Revisions to A closed-form expression for the inverse of a block-matrix

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edited Sep 18, 2019 at 14:21

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Let $\bf A$ be an $n \times n$ non-singular and circulant matrix 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.

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edited Sep 18, 2019 at 8:53

Denis Serre

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Let $\bf A$ be an $n \times n$ non-singular and circulant 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 $\mathbb{F}$. $$ {\bf B}= \left( \begin{array}{cc} x &{\bf 1}_{n} \\ {\bf 1}^T_{n} & {\bf A} \end{array} \right). $$$$ {\bf B}= \left( \begin{array}{cc} x &{\bf 1}^T_{n} \\ {\bf 1}_{n} & {\bf A} \end{array} \right). $$