I was going through the following paper on perturbation analysis of the discrete Riccati equation. https://dml.cz/bitstream/handle/10338.dmlcz/124552/Kybernetika_29-1993-1_2.pdf.
The perturbation analysis involves the norm of the inverse of $\mathcal{T}:\mathbb{R}^{n\times n} \rightarrow \mathbb{R}^{n\times n}$. To be more specific, $\mathcal{T}(X)= X + \gamma A_c' XA_c$ with $0<\gamma\leq 1$, where the eigenvalues of $A_c\in\mathbb{R}^{n\times n}$ (i.e., $\lambda_i,i=1,2,\cdots,n$) lies inside the unique circle in the complex plane. That is $A_c$ is stable (discrete-time). Let $\mathcal{L}(\mathbb{R}^{n\times n},\mathbb{R}^{n\times n})$ be the space of linear operators $\mathbb{R}^{n\times n}\rightarrow \mathbb{R}^{n\times n}$ with the following induced norm
$$
\Vert \mathcal{T}\Vert_\mathcal{L} = \max \{\Vert \mathcal{\mathcal{T}}(X)\Vert:\Vert X\Vert =1 \}, \ \ \ \textrm{for } \mathcal{T}\in \mathcal{L}(\mathbb{R}^{n\times n},\mathbb{R}^{n\times n}).
$$
Note that the eigenvalues of $\mathcal{T}$ are $\mu_{ij}=1-\lambda_i\lambda_j$. Hence, $\mathcal{T}$ is invertible. We are interested in $\Vert \mathcal{T}^{-1}\Vert_{\mathcal{L}}$ or other well-defined norms of $\mathcal{T}^{-1}$.
Since we know $A_c$, I was wondering if we can write $\Vert \mathcal{T}^{-1} \Vert$ explicitly as a function of eigenvalues of $A_c$ or provide an upper bound on $\Vert \mathcal{T}^{-1} \Vert$ that depends on $A_c$.
My attempt:
Since $A_c$ is stable, there exists $L$ such that $L = X – \gamma A_c' X A_c$, which is a Lyapunov equation given $L$ and $A_c$. Then, $X = \sum_{m=0}^\infty \gamma^m (A_c')^m L (A_c)^m$. Hence, we have $\Vert \mathcal{T}^{-1}(L)\Vert = \Vert \sum_{m=0}^\infty \gamma^m (A_c')^m L (A_c)^m \Vert \leq \sum_{m=0}^\infty \gamma^m\Vert A_c^m\Vert^{2}\Vert L\Vert$. Then, $\Vert \mathcal{T}^{-1}\Vert\leq \sum_{m=0}^{\infty}\gamma^m \Vert A_c^m\Vert^{2}$.
The special radius of a matrix is bounded by a norm of the matrix. Can we apply this fact before the triangular inequality is applied and derive a tighter bound on $\Vert \mathcal{T}^{-1}\Vert$ using the eigenvalues of $A_c$?
Best Answer
$\def\vc{\operatorname{vec}}$A possibly tighter bound could be obtained by vectorization operation
$$ \vc(L) = \left( I - \gamma \left(A_c' \otimes A_c'\right) \right) \vc(X) $$
Then,
$$\begin{align} \lVert T^{-1} \rVert &\leq \lVert T^{-1} \rVert_F \\ &= \lVert \vc(X) \rVert \\ &= \lVert \left( I - \gamma \left(A_c' \otimes A_c'\right) \right)^{-1} \vc(L) \rVert \\ &\leq \lVert \left( I - \gamma \left(A_c' \otimes A_c'\right) \right)^{-1} \rVert \lVert \vc(L) \rVert \\ &= \lVert \left( I - \gamma \left(A_c' \otimes A_c'\right) \right)^{-1} \rVert \lVert L \rVert_F \\ &\leq \sqrt{n} \lVert \left( I - \gamma \left(A_c' \otimes A_c'\right) \right)^{-1} \rVert \end{align}$$
where $\lVert \cdot \rVert_F$ is the Frobenius norm. I think even better bounds could be found by somehow converting
$$ \left( I - \gamma \left(A_c' \otimes A_c'\right) \right)^{-1} \vc(L) $$
back into matrix multiplication form.