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There are $k$ matrix equations with the same unknown $\mathbf{X}$:

$\mathbf{A}_i(\mathbf{D}_i-\mathbf{X})^{-1}\mathbf{B}_i=\mathbf{C}_i$

where $i=1,2,...,k$. $\mathbf{A}_i$ is a $m\times n$ matrix. $\mathbf{D}_i$ and $\mathbf{X}$ are $n\times n$ matrices. $\mathbf{B}_i$ is a $n\times m$ matrix. $\mathbf{C}_i$ is a $m\times m$ matrix.

I have $m<n$ which indicates that each matrix euqation alone is underdetermined. According to [Penrose, 1955], and assuming that the inverses exist, we have the general solution for each equation

$(\mathbf{D}_i-\mathbf{X})^{-1}=\mathbf{A}{_i}^{\dagger}\mathbf{C}_i\mathbf{B}{_i}^{\dagger}+\mathbf{Y}_i-\mathbf{A}{_i}^{\dagger}\mathbf{A}{_i}\mathbf{Y}_i\mathbf{B}{_i}\mathbf{B}{_i}^{\dagger}$

$\Rightarrow\mathbf{X}=\mathbf{D}{_i}-(\mathbf{A}{_i}^{\dagger}\mathbf{C}_i\mathbf{B}{_i}^{\dagger}+\mathbf{Y}_i-\mathbf{A}{_i}^{\dagger}\mathbf{A}{_i}\mathbf{Y}_i\mathbf{B}{_i}\mathbf{B}{_i}^{\dagger})^{-1}$

where $\mathbf{Y}_i$ is an arbitrary matrix with the proper dimensions and $^{\dagger}$ denotes the Moore–Penrose pseudoinverse.

I sucessfully found some $\mathbf{X}$ to satisfy or best satisfy all the $k$ equations, using an optimizer. However, the obtained solution may show an undesired physical meaning. I wonder if this could be done in an analytical way? I think the analytical method is likely to provide a more meaningful result.

Thanks very much for taking time to read this :)

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  • $\begingroup$ $C-A(D−X)^{−1}B$ is the Schur complement for $D-X$ in $\begin{bmatrix}C&A\\B&D-X\end{bmatrix}$ (or $A$ and $B$ switched), so that your equation is equivalent to a rank condition on this matrix. $\endgroup$ – LutzL Jan 10 '14 at 14:09
  • $\begingroup$ Thanks very much for the comment :-) Do you mean that the solving the equation $A(D-X)^{-1}B=C$ is equivalent to finding $X$ which makes $\begin{bmatrix}C & A\\B &D-X\end{bmatrix}$ singular? We may easy to find that $A(D-X)^{-1}B=C \Rightarrow \begin{bmatrix}C & A\\B &D-X\end{bmatrix} $is singular, but how to proof that $\begin{bmatrix}C & A\\B &D-X\end{bmatrix}$ is singluar $ \Rightarrow A(D-X)^{-1}B=C$ ? We might just be able to conclude that $det(C-A(D-X)^{-1}B)=0$. In addition, it still seems difficult to solve the $k$ euqations even if such equivalent problem are adopted. $\endgroup$ – Simon Wong Jan 10 '14 at 14:51
  • $\begingroup$ Not only singular, the rank of the big matrix has to be equal to the rank of $D-X$. Which at second glance does not look like a nice condition. $\endgroup$ – LutzL Jan 10 '14 at 15:38

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