院解7 京大数学系H28 基礎II 6 行列値正則関数

$\widetilde{a_{jk}}$で$A(z)$の$(j,k)$成分の余因子を表し$ϵ(\sigma )$で置換$\sigma$の符号を表すとする.各$a_{jk}(z)$の正則性から$\text{det}A(z)$,$A(z)$は正則であり
${\displaystyle \frac{d}{dz}}\text{det}A(z)={\displaystyle \frac{d}{dz}}{\displaystyle \sum _{\sigma \in S_N}ϵ(\sigma )a_{1\sigma (1)}(z)a_{2\sigma (2)}(z)\cdots a_{N\sigma (N)}(z)}$
$={\displaystyle \sum _{k=1}^N\sum _{\sigma \in S_N}ϵ(\sigma )a_{1\sigma (1)}(z)\cdots a_{k\sigma (k)}^{\prime }(z)\cdots a_{N\sigma (N)}(z)}$
$={\displaystyle \sum _{k=1}^N\left|\begin{array}{ccc}{a}_{11}(z)& \cdots & a_{1N}(z)\\ ⋮& & ⋮\\ a_{k1}^{\prime }(z)& \cdots & a_{kN}^{\prime }(z)\\ ⋮& & ⋮\\ a_{N1}(z)& \cdots & a_{NN}(z)\end{array}\right|}$
$={\displaystyle \sum _{k=1}^N\sum _{i=1}^N{a}_{ki}^{\prime }(z)\widetilde{a_{ki}(z)}=\text{tr}((a_{ij}^{\prime }(z))\widetilde{A(z)})}$
(ただし$\widetilde{A(z)}$は$A(z)$の余因子行列)
$=\text{tr}(A^{\prime }(z)(\text{det}A(z))A(z{)}^{-1})=\text{det}A(z)\text{tr}(A(z{)}^{-1}{\displaystyle \frac{d}{dz}}A(z))$.
よって${\displaystyle \frac{(\text{det}A(z){)}^{\prime }}{\text{det}A(z)}}=\text{tr}(A(z{)}^{-1}{\displaystyle \frac{d}{dz}}A(z))$.ここで$\text{det}A(z)$の正則性と正則関数の一般論から${\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{C_{ϵ}}{\displaystyle \frac{(\text{det}A(z){)}^{\prime }}{\text{det}A(z)}}dz}$の値は$\text{det}A(z)$の零点$z=0$の位数である.