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Given a matrix $A$ and an identity matrix $E$, we can get the resultant matrix $X={(E-A)}^{-1}$. Now for a given diagonal matrix $D$, we would like to compute the matrix $Y={(E-DA)}^{-1}$. Is there any effective method for us to use the resultant matrix $X$ and $D$ to compute $Y$ ?

In short, can we compute the matrix ${(E-DA)}^{-1}$ by utilizing the result of ${(E-A)}^{-1}$ ?

Any suggestions are welcome!

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use Neumann series – chaohuang Jul 18 '12 at 15:30
up vote 1 down vote accepted

Here's an expression that may be useful:.

$$ \begin{align} Y &= (I-DA)^{-1} \\&= D^{-1}(D^{-1}-A)^{-1} \\&= D^{-1}((I-A) - (I-D^{-1}))^{-1} \\&= D^{-1}X(I - (I-D^{-1})X)^{-1} \end{align} $$

This still requires a matrix inversion, but if $D$ is close to the identity then it may be easier to compute $(I - (I-D^{-1})X)^{-1}$ approximately. Specifically, if $\rho( (I-D^{-1})X ) < 1$, then:

$$ Y= D^{-1}X\sum_{k=0}^\infty\left((I-D^{-1})X\right)^k $$

There are also a lot of different formulas you could get by manipulating the expression and applying the Woodbury formula, but like the above, it would involve taking the inverse of a matrix.

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Is there any approach that can avoid the inverse of a matrix by using the result of ${(I-A)}^{-1}$? – John Smith Jul 19 '12 at 2:34
Maybe if $A$ has very special structure, but for general matrices probably not. I don't think it's possible even in one dimension. Can one compute $(1 - da)^{-1}$ from the value $(1-a)^{-1}$ without doing a division? – p.s. Jul 19 '12 at 16:26

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