2
$\begingroup$

This is a continuation of my previous question here, but I previously did not fully specify some of the restrictions on the matrices and so the solutions posted there were not that helpful.

I'm looking to find a Hermitian solution $X$ to the pair of equations,

I'm trying to solve a pair of simultaneous matrix equations of the form,

$$AX + XA^\dagger + BJB^\dagger = 0,\ \ X C^\dagger + B J D^\dagger = 0,$$ where, $$J = \text{diag}\left(\begin{bmatrix}1&0\\0&-1\end{bmatrix},\dots,\begin{bmatrix}1&0\\0&-1\end{bmatrix}\right). $$

and $J \in \mathbb{R}^{m \times m}$.

Here $A \in \mathbb{C}^{n\times n}$, $B \in \mathbb{C}^{n\times m}$, $C \in \mathbb{C}^{m\times n}$, $D \in \mathbb{C}^{m\times m}$ where $n$ and $m$ are even

The former is the Lyapunov equation, but the solution $X$ may not be unique since $A$ may be unstable. Furthermore, the matrices $A, B, C$ may all be singular. So there are many solutions, but I need just one which is Hermitian. Does anyone know if there is a general way to find this (or the non-existence of this) for matrices with symbolic elements?

I tried something like this:

X=Array[x,{n,n}];
X=Solve[Simplify[c.a.X==-c.b.j.ConjugateTranspose[b]+d.j.ConjugateTranspose[b].ConjugateTranspose[a]]
&&Simplify[X==ConjugateTranspose[X]],Flatten[X]];

but it takes far too long.

Thanks!

EDIT: here is a concrete example as requested, here $\gamma > 0$

$$ A=\left( \begin{array}{cc} -\frac{\gamma }{2 \gamma ^2+1} & \frac{\gamma \sqrt{4 \gamma ^2+1}}{2 \gamma ^2+1} \\ \frac{\gamma \sqrt{4 \gamma ^2+1}}{2 \gamma ^2+1} & -\frac{4 \gamma ^3+\gamma }{2 \gamma ^2+1} \\ \end{array} \right), $$

$$ B=\left( \begin{array}{cc} 0 & \frac{1}{\sqrt{2 \gamma ^2+1}} \\ \sqrt{\frac{2 \gamma ^2+1}{4 \gamma ^2+1}} & -\frac{2 \gamma ^2}{\sqrt{8 \gamma ^4+6 \gamma ^2+1}} \\ \end{array} \right),$$

$$ C=\left( \begin{array}{cc} -\frac{4 \gamma ^2}{\sqrt{\frac{1}{\gamma ^2}+2}} & -\frac{2 \left(4 \gamma ^3+\gamma \right)}{\sqrt{8 \gamma ^4+6 \gamma ^2+1}} \\ -2 \gamma \sqrt{2 \gamma ^2+1} & 0 \\ \end{array} \right),$$

$$D=I_{2\times 2}.$$

Then a Hermitian solution is

$$ X=\left( \begin{array}{cc} -\frac{1}{4 \gamma ^3+2 \gamma } & \frac{\gamma }{\left(2 \gamma ^2+1\right) \sqrt{4 \gamma ^2+1}} \\ \frac{\gamma }{\left(2 \gamma ^2+1\right) \sqrt{4 \gamma ^2+1}} & \frac{1}{4 \gamma ^3+2 \gamma } \\ \end{array} \right). $$

The above equations in mathematica code

$Assumptions = {\[Gamma] > 0};
{a,b,c,d}={{{-(\[Gamma]/(1 + 2 \[Gamma]^2)), (\[Gamma] Sqrt[1 + 4 \[Gamma]^2])/(
   1 + 2 \[Gamma]^2)}, {(\[Gamma] Sqrt[1 + 4 \[Gamma]^2])/(
   1 + 2 \[Gamma]^2), -((\[Gamma] + 4 \[Gamma]^3)/(
    1 + 2 \[Gamma]^2))}}, {{0, 1/Sqrt[1 + 2 \[Gamma]^2]}, {Sqrt[(
   1 + 2 \[Gamma]^2)/(
   1 + 4 \[Gamma]^2)], -((2 \[Gamma]^2)/Sqrt[
    1 + 6 \[Gamma]^2 + 8 \[Gamma]^4])}}, {{-((4 \[Gamma]^2)/Sqrt[
    2 + 1/\[Gamma]^2]), -((2 (\[Gamma] + 4 \[Gamma]^3))/Sqrt[
    1 + 6 \[Gamma]^2 + 8 \[Gamma]^4])}, {-2 \[Gamma] Sqrt[
    1 + 2 \[Gamma]^2], 0}}, {{1, 0}, {0, 1}}};
X={{-(1/(2 \[Gamma]+4 \[Gamma]^3)),\[Gamma]/((1+2 \[Gamma]^2) Sqrt[1+4 \[Gamma]^2])},{\[Gamma]/((1+2 \[Gamma]^2) Sqrt[1+4 \[Gamma]^2]),1/(2 \[Gamma]+4 \[Gamma]^3)}}

EDIT 2: Here is a slightly more complex example which takes long to compute

$Assumptions = {\[Gamma] > 0, s0 > 0};
{a, b, c, d} =
{{{-((\[Gamma] (1 - s0 \[Gamma] + \[Gamma]^2))/(
    1 + s0 \[Gamma] + \[Gamma]^2)), Sqrt[
   s0 \[Gamma] (s0^2 \[Gamma]^2 - 
      2 s0 \[Gamma] (-1 + \[Gamma]^2) + (1 + \[Gamma]^2)^2)]/(
   1 + s0 \[Gamma] + \[Gamma]^2)}, {(
   Sqrt[(s0^5 \[Gamma]^3)/(1 + s0 \[Gamma] + \[Gamma]^2)] + 
    2 Sqrt[(s0^7 \[Gamma]^5)/(1 + s0 \[Gamma] + \[Gamma]^2)] + 
    2 Sqrt[(s0^5 \[Gamma]^7)/(1 + s0 \[Gamma] + \[Gamma]^2)] + Sqrt[(
    s0^9 \[Gamma]^7)/(1 + s0 \[Gamma] + \[Gamma]^2)] - 
    2 Sqrt[(s0^7 \[Gamma]^9)/(1 + s0 \[Gamma] + \[Gamma]^2)] + Sqrt[(
    s0^5 \[Gamma]^11)/(1 + s0 \[Gamma] + \[Gamma]^2)])/(
   s0^2 \[Gamma] Sqrt[(1 + 
       s0 \[Gamma] + \[Gamma]^2) (s0^2 \[Gamma]^2 - 
       2 s0 \[Gamma] (-1 + \[Gamma]^2) + (1 + \[Gamma]^2)^2)]), \
\[Gamma] (-1 - (2 s0 \[Gamma])/(1 + s0 \[Gamma] + \[Gamma]^2))}}, {{0,
    1/Sqrt[1 + s0 \[Gamma] + \[Gamma]^2]}, {Sqrt[(
   1 + s0 \[Gamma] + \[Gamma]^2)/(
   s0^2 \[Gamma]^2 - 
    2 s0 \[Gamma] (-1 + \[Gamma]^2) + (1 + \[Gamma]^2)^2)], -2 \
\[Gamma] Sqrt[(
    s0 \[Gamma])/((1 + s0 \[Gamma] + \[Gamma]^2) (s0^2 \[Gamma]^2 - 
       2 s0 \[Gamma] (-1 + \[Gamma]^2) + (1 + \[Gamma]^2)^2))]}}, \
{{-4 Sqrt[(s0 \[Gamma]^5)/(
    1 + s0 \[Gamma] + \[Gamma]^2)], -2 \[Gamma] Sqrt[(
    s0^2 \[Gamma]^2 - 
     2 s0 \[Gamma] (-1 + \[Gamma]^2) + (1 + \[Gamma]^2)^2)/(
    1 + s0 \[Gamma] + \[Gamma]^2)]}, {-2 \[Gamma] Sqrt[
    1 + s0 \[Gamma] + \[Gamma]^2], 0}}, {{1, 0}, {0, 1}}};

A Hermitian solution is

X = {{-(1/(2 \[Gamma] (1+s0 \[Gamma]+\[Gamma]^2))),Sqrt[(s0 \[Gamma])/(s0^2 \[Gamma]^2-2 s0 \[Gamma] (-1+\[Gamma]^2)+(1+\[Gamma]^2)^2)]/(1+s0 \[Gamma]+\[Gamma]^2)},{Sqrt[(s0 \[Gamma])/(s0^2 \[Gamma]^2-2 s0 \[Gamma] (-1+\[Gamma]^2)+(1+\[Gamma]^2)^2)]/(1+s0 \[Gamma]+\[Gamma]^2),1/(2 \[Gamma]+2 s0 \[Gamma]^2+2 \[Gamma]^3)}};
$\endgroup$
  • $\begingroup$ Now I see what you mean. This is a nice problem. But you need to formulate it better focusing on the essential part, namely, on analytically solving linear equations with complicated coefficients. I think the problem is that in the course of solution MA does not simplify intermediate expressions and, therefore, cannot decide if certain terms are zero. $\endgroup$ – yarchik Jul 8 at 19:36
  • $\begingroup$ @yarchik Yes that would seem so, I worry that maybe there isn't a neat and fast solution! Which would be a shame as I need this for my quantum network synthesis algorithm. I think I'll have to solve it on a case-by-case basis for now $\endgroup$ – Joe Bentley Jul 9 at 8:05
  • $\begingroup$ It seems so. MA can still be helpful in performing intermediate simplifications. $\endgroup$ – yarchik Jul 9 at 11:35
2
$\begingroup$

Combining equations we obtain a conventional linear equation in fully unique way without any constraints on the matrices:

$$\left\{ \begin{array}{l}XA^\dagger C^\dagger=ABJD^\dagger-BJB^\dagger C^\dagger,\\ X C^\dagger + B J D^\dagger = 0;\end{array}\right.$$ or transposing

$$\left\{ \begin{array}{l}CAX^\dagger=D J B^\dagger A^\dagger-C B JB^\dagger,\\ C X^\dagger =- D J B^\dagger; \end{array}\right.$$

Just solve it using LinearSolve when matrices are not singular:

xd=LinearSolve[C.A, D.J.ConjugateTranspose[B.A]-C.B.J.ConjugateTranspose[B]]
x=ConjugateTranspose[xd]

It works also for symbolic matrices.

Notice that without any additional information on $A,\ldots, D$ it is impossible to tell whether $X$ is Hermitian.

In order to obtain the solution for the modified question we combine two equations and use Solve as in OP:

$Assumptions = {g > 0};

j={{1,0},{0,-1}};
a={{-(g/(1 + 2 g^2)), (g Sqrt[1 + 4 g^2])/(1 + 2 g^2)}, {(g Sqrt[1 + 4 g^2])/(1 + 2 g^2), -((g + 4 g^3)/(1 + 2 g^2))}};
b={{0, 1/Sqrt[1 + 2 g^2]}, {Sqrt[(1 + 2 g^2)/( 1 + 4 g^2)], -((2 g^2)/Sqrt[1 + 6 g^2 + 8 g^4])}};
c={{-((4 g^2)/Sqrt[ 2 + 1/g^2]), -((2 (g + 4 g^3))/Sqrt[ 1 + 6 g^2 + 8 g^4])}, {-2 g Sqrt[1 + 2 g^2], 0}};
d={{1, 0}, {0, 1}};

u=Join[c,c.a//FullSimplify]
v=Join[-d.j.ConjugateTranspose[b]//FullSimplify,
        d.j.ConjugateTranspose[a.b] - c.b.j.ConjugateTranspose[b]//FullSimplify]
X=Array[x,{2,2}]

Solve[u.X==v]//FullSimplify
| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 1. It is not unique since CA is not invertible 2. This does not guarantee X to be Hermitian $\endgroup$ – Joe Bentley Jul 8 at 12:26
  • $\begingroup$ I added a concrete example. It is not clear to me that they are superfluous, especially since it is possible to find a solution to the former equation which is not a solution to the latter using "LyapunovSolve" $\endgroup$ – Joe Bentley Jul 8 at 12:37
  • $\begingroup$ Unfortunately this becomes really slow on slightly more complicated examples, see the second edit on my question. I left it running for about an hour and solve was still running $\endgroup$ – Joe Bentley Jul 8 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.