Hello all,

Assume matrix of complex numbers described as a sum of real matrices $A$ which is diagonal and $B$ which is symmetric (and block symmetric if the term is correct):

$A+ Bi$

I want to calculate the inverse of the above for various values of a real parameter $\lambda$ which is:

$(A+ \lambda Bi)^{-1}$

Can you think of any trick to avoid multiply inversions? I am looking for a numerical solution since it is a programming task. Also, I am not looking for the actual inverse but for the some of its elements (an issue where I got some good answers and I consider solved)


  • 1
    $\begingroup$ Once you have the result for a certain value of $\lambda$, expand in a Taylor series for some radius of convergence about that value. Do A and B have any special properties? Sparse? Symmetric or antisymmetric? $\endgroup$ – Ben Crowell Mar 22 '13 at 15:15
  • $\begingroup$ ... or maybe low rank? $\endgroup$ – Federico Poloni Mar 22 '13 at 15:59
  • $\begingroup$ $A$ is diagonal and $B$ is symmetric (and block symmetric if the term is correct). Their size is approx 1000x1000 $\endgroup$ – Dimitris Dakopoulos Mar 26 '13 at 9:16
  • $\begingroup$ There might be some saving in exploiting that real/complex pattern, but do not expect much better than the cost of a full real inversion (or, I should rather say, the cost of solving a linear system --- you don't really want the inverse, do you? See johndcook.com/blog/2010/01/19/dont-invert-that-matrix). Diagonal updates show up in many problems; there is abundant ongoing research on recycling Krylov subspaces in iterative methods (google for those three words to learn more), but, AFAIK, there exists no holy grail method to get the subsequent inverses "for free". $\endgroup$ – Federico Poloni Mar 26 '13 at 17:42
  • $\begingroup$ I decided to try the Taylor series but it right now seems that I cannot converge maybe because the range of $\lambda$s is too big... I am investigating. Oh and I do not invert. I solve! $\endgroup$ – Dimitris Dakopoulos Mar 28 '13 at 15:56

This is not an answer, but it's too long for comments.

First of all it would be really helpful if you provided some background. Is this a numerical problem or algebraic problem?

Two ideas come to mind.

First, let's say that $A$ is regular, then $$ (A+\lambda \imath B)^{-1} = (A(I + \lambda \imath A^{-1}B)^{-1} = (I + \lambda \imath A^{-1}B)^{-1}A^{-1}. $$

Now if all eigenvalues of $\lambda A^{-1}B$ are smaller than $1$, the converges and you can compute the inverse as $$ (I + \lambda \imath A^{-1}B)^{-1} = \sum_{k=0}^\infty (\lambda\imath)^k (A^{-1}B)^k. $$ If $A^{-1}B$ is nilpotent, then your inverse is just a matrix polynomial in $\lambda$. Otherwise this can be used at least for approximation of the inverse.

Second idea.

If your matrices are not too large you can try to work with analytical inversion which works also for matrices of polynomials. I.e. you can write down the using subdeterminants of $A+ \lambda\imath B$ treating $\lambda$ as a variable.

OK. So the problem seems to be:

Compute (numerically) the sum of elements of the inverse of $A+\lambda\imath B$ for real matrices $A,B$, where $A$ is diagonal, $B$ is symmetric and $\lambda$ is a real parameter. The matrices are of order $1000 \times 1000$.

As was explained in the answer to another problem of OP, the sum of elements of the inverse can be obtained as follows. Let $e$ denote the vector of all ones. Then the sought sum equals to the scalar product of $e$ and the solution $x$ of $(A+\lambda\imath B)x = e$. Without any further information it seems that the best possible course is to rewrite the equation as $$ \Bigl(\frac{-\imath }{\lambda} A + B \Bigr)x = \frac{-\imath }{\lambda}e $$ and google for diagonal updates for iterative methods (e.g. Krylov subspaces method) as Federico Poloni suggested in a comment.

  • $\begingroup$ Thank you for your input! I will do some work on it and write my comments but here is some more info that I missed to mention: - Numerical problem. - Matrix $A$ is always diagonal. - Matrix $B$ is symmetric. $\endgroup$ – Dimitris Dakopoulos Mar 26 '13 at 9:11
  • $\begingroup$ Also, my matrices are square, approximately 1000 x 1000 $\endgroup$ – Dimitris Dakopoulos Mar 26 '13 at 9:13

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