Given a symmetric matrix $A$ of the form $$ A=\begin{bmatrix}B & C^T \\ C & D\end{bmatrix} $$ If we know $A^{-1}$, is there a clever trick to determine $B^{-1}$?
Is it possible to determine the inverse of the upper block of a matrix when the whole inverse is known?
88 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtThere are 2 best solutions below
On
I have had a look at the correct answer by @Arin Chaudhuri using Woodbury-Sherman-Morrison formula which is a good guideline, but the calculations are tedious. One can bypass this by expanding relationship $A^{-1}A=I$ under the following form:
$$\underbrace{\left(\begin{array}{cc} U & V \\ V^T & W \end{array}\right)}_{A^{-1}}\underbrace{\left(\begin{array}{cc} B & C^T \\ C & D \end{array}\right)}_{A}=\left(\begin{array}{cc} I & 0 \\ 0 & I \end{array}\right),$$
giving the equivalent set of equations:
$$\begin{cases}(1)&UB+VC&=&I \\ (2)&UC^T+VD&=&0 \\ (3)&V^TB+WC&=&0 \\ (4)&V^TC^T+WD&=&I \\ \end{cases}$$
Now, starting from identity:
$$(I-VC)+(-VW^{-1})(-WC)=I,$$
using equations (1) and (3):
$$UB+(-VW^{-1})(V^TB)=I \ \ \iff \ \ (U-VW^{-1}V^T)B=I,$$
one gets, by unicity of the inverse :
$$B=(U-VW^{-1}V^T)^{-1}$$
Remark 0 : Formulas (2) and (4) haven't been used.
Remark 1 : $U-VW^{-1}V^T$ is what is called a Schur complement.
Remark 2 : (in connection with Remark 2) This question has been treated (and enlarged) by Field medalist Terry Tao in his wonderful blog:
https://terrytao.wordpress.com/tag/schur-complement/
Remark 3 : Symbol $I$ stands for any identity matrix. The context indicates the adequate dimensions.
Assume, $$ \begin{align} A^{-1} &= \begin{pmatrix} U & V \\ V^{\top} & W \end{pmatrix} \text{ (known)}. \end{align} $$
And we intend to express $B^{-1}$ in terms of $U,V,$ and $W$.
We know from Woodbury-Sherman-Morrison $$ \begin{align} A^{-1} &=\begin{pmatrix}B^{-1} + B^{-1}C^{\top}(D-CB^{-1}C^{\top})^{-1}CB^{-1} & -B^{-1}C^{\top}(D-CB^{-1}C^{\top})^{-1}\\ -(D-CB^{-1}C^{\top})^{-1}CB^{-1} & (D-CB^{-1}C^{\top})^{-1} \\ \end{pmatrix} \end{align} $$.
And it is easy to verify by direct substitution that $U - VW^{-1}V^{\top} = B^{-1}$.