关于矩阵导数的几个性质的证明
其他
2019-02-23 20:17:18
阅读次数: 0
(1)
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\nabla_Atr(AB)=B^T
∇ A t r ( A B ) = B T
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\nabla_Atr(ABA^TC)=CAB+C^TAB^T
∇ A t r ( A B A T C ) = C A B + C T A B T
(3)
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\nabla_A|A|=|A|(A^{-1})^T
∇ A ∣ A ∣ = ∣ A ∣ ( A − 1 ) T
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A=[a_1,\cdots,a_n],B=\begin{bmatrix}b_1'\\\vdots\\b_n'\end{bmatrix}
A = [ a 1 , ⋯ , a n ] , B = ⎣ ⎢ ⎡ b 1 ′ ⋮ b n ′ ⎦ ⎥ ⎤
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\nabla_Atr(AB)=\nabla_Atr(\sum_{i=1}^{n}a_ib_i')=\sum_{i=1}^{n}\nabla_Atr(a_ib_i')
∇ A t r ( A B ) = ∇ A t r ( ∑ i = 1 n a i b i ′ ) = ∑ i = 1 n ∇ A t r ( a i b i ′ )
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\nabla_Atr(a_ib_i')=\nabla_Atr(\begin{bmatrix}a_{1i}\\\vdots\\a_{mi}\end{bmatrix}[b_{i1},\cdots,b_{im}])=\nabla_A\sum_{j=1}^ma_{ji}b_{ij}=[0,\cdots,b_i,\cdots,0]
∇ A t r ( a i b i ′ ) = ∇ A t r ( ⎣ ⎢ ⎡ a 1 i ⋮ a m i ⎦ ⎥ ⎤ [ b i 1 , ⋯ , b i m ] ) = ∇ A ∑ j = 1 m a j i b i j = [ 0 , ⋯ , b i , ⋯ , 0 ]
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\nabla_Atr(AB)=\sum_{i=1}^n\nabla_Atr(a_ib_i')=[b_1,\cdots,b_n]=B^T
∇ A t r ( A B ) = ∑ i = 1 n ∇ A t r ( a i b i ′ ) = [ b 1 , ⋯ , b n ] = B T
∣
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|A|=\sum_jA_{ij}A_{ij}'
∣ A ∣ = ∑ j A i j A i j ′
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A_{ij}'
A i j ′
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\frac{\partial |A|}{\partial A_{ij}}=A_{ij}'
∂ A i j ∂ ∣ A ∣ = A i j ′
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\nabla_A|A|=A'=(A^*)^T=(|A|A^{-1})^T=|A|(A^{-1})^T
∇ A ∣ A ∣ = A ′ = ( A ∗ ) T = ( ∣ A ∣ A − 1 ) T = ∣ A ∣ ( A − 1 ) T
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AA^*=|A|E_n
A A ∗ = ∣ A ∣ E n
转载自 blog.csdn.net/a1109885671/article/details/81226176
