Bound the norm of a matrix function related to discrete algebraic Riccati equation

Question

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 $A_c$?

Answer 1

$\def\vc{\operatorname{vec}}$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.

Answer 2

Your approach is the correct one but you will have to show that the infinite sum is convergent.

Regarding obtaining a better bound, it is often interesting to check what happens in the scalar case or in the diagonal case, just to get some insights. While those cases may suggest that you could do something with eigenvalues, this is not the case in general because eigenvalues are not good for norms; e.g. the matrix $A=\begin{bmatrix} 0 & 1\\0 & 0\end{bmatrix}$ has zero eigenvalues but because of the positive off-diagonal term, the norm will not (and should not) be 0. In this case, I fear that you will not be able to do what you want.