PRML 付録C 行列式の対数の微分の証明

問題
証明
解説

問題

$$\frac{\partial}{\partial x}\ln{|\mathbf{A}|}=\mathrm{Tr} \left({\mathbf{A}^{-1}\frac{\partial\mathbf{A}}{\partial x}}\right)\tag{1}$$

を証明せよ。ただし、

$$\mathbf{u}_{i}^{T}\mathbf{u}_{j}=I_{ij}\tag{2}$$
$$\mathbf{A}=\sum_{i=1}^{M}\lambda_{i}\mathbf{u}_{i}\mathbf{u}_{i}^{T}\tag{3}$$
$$\mathbf{A}=\sum_{i=1}^{M}\frac{1}{\lambda_{i}}\mathbf{u}_{i}\mathbf{u}_{i}^{T}\tag{4}$$
$$|\mathbf{A}|=\prod_{i=1}^{M}\lambda_{i}\tag{5}$$

を用いてよい。ここで、$\lambda_{i}, \mathbf{u}_{i}$ はそれぞれ、行列 $\mathbf{A}$ の $i$ 番目の固有値と固有ベクトルを表す。

証明

$(2)$ 式の両辺を $x$ で微分して

$$\frac{\partial\mathbf{u}_{i}^{T}}{\partial x}\mathbf{u_{j}}+\mathbf{u}_{i}^{T}\frac{\partial\mathbf{u}_{j}}{\partial x}=0\tag{6}$$

$(1)$ 式の左辺を $(5)$ 式を用いて変形して

$$\frac{\partial}{\partial x}\ln{|A|}=\frac{\partial}{\partial x}\ln{\prod_{i=1}^{M}\lambda_{i}}$$
$$=\frac{\partial}{\partial x}\sum_{i=1}^{M}\ln{\lambda_{i}}$$
$$=\sum_{i=1}^{M}\frac{1}{\lambda_{i}}\frac{\partial\lambda_{i}}{\partial x}\tag{7}$$

また、$(3)$ 式の両辺を $x$ で微分して

$$\frac{\partial \mathbf{A}}{\partial x}=\sum_{j=1}^{M}\left\{\frac{\partial\lambda_{j}}{\partial x}\mathbf{u}_{j}\mathbf{u}_{j}^{T}+\lambda_{j}\frac{\partial(\mathbf{u}_{j}\mathbf{u}_{j}^{T})}{\partial x}\right\}$$
$$=\sum_{j=1}^{M}\left\{\frac{\partial\lambda_{j}}{\partial x}\mathbf{u}_{j}\mathbf{u}_{j}^{T}+\lambda_{j}\left(\frac{\partial\mathbf{u}_{j}}{\partial x}\mathbf{u}_{j}^{T}+\mathbf{u}_{j}\frac{\partial\mathbf{u}_{j}^{T}}{\partial x}\right)\right\}$$

辺々に $(4)$ を左からかけると

$$\mathbf{A}^{-1}\frac{\partial \mathbf{A}}{\partial x}=\left(\sum_{i=1}^{M}\frac{1}{\lambda_{i}}\mathbf{u}_{i}\mathbf{u}_{i}^{T}\right)\left[\sum_{j=1}^{M}\left\{\frac{\partial\lambda_{j}}{\partial x}\mathbf{u}_{j}\mathbf{u}_{j}^{T}+\lambda_{j}\left(\frac{\partial\mathbf{u}_{j}}{\partial x}\mathbf{u}_{j}^{T}+\mathbf{u}_{j}\frac{\partial\mathbf{u}_{j}^{T}}{\partial x}\right)\right\}\right]$$
$$=\sum_{i=1}^{M}\sum_{j=1}^{M}\left\{\frac{1}{\lambda_{i}}\frac{\partial\lambda_{j}}{\partial x}\mathbf{u}_{i}\mathbf{u}_{i}^{T}\mathbf{u}_{j}\mathbf{u}_{j}^{T}+\frac{\lambda_{j}}{\lambda_{i}}\mathbf{u}_{i}\mathbf{u}_{i}^{T}\left(\frac{\partial\mathbf{u}_{j}}{\partial x}\mathbf{u}_{j}^{T}+\mathbf{u}_{j}\frac{\partial\mathbf{u}_{j}^{T}}{\partial x}\right)\right\}$$

上式の両辺のトレースを取ることを考え、第１、第２項のベクトルを含む部分について、それぞれ変形すると

$$\mathrm{Tr}\left[\mathbf{u}_{i}\mathbf{u}_{i}^{T}\mathbf{u}_{j}\mathbf{u}_{j}^{T}\right]=\mathrm{Tr}\left[\mathbf{u}_{i}^{T}\mathbf{u}_{j}\mathbf{u}_{j}^{T}\mathbf{u}_{i}\right]$$
$$=\left(\mathbf{u}_{i}^{T}\mathbf{u}_{j}\right)\left(\mathbf{u}_{j}^{T}\mathbf{u}_{i}\right)=I_{ij}I_{ji}=I_{ij}$$
$$\mathrm{Tr}\left[\mathbf{u}_{i}\mathbf{u}_{i}^{T}\left(\frac{\partial\mathbf{u}_{j}}{\partial x}\mathbf{u}_{j}^{T}+\mathbf{u}_{j}\frac{\partial\mathbf{u}_{j}^{T}}{\partial x}\right)\right]=\mathrm{Tr}\left[\mathbf{u}_{i}^{T}\left(\frac{\partial\mathbf{u}_{j}}{\partial x}\mathbf{u}_{j}^{T}+\mathbf{u}_{j}\frac{\partial\mathbf{u}_{j}^{T}}{\partial x}\right)\mathbf{u}_{i}\right]$$
$$=\mathrm{Tr}\left[\mathbf{u}_{i}^{T}\frac{\partial\mathbf{u}_{j}}{\partial x}\mathbf{u}_{j}^{T}\mathbf{u}_{i}+\mathbf{u}_{i}^{T}\mathbf{u}_{j}\frac{\partial\mathbf{u}_{j}^{T}}{\partial x}\mathbf{u}_{i}\right]=\mathrm{Tr}\left[\mathbf{u}_{i}^{T}\frac{\partial\mathbf{u}_{j}}{\partial x}I_{ji}+I_{ij}\frac{\partial\mathbf{u}_{j}^{T}}{\partial x}\mathbf{u}_{i}\right]$$
$$\mathrm{Tr}\left[\mathbf{u}_{i}^{T}\frac{\partial\mathbf{u}_{j}}{\partial x}+\frac{\partial\mathbf{u}_{j}^{T}}{\partial x}\mathbf{u}_{i}\right]I_{ij}=0\qquad(\because (6))$$

より、トレースを取ると第２項は $0$ になり、第１項もトレースを外せる。したがって

$$\mathrm{Tr}\left({\mathbf{A}^{-1}\frac{\partial\mathbf{A}}{\partial x}}\right)=\sum_{i=1}^{M}\sum_{j=1}^{M}I_{ij}\frac{1}{\lambda_{i}}\frac{\partial\lambda_{j}}{\partial x}=\sum_{i=1}^{M}\frac{1}{\lambda_{i}}\frac{\partial\lambda_{i}}{\partial x}\tag{8}$$

$(7), (8)$ 式より

$$\frac{\partial}{\partial x}\ln{|\mathbf{A}|}=\mathrm{Tr} \left({\mathbf{A}^{-1}\frac{\partial\mathbf{A}}{\partial x}}\right)$$

が成り立つ。