Let $A$ be a $2n\times 2n$ real matrix with the following structure \begin{equation} A = \left(\begin{matrix} 0 & -I \\ K & S \end{matrix}\right) \end{equation} with all sub-matrices of size $n\times n$: $I$ is the identity matrix, $K$ is symmetric positive definite and $S$ is diagonal but singular. I am interested in the (numerical) solution of the continuous time Lyapunov equation for the $2n\times 2n$ matrix R: \begin{equation} R A^\text{T} + A R = \left(\begin{matrix} 0 & 0 \\ 0 & \Gamma \end{matrix}\right) \end{equation} where $\Gamma$ is a diagonal and singular $n\times n$ real matrix.

However, I only need a few elements of $R$. More specifically, writing

\begin{equation} R = \left(\begin{matrix} X & C \\ C^\text{T} & V \end{matrix}\right) \end{equation} all I really want are the diagonal entries of $V$ (also $n\times n$).

Is there anyway I could reduce this problem to some other (probably Sylvester) equation for $V$ or, better yet, only it's diagonal? I don't really know how to approach this problem.

  • $\begingroup$ what is $C$?what kind of matrix $\endgroup$ – dato datuashvili Aug 9 '12 at 12:45
  • $\begingroup$ The solution of the Lyapunov equation, $R$, will be symmetric and positive definite (it is actually a covariance matrix). What I can prove is that $C$ is anti-symmetric ($C^\text{T} = -C$) $\endgroup$ – Gabriel Landi Aug 9 '12 at 13:00
  • $\begingroup$ put instead of $C^T$ ;$-C$ if you get identity ,then it should solve problem $\endgroup$ – dato datuashvili Aug 9 '12 at 13:14
  • $\begingroup$ @dato, Sorry I don't understand. $\endgroup$ – Gabriel Landi Aug 9 '12 at 13:22

Just doing the matrix multiplications gives four equations: $$\begin{align} C^T = -C \\ XK + CS &= V \\ KX - SC &= V \\ KC - CK + VS + SV &= \Gamma \end{align}$$ the last equation of course gives $VS + SV = \Gamma - KC + CK$, so if we know what $C$ is we get a Sylvester equation for $V$. If we substitute the LHS of the second and third equation in place for $V$ in the fourth equation we get: $$K(XS + C) + (SX - C)K=\Gamma$$ and assuming $X$ is symmetric, we get $(XS + C)^T = SX - C$, so setting $B = XS + C$, the equation turns into $$KB + B^TK = \Gamma$$ which we want to solve for $B$. Now, this is not really a standard Sylvester equations since the matrix we seek is transposed, but there are methods to solve these kinds of equations too.

Once $B$ has been found we get $$C = \frac{1}{2}(B-B^T).$$ Since $B$ is the sum of one antisymmetric ($C$) and one symmetric matrix ($XS$), the antisymmetric part of $B$ has to be $C$.

Thus, one can now calculate the RHS in $$VS + SV = \Gamma - KC + CK$$ and then solve for $V$ as a standard Sylvester equation.

So, using this method you just solve two $n \times n$ Sylvester-ish equations instead of one $2n \times 2n$ equation. There might be something smarter lurking in here but I have not found it yet.

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  • $\begingroup$ Wow, this is great. Thank you! I'll start testing this idea right away and post some results as soon as they come out. $\endgroup$ – Gabriel Landi Aug 14 '12 at 0:27
  • $\begingroup$ Actually, I have been looking at the web and haven't yet found a method for solving the eq. for B. Any ideas on where to look? $\endgroup$ – Gabriel Landi Aug 14 '12 at 15:03
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    $\begingroup$ Try this article: H. W. Braden, The Equations $A^TX \pm X^TA = B$, SIAM J. Matrix Anal. Appl. Vol. 20, No. 2, pp. 295-302. $\endgroup$ – Calle Aug 14 '12 at 15:38

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