Upper bounds for Lyapunov equation

Question

Consider the Lyapunov equation $$ AX+XA^T+I=0 $$ for the given stable matrix $A$.

Is it possible to obtain the upper estimation for $\|X\|$ (in spectral or Frobenius norm) in terms of $A$?

Answer 1

For the spectral norm: it is well known that the unique solution to this equation can be expressed as $$ X = \int_0^\infty e^{At}e^{A^Tt}\,dt. $$ We note that $$ \|X\| = \left\|\int_0^\infty e^{At}e^{A^Tt}\,dt \right\| \leq \int_0^\infty \left\|e^{At}e^{A^Tt}\right\|\,dt = \int_0^\infty \|e^{At}\|^2\,dt $$ Let $\alpha$ denote the maximum value of $\operatorname{Re}(\lambda)$ among the eigenvalues $\lambda$ of $A$. There exists a constant $C > 0$ such that $\|e^{At}\| \leq C \cdot e^{\alpha t}$ (in particular, if $A$ is diagonalizable and $SAS^{-1}$ is diagonal, then $C$ can be taken to be the condition number of $S$). By the stability of $A$, we must have $\alpha < 0$. With that, we have $$ \| X\| \leq \int_0^\infty \|e^{At}\|^2\,dt \leq C\int_0^\infty e^{2 \alpha t}\,dt = \frac C{2|\alpha|}. $$

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For the Frobenius norm: we can write $$ \operatorname{vec}(X) = -(I_n \otimes A + A \otimes I_n)^{-1} \operatorname{vec}(I). $$ It follows that $$ \|X\|_F = \|\operatorname{vec}(X)\| = \|(I_n \otimes A + A \otimes I_n)^{-1} \operatorname{vec}(I_n)\| \\ \leq \|(I_n \otimes A + A \otimes I_n)^{-1}\| \cdot \|\operatorname{vec}(I_n)\| \\ \leq \|(I_n \otimes A + A \otimes I_n)^{-1}\| \cdot \sqrt{n}. $$ Let $S$ be such that $D = S^{-1}AS$ is diagonal. We note that $$ \|(I_n \otimes A + A \otimes I)^{-1}\| = \\ (S \otimes S)^{-1}(I \otimes D + D \otimes I)^{-1}(S \otimes S) \leq \\ \|(S \otimes S)^{-1}\|\cdot\|(I \otimes D + D \otimes I)^{-1}\| \cdot \|S \otimes S\| =\\ \|S^{-1}\|^2 \cdot \|(I \otimes D + D \otimes I)^{-1}\| \cdot \|S\|^2. $$ Let $\alpha$ denote the maximum value of $\operatorname{Re}(\lambda)$ among the eigenvalues $\lambda$ of $A$. Let $C$ denote the condition number of $S$, i.e. $C = \|S\|\cdot\|S\|^{-1}$. We have $$ \|(I \otimes D + D \otimes I)^{-1}\| \leq \frac 1{2|\alpha|}, $$ so that $\|X\|_F \leq \frac{C^2}{2|\alpha|}$.