# Inverse of sum of two marices, one being diagonal and other unitary.

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

Edit 2: (important edit) Iam interested in this question, because my matrix $$A$$ is huge and so is $$C$$. So computing inverse of $$C$$ is not practical, but luckily the matrix $$A$$ is unitary, so $$A^{-1} = A^*$$, so I easily have $$A^{-1}$$, and finding ways to use it to get $$C^{-1}$$.

• $C$ need not be even invertible. Take $A=I$ and $D=-I$. – Anurag A Oct 30 '18 at 5:47
• @AnuragA : I am not talking about $A = I$. just a general case. Just asume $C$ is invertible, is there a way to compute it faster from knowledge of $A^{-1}$ – Rajesh Dachiraju Oct 30 '18 at 5:52
• In general also there is no guarantee that $C$ is invertible. For example, take $A=\begin{bmatrix}1&2\\0&-2\end{bmatrix}$ and $D=\begin{bmatrix}-1&0\\0&2\end{bmatrix}$. Then $C=\begin{bmatrix}0&2\\0&0\end{bmatrix}$ is NOT invertible. – Anurag A Oct 30 '18 at 5:54

I think that you cannot expect better than a complexity $$\sim n^3$$.

Indeed i) $$(A+D)^{-1}=D^{-1}(I+AD^{-1})^{-1}$$. (it's not better using the Woodbury identity).

All the calculations are in $$O(n^2)$$, except the calculation of $$(I+U)^{-1}$$ where $$U=AD^{-1}$$.

or ii) $$(A+D)^{-1}=A^*(I+DA^*)^{-1}$$. Here all is in $$O(n^2)$$ except the calculations of $$(I+V)^{-1}$$, where $$V=DA^*$$, and of the product of the result by $$A^*$$.

Then the problem reduces to the calculation of $$(I+W)^{-1}$$ where $$W$$ is, roughly speaking, a polar form. Then $$W$$, a priori, has no particularity. Then the complexity of the previous calculation is $$\sim n^3$$.

Remark. The hypothesis $$||W||<1$$ is absolutely useless; to believe the opposite is an urban legend. Indeed $$(I+W)^{-1}\approx I-W+W^2-W^3$$ has already a complexity $$\sim 2n^3$$.

• Its not evident that you have tried to use the fact that $A$ is unitary. Please see my full question, where I have mentioned $A$ is unitary. Hope it will change the answer. – Rajesh Dachiraju Oct 30 '18 at 14:29