Inverting a special matrix

Question

Consider matrices $A$ and $B$ of the forms below:

$$A = \lambda \cdot I$$ $$B = \beta \cdot \pmatrix{ 1 & 1 & \cdots & 1\\ 1 & 1 & \cdots & 1\\ \vdots & \vdots & \ddots&\vdots\\ 1 & 1 & \cdots &1 }$$

In other words, $A$ is a diagonal matrix with all elements of the main diagonal equal and $B$ is a matrix of the same size as $A$ but with all of its elements equal to $\beta$.

Now, consider the matrix $C = A + B$. Is there any formula for $C^{-1}$ based on $A$, $A^{-1}$, $B$?

Thanks.

You can also assume $A=I$. Then if the inverse to $C$ exists, then you should have $C^{-1}=I-B+B^2-B^3+\dots$ — Sasha Patotski, Jul 24 '13 at 22:15
Whoa, boy! Watch out! $B^{-1}$ can't exist, since all the rows and columns of $B$ are equal, $\det(B) = 0$! — Robert Lewis, Jul 24 '13 at 22:18

score 4 · Accepted Answer · edited Apr 13 '17 at 12:20

4

Probably, this paper answers to your question. See also this topic

edited Apr 13 '17 at 12:20

Community

1

answered Jul 24 '13 at 22:16

giorgi

128

score 1 · Answer 2 · answered Jul 24 '13 at 22:20

1

I tried some examples with WolframAlpha:

It looks like:

$$A + B \in \mathbb{R}^{n\times n} \Rightarrow (A+B)^{-1} = \frac{1}{n\lambda\beta+\beta^2}\left(A-B+n\beta I\right)$$

answered Jul 24 '13 at 22:20

Jakube

1,847

what is $\lambda$ ? – Sam Jul 24 '13 at 22:21
@Sam: Read the question. – dfeuer Jul 24 '13 at 22:22
Sam wrote the OP himself. $\lambda$ are the entries in $A$ ($A = \lambda I$). – Jakube Jul 24 '13 at 22:24
@dfeuer: funny!!! I am the one who posted the question ! :)))) For some reasons I read $\beta$ as B then I thought it's for general A and B ! You are right ! – Sam Jul 24 '13 at 22:25

Inverting a special matrix

2 Answers2