矩阵求逆引理与矩阵分块求逆推导

矩阵求逆引理要解决的问题是：已知一个矩阵的逆矩阵，当矩阵产生了变化时，能不能根据已知的逆矩阵，求产生变化后的矩阵的逆。这里说的变化量，指的是${m{B}}{m{D}}^{-1}{m{C}}$

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

设矩阵${m{A}}$的变化量为${m{B}}{m{D}}^{-1}{m{C}}$，逆矩阵的变化量为${m{X}}$，则
egin{equation}
egin{split}
{m{A}}^{-1} + {m{X}} = left({m{A}} + {m{B}}{m{D}}^{-1}{m{C}} ight)^{-1} \
left({m{A}} + {m{B}}{m{D}}^{-1}{m{C}} ight)left({m{A}}^{-1} + {m{X}} ight) = {m{I}} \
{m{I}} + {m{A}}{m{X}} + {m{B}}{m{D}}^{-1}{m{C}}{m{A}}^{-1} + {m{B}}{m{D}}^{-1}{m{C}}{m{X}} = {m{I}} \
{m{A}}{m{X}} + {m{B}}{m{D}}^{-1}{m{C}}{m{A}}^{-1} + {m{B}}{m{D}}^{-1}{m{C}}{m{X}} = {m{0}} \
left({m{A}} + {m{B}}{m{D}}^{-1}{m{C}} ight){m{X}} + {m{B}}{m{D}}^{-1}{m{C}}{m{A}}^{-1} = {m{0}}
end{split}
end{equation}

所以逆矩阵的变化为：
egin{equation}
egin{split}
{m{X}} &= -left({m{A}} + {m{B}}{m{D}}^{-1}{m{C}} ight)^{-1}{m{B}}{m{D}}^{-1}{m{C}}{m{A}}^{-1} \
&= -left[{m{B}}left({m{B}}^{-1}{m{A}}+{m{D}}^{-1}{m{C}} ight) ight]^{-1} {m{B}}{m{D}}^{-1}{m{C}}{m{A}}^{-1} \
&= -left[{m{B}}^{-1}{m{A}}+{m{D}}^{-1}{m{C}} ight]^{-1} {m{B}}{m{D}}^{-1}{m{C}}{m{A}}^{-1} \
&= -left[{m{D}}^{-1}left({m{D}}{m{B}}^{-1}{m{A}}+{m{C}} ight) ight]^{-1}{m{D}}^{-1}{m{C}}{m{A}}^{-1} \
&= -left[{m{D}}{m{B}}^{-1}{m{A}}+{m{C}} ight]^{-1}{m{C}}{m{A}}^{-1} \
&= -left[left({m{D}}{m{B}}^{-1}+{m{C}}{m{A}}^{-1} ight){m{A}} ight]^{-1}{m{C}}{m{A}}^{-1} \
&= -{m{A}}^{-1}left[{m{D}}{m{B}}^{-1}+{m{C}}{m{A}}^{-1} ight]^{-1}{m{C}}{m{A}}^{-1} \
&= -{m{A}}^{-1}left[left({m{D}}+{m{C}}{m{A}}^{-1}{m{B}} ight){m{B}}^{-1} ight]^{-1}{m{C}}{m{A}}^{-1} \
&= -{m{A}}^{-1}{m{B}}left[{m{D}}+{m{C}}{m{A}}^{-1}{m{B}} ight]^{-1}{m{C}}{m{A}}^{-1} \
end{split}
end{equation}

根据矩阵求逆引理，可以计算常用的两种场景的逆矩阵，一种是RLS自适应滤波器中的输入向量相关矩阵求逆，具体推导过程请看RLS自适应滤波器中用矩阵求逆引理来避免求逆运算这里不再详细说明。另一种是相关矩阵的递归平均估计${m{R}}(n+1) = alpha {m{R}}(n) + eta {m{x}}(n){{m{x}}^H(n)}$。推导结果如下，大家可自行验证结果是否正确

设：${m{A}} = alpha {m{R}}(n)$、${m{B}} = {m{x}}(n)$、${m{C}} = {m{x}}^H(n)$、${m{D}} = {eta}^{-1} = {(1-alpha)}^{-1}$，根据矩阵求逆引理，可得

egin{align*}
{{m{R}}(n+1)}^{-1} &= {alpha}^{-1}{{m{R}}(n)}^{-1} - frac{{alpha}^{-1}{{m{R}}(n)}^{-1}{m{x}}(n){m{x}}^{H}(n){{m{R}}(n)}^{-1}{alpha}^{-1}}
{{eta}^{-1}+{m{x}}^{H}(n){alpha}^{-1}{{m{R}}(n)}^{-1}{m{x}}(n)} \
&= {alpha}^{-1}{{m{R}}(n)}^{-1} - frac{{alpha}^{-1}{{m{R}}(n)}^{-1}{m{x}}(n){m{x}}^{H}(n){{m{R}}(n)}^{-1}}
{{alpha}{eta}^{-1}+{m{x}}^{H}(n){{m{R}}(n)}^{-1}{m{x}}(n)}
end{align*}

 其它例子就不再举了。总之，记好矩阵求逆的条件（开始时的加粗红字），以方便碰到符合的场景随时使用。