矩陣求逆引理要解決的問題是：已知一個矩陣及其逆矩陣，當矩陣產生了變化時，能不能根據已知的逆矩陣，求產生變化后的矩陣的逆。這里說的變化量，指的是${\bm{B}}{\bm{D}}^{-1}{\bm{C}}$
\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$m\times m$矩陣，${\bm{B}}$是$m\times n$矩陣，${\bm{C}}$是$n\times m$矩陣，${\bm{D}}$是$n\times n$矩陣，${\bm{E}}={\bm{D}} - {\bm{C}}{\bm{A}}^{-1}{\bm{B}}$ 是可逆矩陣
\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$m\times m$矩陣，${\bm{B}}$是$m\times n$矩陣，${\bm{C}}$是$n\times m$矩陣，${\bm{D}}$是$n\times n$矩陣，${\bm{F}}={\bm{A}} - {\bm{B}}{\bm{D}}^{-1}{\bm{C}}$ 是可逆矩陣
\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}
當${\bm{E}}$、${\bm{F}}$均可逆時，上面兩個式子中每個分塊都相等
\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$m\times n$矩陣，${\bm{B}}$是$m\times m$矩陣，${\bm{C}}$是$n\times n$矩陣，${\bm{D}}$是$n\times m$矩陣，${\bm{G}}={\bm{C}}-{\bm{D}}{\bm{B}}^{-1}{\bm{A}}$ 是可逆矩陣
\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$m\times n$矩陣，${\bm{B}}$是$m\times m$矩陣，${\bm{C}}$是$n\times n$矩陣，${\bm{D}}$是$n\times m$矩陣，${\bm{H}}={\bm{B}}-{\bm{A}}{\bm{C}}^{-1}{\bm{D}}$ 是可逆矩陣
\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}
當${\bm{G}}$、${\bm{H}}$均可逆時，上面的兩個式子中的每個分塊都相等
\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}
當${\bm{E}}$、${\bm{F}}$、${\bm{G}}$、${\bm{H}}$均可逆時：
\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}
證明:
設矩陣${\bm{A}}$的變化量為${\bm{B}}{\bm{D}}^{-1}{\bm{C}}$，逆矩陣的變化量為${\bm{X}}$，則
\begin{equation}
\begin{split}
{\bm{A}}^{-1} + {\bm{X}} = \left({\bm{A}} + {\bm{B}}{\bm{D}}^{-1}{\bm{C}} \right)^{-1} \\
\left({\bm{A}} + {\bm{B}}{\bm{D}}^{-1}{\bm{C}} \right)\left({\bm{A}}^{-1} + {\bm{X}}\right) = {\bm{I}} \\
{\bm{I}} + {\bm{A}}{\bm{X}} + {\bm{B}}{\bm{D}}^{-1}{\bm{C}}{\bm{A}}^{-1} + {\bm{B}}{\bm{D}}^{-1}{\bm{C}}{\bm{X}} = {\bm{I}} \\
{\bm{A}}{\bm{X}} + {\bm{B}}{\bm{D}}^{-1}{\bm{C}}{\bm{A}}^{-1} + {\bm{B}}{\bm{D}}^{-1}{\bm{C}}{\bm{X}} = {\bm{0}} \\
\left({\bm{A}} + {\bm{B}}{\bm{D}}^{-1}{\bm{C}}\right){\bm{X}} + {\bm{B}}{\bm{D}}^{-1}{\bm{C}}{\bm{A}}^{-1} = {\bm{0}}
\end{split}
\end{equation}
所以逆矩陣的變化為：
\begin{equation}
\begin{split}
{\bm{X}} &= -\left({\bm{A}} + {\bm{B}}{\bm{D}}^{-1}{\bm{C}}\right)^{-1}{\bm{B}}{\bm{D}}^{-1}{\bm{C}}{\bm{A}}^{-1} \\
&= -\left[{\bm{B}}\left({\bm{B}}^{-1}{\bm{A}}+{\bm{D}}^{-1}{\bm{C}}\right)\right]^{-1} {\bm{B}}{\bm{D}}^{-1}{\bm{C}}{\bm{A}}^{-1} \\
&= -\left[{\bm{B}}^{-1}{\bm{A}}+{\bm{D}}^{-1}{\bm{C}}\right]^{-1} {\bm{B}}{\bm{D}}^{-1}{\bm{C}}{\bm{A}}^{-1} \\
&= -\left[{\bm{D}}^{-1}\left({\bm{D}}{\bm{B}}^{-1}{\bm{A}}+{\bm{C}}\right)\right]^{-1}{\bm{D}}^{-1}{\bm{C}}{\bm{A}}^{-1} \\
&= -\left[{\bm{D}}{\bm{B}}^{-1}{\bm{A}}+{\bm{C}}\right]^{-1}{\bm{C}}{\bm{A}}^{-1} \\
&= -\left[\left({\bm{D}}{\bm{B}}^{-1}+{\bm{C}}{\bm{A}}^{-1}\right){\bm{A}}\right]^{-1}{\bm{C}}{\bm{A}}^{-1} \\
&= -{\bm{A}}^{-1}\left[{\bm{D}}{\bm{B}}^{-1}+{\bm{C}}{\bm{A}}^{-1}\right]^{-1}{\bm{C}}{\bm{A}}^{-1} \\
&= -{\bm{A}}^{-1}\left[\left({\bm{D}}+{\bm{C}}{\bm{A}}^{-1}{\bm{B}}\right){\bm{B}}^{-1}\right]^{-1}{\bm{C}}{\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}
根據矩陣求逆引理，可以計算常用的兩種場景的逆矩陣，一種是RLS自適應濾波器中的輸入向量相關矩陣求逆，具體推導過程請看RLS自適應濾波器中用矩陣求逆引理來避免求逆運算這里不再詳細說明。另一種是相關矩陣的遞歸平均估計${\bm{R}}(n+1) = \alpha {\bm{R}}(n) + \beta {\bm{x}}(n){{\bm{x}}^H(n)}$。推導結果如下，大家可自行驗證結果是否正確
設：${\bm{A}} = \alpha {\bm{R}}(n)$、${\bm{B}} = {\bm{x}}(n)$、${\bm{C}} = {\bm{x}}^H(n)$、${\bm{D}} = {\beta}^{-1} = {(1-\alpha)}^{-1}$，根據矩陣求逆引理，可得
\begin{align*}
{{\bm{R}}(n+1)}^{-1} &= {\alpha}^{-1}{{\bm{R}}(n)}^{-1} - \frac{{\alpha}^{-1}{{\bm{R}}(n)}^{-1}{\bm{x}}(n){\bm{x}}^{H}(n){{\bm{R}}(n)}^{-1}{\alpha}^{-1}}
{{\beta}^{-1}+{\bm{x}}^{H}(n){\alpha}^{-1}{{\bm{R}}(n)}^{-1}{\bm{x}}(n)} \\
&= {\alpha}^{-1}{{\bm{R}}(n)}^{-1} - \frac{{\alpha}^{-1}{{\bm{R}}(n)}^{-1}{\bm{x}}(n){\bm{x}}^{H}(n){{\bm{R}}(n)}^{-1}}
{{\alpha}{\beta}^{-1}+{\bm{x}}^{H}(n){{\bm{R}}(n)}^{-1}{\bm{x}}(n)}
\end{align*}
其它例子就不再舉了。總之，記好矩陣求逆的條件（開始時的加粗紅字），以方便碰到符合的場景隨時使用。