Characterization and properties of (R, S)-symmetric, (R, S)-skew symmetric, and (R, S)-conjugate matrices

Let $R\in \mathbb{C}^{m\times m}$ and $S\in \mathbb{C}^{n\times n}$ be nontrivial involutions; i.e., $R=R^{-1}\ne\pm I_m$ and $S=S^{-1}\ne\pm I_n$. We say that $A\in \mathbb{C}^{m\times n}$ is (R,S)-symmetric ((R,S)-skew symmetric) if RAS=A (RAS=-A). We give an explicit representation of an arbitrar...

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Veröffentlicht in:SIAM journal on matrix analysis and applications 2005, Vol.26 (3), p.748-757
1. Verfasser: TRENCH, William F
Format: Artikel
Sprache:eng
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Zusammenfassung:Let $R\in \mathbb{C}^{m\times m}$ and $S\in \mathbb{C}^{n\times n}$ be nontrivial involutions; i.e., $R=R^{-1}\ne\pm I_m$ and $S=S^{-1}\ne\pm I_n$. We say that $A\in \mathbb{C}^{m\times n}$ is (R,S)-symmetric ((R,S)-skew symmetric) if RAS=A (RAS=-A). We give an explicit representation of an arbitrary (R,S)-symmetric matrix A in terms of matrices P and Q associated with R and U and V associated with S. If R = R*, then the least squares problem for A can be solved by solving the independent least squares problems for $A_{PU}=P^{*}AU \in \mathbb{C}^{r\times k}$ and $A_{QV}=Q^{*}AV \in \mathbb{C}^{s\times \ell}$, where $r+s=m$ and $k+\ell =n$. If, in addition, either $\rank(A)=n$ or $S^*=S$, then $A^\dagger$ can be expressed in terms of $A_{PU}^\dagger$ and $A_{QV}^\dagger$. If R = R* and S = S*, then a singular value decomposition of A can obtained from singular value decompositions of $A_{PU}$ and $A_{QV}$. Similar results hold for (R,S)-skew symmetric matrices. We say that $A\in\mathbb{C}^{m\times n}$ is R-conjugate if $RAS=\overline R$, where $R\in\mathbb{R}^{m\times m}$ and $S\in\mathbb{R}^{n\times n}$, $R=R^{-1}\ne\pm I_m$, and $S=S^{-1}\ne\pm I_n$. In this case $\Re(A)$ is $(R,S)$-symmetric and $\Im(A)$ is (R,S)-skew symmetric, so our results provide explicit representations for (R,S)-conjugate matrices. If RT= R, then the least squares problem for the complex matrix A reduces to two least squares problems for a real matrix K. If, in addition, either $\rank(A)=n$ or ST = S, then $A^\dagger$ can be obtained from $K^\dagger$. If both RT = R and ST = S, a singular value decomposition of A can be obtained from a singular value decomposition of K.
ISSN:0895-4798
1095-7162
DOI:10.1137/S089547980343134X