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I have a Hermitian matrix $A$ with the following block Kronecker structure,

$A = \begin{pmatrix} R_0 \bigotimes S_0 & R_1 \bigotimes S_1\\ R_1^H \bigotimes S_1^H & R_0 \bigotimes S_0 \end{pmatrix} $

in which $S_0$ and $S_1$ are both circulant matrices, and therefore can be diagonalized respectively as

$S_0 = FD_0F^H$

$S_1 = FD_1F^H$

where $F$ stands for the DFT matrix. Besides, $D_1$ is a rotated version of $D_0$, which means

$D_1 = D_0 R$,

where R stands for a diagonal phase-rotation matrix.

Unfortunately, $R_0$ and $R_1$ do not possess the circulant structure. In fact, they are Hermitian only. And their eigen-decompositions are respectively

$R_0 = U_0\Lambda_0 U_0^H$

$R_1 = U_1\Lambda_1 U_1^H$

where $U_0$ is not equal to $U_1$ generally.

However, the size of $R_0$ and $R_1$ is very small, such as 2 or 4.

My question: is there any efficient way to calculate $(A+I)^{-1}$ ?

Any comments and suggestions are welcome. Thanks in advance.

share|improve this question
I hope you have already tried en.wikipedia.org/wiki/Invertible_matrix#Blockwise_inversion --- since the individual blocks are easy to invert, this should be easy. The only tricky part will be the Schur complement... –  Suvrit Oct 12 '13 at 14:10
Thanks for your comment. I tried the block-wise inversion. However, it seems that we cannot invert the Schur complement directly, since the term does not have a product form. Right? –  user38679 Oct 16 '13 at 22:09
Yes, as I suspected, inverting the Schur complement is what breaks this... –  Suvrit Oct 17 '13 at 16:30
I agree with you. Probably there does not exist a convenient way to invert it... –  user38679 Oct 24 '13 at 18:24

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