# How to compute inverse of sum of a unitary matrix and a full rank diagonal matrix?

$$C = A+D$$, $$A$$ 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 $$C^{-1}$$.

• Is $A$ sparse? And why do you need the inverse? If you can avoid explicitly computing $C^{-1}$, you should. – Robert Israel Oct 30 '18 at 17:02

## 1 Answer

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.

• But the matrix multiplications needed to compute a lot of terms of this series may be more time-consuming than matrix inversion. – Robert Israel Oct 30 '18 at 17:04
• @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. – Mark Meckes Oct 30 '18 at 17:12
• Also, this may be more time-consuming than matrix inversion, but potentially more numerically stable. – Mark Meckes Oct 30 '18 at 17:13
• Note that computing two (unstructured) matrix products is already more expensive than one matrix inversion. – Federico Poloni Nov 30 '18 at 13:37