Inverse of the following matrix.

Question

$$M=\left[\begin{array}{cc} (m+2) & 2\cdot1{^\prime}_m \\\ 2\cdot1_m & I_{m}+J_{m,m}\\ \end{array}\right] $$

Is there a closed form of $M^{-1}$?

Where $I_m$ is an identity matrix, $1_m$ is a column vector of ones and $J_{m,m}$ is a matrix of ones.

Answer 1

Since $\left( {{\bf I}_{\,m} + {\bf J}_{\,m} } \right)$ has a quite compact inverse, which is easy to demonstrate to be: $$ \left( {{\bf I}_{\,m} + {\bf J}_{\,m} } \right)^{\,{\bf - 1}} = \left( {{\bf I}_{\,m} - {1 \over {m + 1}}{\bf J}_{\,m} } \right) $$ then we can proceed in two alternative ways:

1. use Blockwise Inversion, which is $$ \left[ {\matrix{ {\bf A} & {\bf B} \cr {\bf C} & {\bf D} \cr } } \right]^{\,\,{\bf - 1}} = \left[ {\matrix{ {\left( {{\bf A - BD}^{\,{\bf - 1}} {\bf C}} \right)^{\,{\bf - 1}} } & {{\bf - }\left( {{\bf A - BD}^{\,{\bf - 1}} {\bf C}} \right)^{\,{\bf - 1}} {\bf BD}^{\,{\bf - 1}} } \cr {{\bf - D}^{\,{\bf - 1}} {\bf C}\left( {{\bf A - BD}^{\,{\bf - 1}} {\bf C}} \right)^{\,{\bf - 1}} } & {{\bf D}^{\,{\bf - 1}} {\bf + D}^{\,{\bf - 1}} {\bf C}\left( {{\bf A - BD}^{\,{\bf - 1}} {\bf C}} \right)^{\,{\bf - 1}} {\bf BD}^{\,{\bf - 1}} } \cr } } \right] $$
2. standing the formula for the inverse of $\left( {{\bf I}_{\,m} + {\bf J}_{\,m} } \right)$ and the fact that $\mathbf M$ is symmetric, we can assume that $$ {\bf M}_{\,m} ^{\,\,{\bf - 1}} = \left[ {\matrix{ a & {b\overline {\bf 1} _{\,m} } \cr {b{\bf 1}_{\,m} } & {c{\bf I}_{\,m} + d{\bf J}_{\,m} } \cr } } \right] $$ then, imposing that $$ \eqalign{ & {\bf I}_{\,m} = {\bf M}_{\,m} {\bf M}_{\,m} ^{\,\,{\bf - 1}} = \cr & = \left[ {\matrix{ {m + 2} & {2\overline {\bf 1} _{\,m} } \cr {2{\bf 1}_{\,m} } & {{\bf I}_{\,m} + {\bf J}_{\,m} } \cr } } \right]\left[ {\matrix{ a & {b\overline {\bf 1} _{\,m} } \cr {b{\bf 1}_{\,m} } & {c{\bf I}_{\,m} + d{\bf J}_{\,m} } \cr } } \right] = \cr & = \left[ {\matrix{ {\left( {m + 2} \right)a + 2mb} & {\left( {\left( {m + 2} \right)b + 2c + 2md} \right)\overline {\bf 1} _{\,m} } \cr {\left( {2a + \left( {m + 1} \right)b} \right){\bf 1}_{\,m} } & {c\,{\bf I}_{\,m} + \left( {2b + \left( {m + 1} \right)d + c} \right){\bf J}_{\,m} } \cr } } \right] \cr} $$ we arrive at: $$ \left\{ \matrix{ \left( {m + 2} \right)a + 2mb = 1 \hfill \cr \left( {m + 2} \right)b + 2c + 2md = 0 \hfill \cr 2a + \left( {m + 1} \right)b = 0 \hfill \cr c = 1 \hfill \cr 2b + c + \left( {m + 1} \right)d = 0 \hfill \cr} \right.\quad \Rightarrow \quad \left\{ \matrix{ a = \left( {m + 1} \right)/\left( {m^{\,2} - m + 2} \right) \hfill \cr b = - 2/\left( {m^{\,2} - m + 2} \right) \hfill \cr c = 1 \hfill \cr d = \left( {2 - m} \right)/\left( {m^{\,2} - m + 2} \right) \hfill \cr} \right. $$