Inverse of generalized Sylvester mapping

Question

Given appropriately-sized matrices $A, B$, the Sylvester mapping is defined by

$$ S_{A,B}: X \mapsto XA - BX, \quad X \in \mathbf{R}^{m \times n} $$

Under certain conditions, this map has an inverse $S_{A,B}^{-1}$; for example, if $A$ and $B$ are diagonal matrices $A = \mathsf{diag}(\alpha_1,\dots,\alpha_n)$ and $B = \mathsf{diag}(\beta_1,\dots,\beta_m)$ and their spectra are disjoint, the inverse map can be verified to be

$$ S_{A,B}^{-1}: Y \mapsto \sum_{i=1}^m \sum_{j=1}^n \frac{1}{\alpha_j - \beta_i} e_i e_j^{\mathsf{T}} \circ Y, $$ where $\circ$ denotes Hadamard product. Clearly, $S_{A,B}(S_{A,B}^{-1}(Y)) = Y$ here.

Question: Is there a similar "nice" analytical expression for the inverse of the generalized Sylvester mapping $T_{A,B}$, defined below? $$ T_{A,B}: (X, Y) \mapsto \begin{pmatrix} X A - B Y \\ Y A - B X \end{pmatrix} $$ As above, we may assume that $A$ and $B$ are diagonal matrices with disjoint spectra.

The generalized Sylvester mapping appears in the study of singular subspace perturbations (see, e.g., here), but all I have been able to find in the literature so far is about bounding $\| T_{A,B}^{-1} \|$. Any ideas or pointers to references are welcome!

Answer 1

The sizes of matrices $X$ and $Y$ are both $m\times n$. The size of $A$ is $n\times n$, and that of $B$ is $m\times m$.

We rewrite $(X,Y)$ as $\mathcal{X}=\begin{pmatrix}X\\Y\end{pmatrix}\in \mathbb{R}^{2m\times n}$. Then $T_{A,B}$ can be written as $$ \mathcal{T}(\mathcal{X})=\mathcal{T}_{A,B} (\mathcal{X})=\mathcal{X}A-\begin{pmatrix}0 & B\\ B&0\end{pmatrix}\mathcal{X}=\begin{pmatrix} XA-BY\\YA-BX\end{pmatrix}. $$ Thus, $\mathcal{T}$ can also be considered as Sylvester mapping with the matrices $A\in \mathbb{R}^{n\times n}$ and $\mathcal{B}=\begin{pmatrix}0 & B\\ B&0\end{pmatrix}\in\mathbb{R}^{2m\times 2m}$.

The matrix $\mathcal{B}$ has spectra $\pm \beta_i$, $i=1,\ldots,m$. Then for $\mathcal{T}$ to be invertible, we need the disjointness of $\{\alpha_i\}$ and $\{\pm \beta_i\}$. If the disjointness above is satisfied, then we may proceed as the usual Sylvester mapping case.

To see this, we show that $\mathcal{B}$ is diagonalizable. $$ \begin{pmatrix} 0&B\\B&0\end{pmatrix}\begin{pmatrix}I&-I\\I&I\end{pmatrix}=\begin{pmatrix}I&-I\\I&I\end{pmatrix}\begin{pmatrix}B&0\\0&-B \end{pmatrix}. $$ Let $Q=\begin{pmatrix}I&-I\\I&I\end{pmatrix}\in\mathbb{R}^{2m\times 2m}$ and $\mathcal{D}=\begin{pmatrix}B&0\\0&-B\end{pmatrix}$. Then we show that $\mathcal{T}$ is similar to $$ \mathcal{U}:\mathbb{R}^{2m\times n}\rightarrow\mathbb{R}^{2m\times n},$$ $$ \mathcal{U}(\mathcal{Y})=\mathcal{Y}A-\mathcal{D}\mathcal{Y}. $$ We have $$ \mathcal{T}(\mathcal{X})=\mathcal{X}A-Q\mathcal{D}Q^{-1}\mathcal{X}. $$ If we define $\mathcal{Q}: \mathcal{Y}\mapsto Q\mathcal{Y}$, then $\mathcal{Q}^{-1}\mathcal{T}\mathcal{Q}=\mathcal{U}$ and $\mathcal{T}=\mathcal{Q}\mathcal{U}\mathcal{Q}^{-1}$. Thus, $\mathcal{T}^{-1}=\mathcal{Q}\mathcal{U}^{-1}\mathcal{Q}^{-1}$. For $\mathcal{U}^{-1}$, the formula you gave works. $$ \mathcal{U}^{-1}(\mathcal{Y})=\sum_{i=1}^{2m}\sum_{j=1}^n\frac1{\alpha_j-\beta_i}e_ie_j^T\circ\mathcal{Y} $$ with $\beta_i=-\beta_{i-m}$ for $i=m+1, \ldots, 2m$.

Hence, $$ \mathcal{T}^{-1}(\mathcal{X})=Q\sum_{i=1}^{2m}\sum_{j=1}^n\frac1{\alpha_j-\beta_i}e_ie_j^T\circ (Q^{-1}\mathcal{X}) $$ with $$ Q^{-1}=\frac1{2}\begin{pmatrix} I&I\\ -I&I \end{pmatrix}. $$