for positively stable $A$, the solution to the above equation can be represented as

\begin{equation*} X = \int_0^\infty e^{-tA}B(e^{-tA})^Tdt \end{equation*}

But that does not seem to be computationally that nice.

If $n$ is largish, one can still solve the linear system written using tensor products by using an iterative algorithm for solving the linear system, as long as the iterative algorithm (e.g., conjugate gradient, or other related methods) depends on just "matrix-vector" products. Because you would need to only compute $(A \otimes I + I \otimes A)x$ several times, and that can be done using matrix multiply without actually forming the tensor products.

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In your closed-form solution, when $B=I$, can we compute $$\int_{0}^{\infty} e^{-tA} \cdot {({e^{-tA}})}^T dt$$ in an easier way ? –  Hellen Sep 5 '12 at 13:02
Let me underline that $X$ is symmetric whenever $B$ is (proof: clear from the integral formula). If I am interpreting it correctly, Hellen meant that $AX$ and $XA^T$ are one the transposed of the other with "$AX$ and $XA^T$ are symmetric". So what she thinks is a special case is in fact the general behaviour for symmetric $B$. –  Federico Poloni Sep 5 '12 at 21:07
Why are there two indices (i and j) in your expression for the entries of the diagonal matrix D? –  Vidit Nanda Jun 13 '14 at 19:06
@ViditNanda: thanks for catching that typo! $D$ is not diagonal (because I already assumed $A$ to be diagonal). –  Suvrit Jun 13 '14 at 20:04

Let $A$ be an invertible $n\times n$ matrix which is antisymmetric: $A^T=-A$, e.g. the symplectic matrix $$\begin{pmatrix} 0&1 \\\\ -1&0 \end{pmatrix}.$$ The equation $AX+XA^T=I$ cannot have a matrix solution $X$ since that would imply $$AX-XA=I,$$ which is impossible since $trace (AX-XA)=0$. Of course that does not contradict the previous answer, but shows that some further conditions should be imposed on $A$.

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