理解矩阵求逆的引理和分块求逆的推导过程

\begin{equation*}
\begin{split}
{\left( {\bm{A}} + {\bm{B}}{\bm{D}}^{-1}{\bm{C}} \right)}^{-1} =
{\bm{A}}^{-1} - {\bm{A}}^{-1}{\bm{B}}{\left({\bm{D}} + {\bm{C}}{\bm{A}}^{-1}{\bm{B}}\right)}^{-1}{\bm{C}}{\bm{A}}^{-1} \\
{\left( {\bm{A}} - {\bm{B}}{\bm{D}}^{-1}{\bm{C}} \right)}^{-1} =
{\bm{A}}^{-1} + {\bm{A}}^{-1}{\bm{B}}{\left({\bm{D}} - {\bm{C}}{\bm{A}}^{-1}{\bm{B}}\right)}^{-1}{\bm{C}}{\bm{A}}^{-1}
\end{split}
\end{equation*}
${\bm{A}}$是$m\times m$矩阵，${\bm{B}}$是$m\times n$矩阵，${\bm{C}}$是$n\times m$矩阵，${\bm{D}}$是$n\times n$矩阵，${\bm{E}}={\bm{D}} - {\bm{C}}{\bm{A}}^{-1}{\bm{B}}$ 是可逆矩阵
\begin{equation}
{\left[
\begin{array}{*{20}{c}}
{\bm{A}} & {\bm{B}} \\
{\bm{C}} & {\bm{D}}
\end{array}
\right]}_{(m+n)(m+n)}^{-1} =
{\left[
\begin{array}{*{20}{c}}
{\bm{A}}^{-1} + {\bm{A}}^{-1}{\bm{B}}{\bm{E}}^{-1}{\bm{C}}{\bm{A}}^{-1} &
-{\bm{A}}^{-1}{\bm{B}}{\bm{E}}^{-1} \\
-{\bm{E}}^{-1}{\bm{C}}{\bm{A}}^{-1} & {\bm{E}}^{-1}
\end{array}
\right]}
\end{equation}
${\bm{A}}$是$m\times m$矩阵，${\bm{B}}$是$m\times n$矩阵，${\bm{C}}$是$n\times m$矩阵，${\bm{D}}$是$n\times n$矩阵，${\bm{F}}={\bm{A}} - {\bm{B}}{\bm{D}}^{-1}{\bm{C}}$ 是可逆矩阵
\begin{equation}
{\left[
\begin{array}{*{20}{c}}
{\bm{A}} & {\bm{B}} \\
{\bm{C}} & {\bm{D}}
\end{array}
\right]}_{(m+n)(m+n)}^{-1} =
{\left[
\begin{array}{*{20}{c}}
{\bm{F}}^{-1} &
-{\bm{F}}^{-1}{\bm{B}}{\bm{D}}^{-1} \\
-{\bm{D}}^{-1}{\bm{C}}{\bm{F}}^{-1} &
{\bm{D}}^{-1} + {\bm{D}}^{-1}{\bm{C}}{\bm{F}}^{-1}{\bm{B}}{\bm{D}}^{-1}
\end{array}
\right]}
\end{equation}
当${\bm{E}}$、${\bm{F}}$均可逆时，上面两个式子中每个分块都相等
\begin{equation}
{\left[
\begin{array}{*{20}{c}}
{\bm{A}}^{-1} + {\bm{A}}^{-1}{\bm{B}}{\bm{E}}^{-1}{\bm{C}}{\bm{A}}^{-1} &
-{\bm{A}}^{-1}{\bm{B}}{\bm{E}}^{-1} \\
-{\bm{E}}^{-1}{\bm{C}}{\bm{A}}^{-1} & {\bm{E}}^{-1}
\end{array}
\right]} =
{\left[
\begin{array}{*{20}{c}}
{\bm{F}}^{-1} &
-{\bm{F}}^{-1}{\bm{B}}{\bm{D}}^{-1} \\
-{\bm{D}}^{-1}{\bm{C}}{\bm{F}}^{-1} &
{\bm{D}}^{-1} + {\bm{D}}^{-1}{\bm{C}}{\bm{F}}^{-1}{\bm{B}}{\bm{D}}^{-1}
\end{array}
\right]}
\end{equation}
${\bm{A}}$是$m\times n$矩阵，${\bm{B}}$是$m\times m$矩阵，${\bm{C}}$是$n\times n$矩阵，${\bm{D}}$是$n\times m$矩阵，${\bm{G}}={\bm{C}}-{\bm{D}}{\bm{B}}^{-1}{\bm{A}}$ 是可逆矩阵
\begin{equation}
{\left[
\begin{array}{*{20}{c}}
{\bm{A}} & {\bm{B}} \\
{\bm{C}} & {\bm{D}}
\end{array}
\right]}_{(m+n)(m+n)}^{-1} =
{\left[
\begin{array}{*{20}{c}}
-{\bm{G}}^{-1}{\bm{D}}{\bm{B}}^{-1} & {\bm{G}}^{-1} \\
{\bm{B}}^{-1} + {\bm{B}}^{-1}{\bm{A}}{\bm{G}}^{-1}{\bm{D}}{\bm{B}}^{-1} &
-{\bm{B}}^{-1}{\bm{A}}{\bm{G}}^{-1}
\end{array}
\right]}
\end{equation}
${\bm{A}}$是$m\times n$矩阵，${\bm{B}}$是$m\times m$矩阵，${\bm{C}}$是$n\times n$矩阵，${\bm{D}}$是$n\times m$矩阵，${\bm{H}}={\bm{B}}-{\bm{A}}{\bm{C}}^{-1}{\bm{D}}$ 是可逆矩阵
\begin{equation}
{\left[
\begin{array}{*{20}{c}}
{\bm{A}} & {\bm{B}} \\
{\bm{C}} & {\bm{D}}
\end{array}
\right]}_{(m+n)(m+n)}^{-1} =
{\left[
\begin{array}{*{20}{c}}
-{\bm{C}}^{-1}{\bm{D}}{\bm{H}}^{-1} &
{\bm{C}}^{-1} + {\bm{C}}^{-1}{\bm{D}}{\bm{H}}^{-1}{\bm{A}}{\bm{C}}^{-1} \\
{\bm{H}}^{-1} &
-{\bm{H}}^{-1}{\bm{A}}{\bm{C}}^{-1}
\end{array}
\right]}
\end{equation}
当${\bm{G}}$、${\bm{H}}$均可逆时，上面的两个式子中的每个分块都相等
\begin{equation}
{\left[
\begin{array}{*{20}{c}}
-{\bm{G}}^{-1}{\bm{D}}{\bm{B}}^{-1} & {\bm{G}}^{-1} \\
{\bm{B}}^{-1} + {\bm{B}}^{-1}{\bm{A}}{\bm{G}}^{-1}{\bm{D}}{\bm{B}}^{-1} &
-{\bm{B}}^{-1}{\bm{A}}{\bm{G}}^{-1}
\end{array}
\right]} =
{\left[
\begin{array}{*{20}{c}}
-{\bm{C}}^{-1}{\bm{D}}{\bm{H}}^{-1} &
{\bm{C}}^{-1} + {\bm{C}}^{-1}{\bm{D}}{\bm{H}}^{-1}{\bm{A}}{\bm{C}}^{-1} \\
{\bm{H}}^{-1} &
-{\bm{H}}^{-1}{\bm{A}}{\bm{C}}^{-1}
\end{array}
\right]}
\end{equation}
当${\bm{E}}$、${\bm{F}}$、${\bm{G}}$、${\bm{H}}$均可逆时：
\begin{equation}
{\left[
\begin{array}{*{20}{c}}
{\bm{A}} & {\bm{B}} \\
{\bm{C}} & {\bm{D}}
\end{array}
\right]}_{(m+n)(m+n)}^{-1} =
{\left[
\begin{array}{*{20}{c}}
{({\bm{A}}-{\bm{B}}{\bm{D}}^{-1}{\bm{C}})}^{-1} &
{({\bm{C}}-{\bm{D}}{\bm{B}}^{-1}{\bm{A}})}^{-1} \\
{({\bm{B}}-{\bm{A}}{\bm{C}}^{-1}{\bm{D}})}^{-1} &
{({\bm{D}}-{\bm{C}}{\bm{A}}^{-1}{\bm{B}})}^{-1}
\end{array}
\right]}
\end{equation}
证明: