1
$\begingroup$

Let $\bf A$ be an $n \times n$ non-singular matrix over $\mathbb{F}$. Let $x$ be a non-zero element of $\mathbb{F}$. Suppose that ${\bf 1}_{n}$ is a symbol for the all-one vector of length $n$ over $\mathbb{F}$. Now consider the following $(n+1) \times (n+1)$ matrix and assume that $\bf B$ is a invertible matrix over $\mathbb{F}$. $$ {\bf B}= \left( \begin{array}{cc} x &{\bf 1}^T_{n} \\ {\bf 1}_{n} & {\bf A} \end{array} \right). $$

My question: Is there a closed-form expression for the inverse of $\bf B$, denoted with ${\bf B}^{-1}$?

Thanks for any help.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

Say that $$B^{-1}=:\begin{pmatrix} b & X^T \\ Y & M \end{pmatrix}.$$ Then using Schur's complement formula (thanks to Nathaniel), $b=(x-{\bf1}^TA^{-1}{\bf1})^{-1}$ and $M=(A-x^{-1}{\bf11}^T)^{-1}$. From this, you can compute the vectors $$Y=-x^{-1}M{\bf1},\qquad X^T=-b{\bf1}^TA^{-1}.$$

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Maybe worth mentioning that this type of result comes immediately from using the Schur complement. $\endgroup$
    – Nathaniel Johnston
    Commented Sep 18, 2019 at 9:35
  • $\begingroup$ @NathanielJohnston. Of course you're right. $\endgroup$
    – Denis Serre
    Commented Sep 18, 2019 at 12:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .