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$C = A+D$, $A$ being a unitary matrix and $D$ a full rank diagonal matrix. Is there any easy way to compute $C^{-1}$ from $A^{-1}$ and $D$, if it exists?

I am interested in this question, because my matrix $A$ is huge and so is $C$. So computing inverse of $C$ from scratch is not practical, but luckily the matrix $A$ is unitary, so $A^{-1} = A^*$, so I easily have $A^{-1}$, and hence finding ways to use it to get $C^{-1}$.

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    $\begingroup$ Is $A$ sparse? And why do you need the inverse? If you can avoid explicitly computing $C^{-1}$, you should. $\endgroup$
    – Robert Israel
    Commented Oct 30, 2018 at 17:02

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With additional assumptions you can get an infinite series expansion, using the fact that $$ (I+B)^{-1} = \sum_{k=0}^\infty (-1)^k B^k $$ whenever $B$ is a square matrix with spectral radius $\rho(B) < 1$. So $$ (A+D)^{-1} = (A(I+A^* D))^{-1} = \sum_{k=0}^\infty (-1)^k (A^* D)^k A^* $$ if $\rho(A^* D) < 1$, and $$ (A+D)^{-1} = (D(I+D^{-1}A))^{-1} = \sum_{k=0}^\infty (-1)^k (D^{-1}A)^k D^{-1} $$ if $\rho(D^{-1} A) < 1$.

In particular, the first expansion works whenever the entries of $D$ all have absolute values smaller than 1, and the second expansion works whenever the entries of $D$ all have absolute values larger than 1.

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    $\begingroup$ But the matrix multiplications needed to compute a lot of terms of this series may be more time-consuming than matrix inversion. $\endgroup$
    – Robert Israel
    Commented Oct 30, 2018 at 17:04
  • $\begingroup$ @RobertIsrael: True. To make this practically useful you would want to truncate the series after a small number of terms. If one of the spectral radii is very small you could justify that. $\endgroup$
    – Mark Meckes
    Commented Oct 30, 2018 at 17:12
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    $\begingroup$ Also, this may be more time-consuming than matrix inversion, but potentially more numerically stable. $\endgroup$
    – Mark Meckes
    Commented Oct 30, 2018 at 17:13
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    $\begingroup$ Note that computing two (unstructured) matrix products is already more expensive than one matrix inversion. $\endgroup$
    – Federico Poloni
    Commented Nov 30, 2018 at 13:37
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    $\begingroup$ Why not to use Cayley Hamilton Theorem?. You should need matrix powers and sums just up to n terms independent of spectral radious. I think this is one of the fastest way to compute such inverse, $\endgroup$
    – Jorge Zuniga
    Commented Apr 13, 2022 at 22:38
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You could use Woodbury matrix identity. In your case, this is $$ (A + D)^{-1} = A^{-1} - A^{-1} (D^{-1} + A^{-1})^{-1} A^{-1}. $$ You still need to compute $(D^{-1} + A^{-1})^{-1}$, which may not be easy but, depending on your problem, maybe that could help.

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