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Suppose I have a matrix $\mathbf{H}$ of size $n\times n$, and that I know its inverse $\mathbf{W}=\mathbf{H}^{-1}$.

Then I add a column and a row to $\mathbf{H}$ to obtain a new matrix $\mathbf{G}$. That is $\mathbf{G}$ is given by $$\mathbf{G}=\left( \begin{array}{c|c} r_1 & \begin{array}{ccc} r_2 & \cdots & r_n \end{array} \\ \hline \begin{array}{c} c_2 \\ \vdots \\ c_n \end{array} & {\Huge{\mathbf{H}}} \end{array} \right) $$ Is there a relation between $\mathbf{W}$ and $\mathbf{G}$ and $\mathbf{G}^{-1}$?

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    $\begingroup$ Nice use of nested arrays to format the expanded matrix! $\endgroup$
    – hardmath
    Commented Aug 9, 2015 at 21:47

1 Answer 1

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The block matrix inverse formula is celebrated: \begin{equation*} G^{-1} = \begin{pmatrix} a & x \\ y & H \end{pmatrix}^{-1} = \begin{pmatrix} \xi^{-1} & -\xi^{-1}xH^{-1} \\ -H^{-1}y\xi^{-1} & H^{-1} + H^{-1}y\xi^{-1}xH^{-1} \end{pmatrix} \end{equation*} where $\xi = a - xH^{-1}y$, known as the Schur complement of $H$.

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  • $\begingroup$ Such is also called a "bordered matrix" in the literature (i.e. adding a "border" of new row and column to an existing matrix $H$). $\endgroup$
    – hardmath
    Commented Aug 9, 2015 at 21:37
  • $\begingroup$ @hardmath Thank you for letting us know the term. $\endgroup$
    – Zhanxiong
    Commented Aug 9, 2015 at 21:39

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