From M you might create another matrix A such that

M = D  I  + A

whereby A = M - D I

where D is the magnitude of the largest diagonal element in M (hence D is a scalar), and I is the identify matrix.

Then use a Taylor series approximation for the inverse of M:

M-1 = I(D I + A)-1 = (1/D) [I + (1/D)A]-1

      = SUM over n = 0 to infinity of [(-1)n (1/D)n+1 An]

The series converges if  ||A||/D < 1

If ||A||/D is small, then I think that the series should converge quickly, and you only need to consider a few terms in the series n=0,1,2... for a good approximation.

I'm not sure that the expressions are simpler.

Ronald

From M you might create another matrix A such that

M = D  I  + A

whereby A = M - D I

where D is the magnitude of the largest diagonal element in M (hence D is a scalar), and I is the identify matrix.

Then use a Taylor series approximation for the inverse of M:

M-1 = I(D I + A)-1 = (1/D) [I + (1/D)A]-1

      = SUM over n = 0 to infinity of [(-1)n (1/D)n+1 An]

The series converges if  ||A||/D < 1

If ||A||/D is small, then I think that the series should converge quickly, and you only need to consider a few terms in the series n=0,1,2... for a good approximation.

I'm not sure that the expressions are simpler.

Ronald

Thanks Ronald ! I will try it !

I was going to suggest this: For M close to identity

M-1=I + (I-M) + (I-M)^2 + (I-M)^3 + ...

where speed of convergence depends on ||I-M||. But I think this is quite similar to Ronalds answer.

Depending on what is easier, the above equation can also be expressed in terms of power of M instead of power of (I-M). After n terms, you would have

M-1 \approx (n+1,1) I - (n+1,2) M + (n+1,3) M2 -+... +(-1)n (n+1,n+1) Mn

where (n,k) is the binomial coefficient (n choose k).

please watch 

https://www.youtube.com/watch?v=z8jHrzceRws

Consider F(M), a differentiable matrix valued function of the matrix M. Let F_km(M)  denote the k,m'th element of F(M) .

We can write the first order Taylor series expansion of F_km(M) around the matrix M0 as follows.

F_km(M)=F_km(M0)+sum_i sum_j dF_km/dM_ij (M_ij-M0_ij)                 (1)

where the derivative dF_km/dM_ij denotes the PARTIAL derivative of  F_km(M) with respect to M_ij, ij'th element of M, EVALUATED at M=M0.

Now we can apply this theory to F(M)=M^-1 as follows.

By equation (60) of The Matrix Cookbook (see the reference below), we have

dF_km/dM_ij= -[M0^-1]_ki [M0^-1]_jm                                                       (2)

where the notation [.]_ij denote the ij'th element of the argument matrix.

Substituting (2) into (1), we get

F_km(M)=F_km(M0) - sum_i sum_j [M0^-1]_ki  (M_ij-M0_ij) [M0^-1]_jm

which can be written as

F_km(M)=F_km(M0) - [ M0^-1 ( M_ij - M0_ij ) M0^-1 ]_km                       (3)

This result is correct for any k,m values. Hence, collecting F_km results in a matrix, we get

F(M)=F(M0) - M0^-1 (M - M0) M0^-1

Since F(M)=M^-1, we have

M^-1 = M0^-1 - M0^-1 (M - M0) M0^-1                                                     (4)

Now let's decompose your M matrix in the following way.

M= D + A 

where D is the diagonal matrix composed of the diagonal entries of M and the A is the rest. Note that D is in general not equal to a constant time identity matrix as opposed to the decomposition of Ronald Kessel above. I am also assuming that D does not contain a diagonal element equal to zero, hence we assume D is invertible.

Now we take M0=D and hence M-M0=A  in (4), which gives

M^-1 = D^-1 - D^-1 A D^-1                                                  (5)                                 

which is slightly different than the earlier results here. This result is the generalized version of the first order Taylor series expansion of the scalar valued function f(m)=1/m around m=d  (m is a scalar) which is given as

1/m = 1/d - a/(d^2)

where a=m-d.

After writing this, I saw the equation written by Peter Breuer above given below.

D^-1/2 M D^-1/2 = I+D^-1/2AD^-1/2                                    (6)

Taking the inverse of both sides gives

D^1/2 M^-1 D^1/2 =inv( I+D^-1/2AD^-1/2)                        (7)

The right hand side can be written approximately using a first order Taylor series expansion as

inv( I+D^-1/2AD^-1/2) = I - D^-1/2AD^-1/2                       (8)

Substituting (8) into (7), we get

D^1/2 M^-1 D^1/2 = I - D^-1/2AD^-1/2  

Multiplying both sides from left and right by D^-1/2 gives (5) as well.

Hope this is useful.

[1] The Matrix Cookbook by Kaare Brandt Petersen and  Michael Syskind Pedersen,

http://www2.imm.dtu.dk/pubdb/views/edoc_download.php/3274/pdf/imm3274.pdf