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Given the discrete Lyapunov equation $$AXA^T - X + Q = 0$$ how can I solve for $X$ as a function of the eigenvectors of some matrix $H$?

More precisely, in the case of the continuous Lyapunov equation $$AX + XA^T + Q = 0$$ This equation can be re-written as \begin{align} AX + XA^T + Q &= 0 \\ \begin{bmatrix} X & -I\end{bmatrix} \begin{bmatrix} A^T \\ -Q - AX\end{bmatrix} &= 0 \\ \begin{bmatrix} X & -I\end{bmatrix} \begin{bmatrix} A^T & 0 \\ -Q & -A\end{bmatrix} \begin{bmatrix} I \\ X\end{bmatrix} &= 0 \\ \begin{bmatrix} X & -I\end{bmatrix} H \begin{bmatrix} I \\ X\end{bmatrix} &= 0 \tag{1} \label{eq:cont_lyap} \end{align} where $$H = \begin{bmatrix} A^T & 0 \\ -Q & -A\end{bmatrix}$$ is a $2n \times 2n$ Hamiltonian matrix. From \eqref{eq:cont_lyap}, we see that $H \begin{bmatrix} I \\ X\end{bmatrix}$ is in the null-space of $\begin{bmatrix} X & -I\end{bmatrix}$, which can be used to show that there exists a matrix $W$, such that its columns form a basis for the image of $\begin{bmatrix} I \\ X\end{bmatrix}$, so that $$H \begin{bmatrix} I \\ X\end{bmatrix} = \begin{bmatrix} I \\ X\end{bmatrix}W$$ By diagonalizing $W$, so that $W = X_1 \Lambda X_1^{-1}$, we get the equation $$H \begin{bmatrix} X_1 \\ X_2\end{bmatrix} = \begin{bmatrix} X_1 \\ X_2\end{bmatrix}\Lambda$$ where $X = X_2X_1^{-1}$. This means that we can compute $X$ by computing the $2n$ eigenvectors of $H$, picking out $n$ of them and forming the matrix $\begin{bmatrix} X_1 \\ X_2\end{bmatrix}$, and then computing $X = X_2X_1^{-1}$.

What I am looking for is a similar derivation to the one above, but for the discrete (not continuous) Lyapunov equation.

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2 Answers 2

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You can do it with symplectic matrix pencils instead of Hamiltonian matrices, even in the more general case of discrete-time algebraic Riccati equations. $$ \begin{bmatrix} A & 0\\ -Q & I \end{bmatrix} \begin{bmatrix} I\\X \end{bmatrix} = \begin{bmatrix} I & G\\ 0 & A^T \end{bmatrix} \begin{bmatrix} I\\X \end{bmatrix}W $$ is equivalent to $X - Q = A^TX(I+GX)^{-1}A$, after eliminating $W$ from the two resulting block equations. Set $G=0$ to recover the case of a discrete-time Lyapunov equation.

This is all "folklore" in numerical linear algebra / control theory circles, but if you wish to have a reference you can take my review paper https://dx.doi.org/10.1002/gamm.202000018 (it's Equation 25 there).

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    $\begingroup$ Thank you very much for this. Your reference in particular is brilliant. For others who don't have paid access to Wiley, the arXiv version of the paper is here. $\endgroup$
    – mhdadk
    Commented Nov 15, 2022 at 14:48
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    $\begingroup$ Just one question: based on what you wrote in section 4.1 of the paper, it seems to me that the generalized eigenvalue problem in eq. 25 can only be derived as long as $A$ is invertible. You did mention that the results are still valid if $A$ is singular, but I'm wondering if it would still be possible to derive this eigenvalue problem if $A$ was singular. That is, we don't need to left-multiply both sides of eq. 25 by $$\begin{bmatrix} I & G\\0 & A^T\end{bmatrix}^{-1}$$ $\endgroup$
    – mhdadk
    Commented Nov 15, 2022 at 14:50
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    $\begingroup$ @mhdadk Yes, for a suitable generalization of "eigenvalue problem". One can define and solve generalized eigenvalue problems of the form $\det(M-\lambda N) = 0$, without the need to invert $N$. In Matlab, it's eig(M, N), for instance; in Python, scipy.linalg.eig(M, N). $\endgroup$ Commented Nov 15, 2022 at 18:28
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    $\begingroup$ @mh Alternatively, you can use Cayley transforms, i.e., a change of variable of the form $\lambda = \frac{\mu + \gamma}{\mu - \gamma}$, for a suitable $\gamma > 0$; this should transform your problem into a standard eigenvalue problem (assuming a certain matrix depending on $\gamma$ is invertible) but that's trickier to handle as it changes the location of the eigenvalues wrt the unit circle, and it is less numerically stable. $\endgroup$ Commented Nov 15, 2022 at 18:32
  • $\begingroup$ Hey Federico, if you have the time, could you possibly have a look at my question Is it possible to express the solution of a matrix Riccati differential equation as an eigenvalue problem?? I think it is related to your answer here, although I could be wrong. $\endgroup$
    – mhdadk
    Commented Dec 18, 2022 at 7:52
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For completeness, I'm including a full derivation for the solution of the discrete Lyapunov equation that is adapted from @FedericoPoloni's paper (specifically section 4.1). For the derivation for the more general discrete algebraic Riccati equation, see the paper.

Given the discrete Lyapunov equation $$AXA^T - X + Q = 0$$ Suppose that $A$ is invertible, such that $A^{-1}$ exists. As @FedericoPoloni mentioned in section 4.1 in their paper, this is necessary for the following derivation, although the results in section 4.1 are valid even if $A$ is singular.

Note that the discrete Lyapunov equation can be re-written as $$ \begin{align} AXA^T - X + Q &= 0 \\ A^{-1}AXA^T - A^{-1}X + A^{-1}Q &= 0 \\ XA^T - A^{-1}X + A^{-1}Q &= 0 \\ \begin{bmatrix} X & -I\end{bmatrix} \begin{bmatrix} A^T \\ A^{-1}X - A^{-1}Q\end{bmatrix} &= 0 \\ \begin{bmatrix} X & -I\end{bmatrix} \begin{bmatrix} A^T & 0 \\ -A^{-1}Q & A^{-1}\end{bmatrix} \begin{bmatrix} I \\ X\end{bmatrix} &= 0 \\ \begin{bmatrix} X & -I\end{bmatrix} H \begin{bmatrix} I \\ X\end{bmatrix} &= 0 \tag{1} \label{eq:sympl_eig1} \end{align} $$ where we defined $$H = \begin{bmatrix} A^T & 0 \\ -A^{-1}Q & A^{-1}\end{bmatrix}$$ which is a symplectic matrix that has the property that (lemma 2 in section 4.1 of @FedericoPoloni's paper), if all the eigenvalues of $A$ lie strictly inside the unit circle in the complex plane, then the $2n$ eigenvalues of $H$ consist of $n$ eigenvalues inside the unit circle, and $n$ eigenvalues outside the unit circle (to quickly see why this is true, note that $H$ is block lower-triangular, so its eigenvalues are the union of the eigenvalues of the matrices on its diagonal). This property will be useful later.

Furthermore, note from \eqref{eq:sympl_eig1} that $H \begin{bmatrix} I \\ X\end{bmatrix}$ is in the null space of $\begin{bmatrix} X & -I\end{bmatrix}$. Equivalently, because the null space of $\begin{bmatrix} X & -I\end{bmatrix}$ is orthogonal to the image of $$\begin{bmatrix} X & -I\end{bmatrix}^T = \begin{bmatrix} X \\ -I\end{bmatrix}$$ (see orthogonal complementarity) then $H \begin{bmatrix} I \\ X\end{bmatrix}$ is in the image of \begin{align} \begin{bmatrix} 0 & -I \\ I & 0\end{bmatrix} \begin{bmatrix} X \\ -I\end{bmatrix} &= \begin{bmatrix} I \\ X\end{bmatrix} \end{align} where the matrix $$\begin{bmatrix} 0 & -I \\ I & 0\end{bmatrix}$$ represents a 90-degree counter-clockwise rotation in $2n$-dimensional space. Let $W$ be an $n \times n$ matrix such that each of its columns span the $n$-dimensional image of $\begin{bmatrix} I \\ X\end{bmatrix}$. Then, because $H \begin{bmatrix} I \\ X\end{bmatrix}$ is in the image of $\begin{bmatrix} I \\ X\end{bmatrix}$, then $$H \begin{bmatrix} I \\ X\end{bmatrix} = \begin{bmatrix} I \\ X\end{bmatrix}W$$ which almost looks like an eigenvalue problem. The problem is that $W$ is not necessarily diagonal (or in Jordan normal form). Before we solve this problem, we can first determine what $W$ corresponds to. Recall that \begin{align} H \begin{bmatrix} I \\ X\end{bmatrix} &= \begin{bmatrix} A^T & 0 \\ -A^{-1}Q & A^{-1}\end{bmatrix} \begin{bmatrix} I \\ X\end{bmatrix} \\ &= \begin{bmatrix} A^T \\ A^{-1}X - A^{-1}Q\end{bmatrix} \end{align} Then, \begin{align} H \begin{bmatrix} I \\ X\end{bmatrix} &= \begin{bmatrix} I \\ X\end{bmatrix}W \\ \begin{bmatrix} A^T \\ A^{-1}X - A^{-1}Q\end{bmatrix} &= \begin{bmatrix} W \\ XW\end{bmatrix} \end{align} which implies that $W = A^T$.

Going back to our problem where $W$ (or $A^T$) is not necessarily diagonal, suppose that $W$ is diagonalizable, such that $W = X_1 \Lambda X_1^{-1}$, where $\Lambda$ is diagonal and contains the eigenvalues of $A^T$ (which are the same as the eigenvalues of $A$). Then, \begin{align} H \begin{bmatrix} I \\ X\end{bmatrix} &= \begin{bmatrix} I \\ X\end{bmatrix}W \\ H \begin{bmatrix} I \\ X\end{bmatrix} &= \begin{bmatrix} I \\ X\end{bmatrix}X_1 \Lambda X_1^{-1} \\ H \begin{bmatrix} I \\ X\end{bmatrix} &= \begin{bmatrix} X_1 \\ XX_1\end{bmatrix} \Lambda X_1^{-1} \\ H \begin{bmatrix} X_1 \\ XX_1\end{bmatrix} &= \begin{bmatrix} X_1 \\ XX_1\end{bmatrix} \Lambda \end{align} Moreover, let $X_2 = XX_1$, such that $X = X_2X_1^{-1}$. Then, \begin{align} H \begin{bmatrix} X_1 \\ X_2\end{bmatrix} &= \begin{bmatrix} X_1 \\ X_2\end{bmatrix} \Lambda \end{align} which represents an eigenvalue problem. Note that the $n$ eigenvalues of $W$ (or $A^T$) are a subset of the $2n$ eigenvalues of $H$. Also, note that $\begin{bmatrix} X_1 \\ X_2\end{bmatrix}$ has $2n$ rows and $n$ columns, while $H$ has $2n$ eigenvalues and eigenvectors. This implies that we will have to pick out $n$ eigenvalues and eigenvectors out of the $2n$ possible ones of $H$. Because $W = A^T$ and $\Lambda$ consists of its eigenvalues, and because $A$ is already given and assumed to be stable, then we should pick the $n$ eigenvectors that correspond to the $n$ eigenvalues in $\Lambda$ to form the matrix $\begin{bmatrix} X_1 \\ X_2\end{bmatrix}$. Then, we can compute $X$ as $X = X_2X_1^{-1}$.

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  • $\begingroup$ Maybe you should switch the accepted answer to yours; there is much more detail here. $\endgroup$ Commented Nov 16, 2022 at 7:58
  • $\begingroup$ @FedericoPoloni this answer would not exist without your help :-) $\endgroup$
    – mhdadk
    Commented Nov 16, 2022 at 11:00
  • $\begingroup$ Note that if you only care about getting $H\begin{bmatrix}I\\X\end{bmatrix} = \begin{bmatrix}I\\X\end{bmatrix} W$ (or the equivalent equation in my answer without inverting the 2x2 block matrix in the RHS), then this relation is easier to prove directly: just write down the equations corresponding to the two blocks, and check that they are verified if $X$ solves the discrete Lyapunov equation and $W=A^T$. $\endgroup$ Commented Nov 16, 2022 at 18:22
  • $\begingroup$ @FedericoPoloni not sure I understand. Do you mean this? $$\begin{align} H \begin{bmatrix} I \\ X\end{bmatrix} &= \begin{bmatrix} I \\ X\end{bmatrix}W \\ \begin{bmatrix} A^T \\ A^{-1}X - A^{-1}Q\end{bmatrix} &= \begin{bmatrix} W \\ XW\end{bmatrix} \end{align}$$ $\endgroup$
    – mhdadk
    Commented Nov 16, 2022 at 19:33
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    $\begingroup$ Yes; from the first block you get $A^T=W$, and if you multiply the second one by $A$ you get $X-Q = AXA^T$. Since $X$ solves the Lyapunov equation this equality is verified; this proves that $H\begin{bmatrix}I\\X\end{bmatrix} = \begin{bmatrix}I\\X\end{bmatrix} W$. $\endgroup$ Commented Nov 16, 2022 at 21:26

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