# What we can say about block inverse besides the Schur complement?

Suppose I have a matrix $M$, which has a block structure $% \begin{bmatrix} A & B \\ B^{T} & C% \end{bmatrix}%$, where A has the inverse. How can I better numerically calculate $A^{-1}B$ ? I saw somewhere that I may first do cholesky factorization, say $RR^{\ast }=$ $M$. Then write R also as block structure $% \begin{bmatrix} R_{1} & R_{2} \\ 0 & R_{4}% \end{bmatrix}% ,$ then $A^{-1}B$ would equal to $R_{1}^{-1}R_{2}$.

Is that true? I tried with schur complement, but it doesn't seem a proper tool to answer that.

Thanks a lot.

• @SoManyProb_for_a_brol Why do you think that the Schur complement and block inverse formula will help you in calculating $A^{-1}B$? – MathOverview Apr 2 '13 at 19:06

Presumably $M$ is real and positive definite. Then what you read is mostly true, except that the Cholesky decomposition should be $M=R^T R$. You may verify that $$\pmatrix{A&B\\ B^T&C} =\pmatrix{R_1^T&0\\ R_2^T&R_4^T} \pmatrix{R_1&R_2\\ 0&R_4} =\pmatrix{R_1^T R_1&R_1^T R_2\\ R_2^T R_1&R_2^T R_2 + R_1^T R_4}.$$ Therfore $A^{-1}B=(R_1^TR_1)^{-1}(R_1^TR_2)=R_1^{-1}R_2$.
However, if all you need is $A^{-1}B$, I think it's more efficient to decompose $A$ as $LL^T$ and solve $LL^TX=B$, rather than decomposing $M$ and solve $R_1X=R_2$. But both ways for finding $A^{-1}B$ are backward stable, anyway.