矩阵的迹和导数的主要性质

今天药忘吃喽~ 2022-05-18 08:30 237阅读 0赞

## 主要公式： ##

### Trace: ###

trAtrAtraAtrA\+BtrABtrABCtrABCD=======∑i=1naiitrATatrAtrA\+trBtrBAtrBCA=trCABtrDABC=trCDAB=trBCDA(26)(27)(28)(29)(30)(31)(32)  (26) t r A = ∑ i = 1 n a i i (27) t r A = t r A T (28) t r a A = a t r A (29) t r A + B = t r A + t r B (30) t r A B = t r B A (31) t r A B C = t r B C A = t r C A B (32) t r A B C D = t r D A B C = t r C D A B = t r B C D A

### [Derivative:][Derivative] ###

dXTdX=dXdXTd(AX)TdXd(AX)dXTd(XTX)dXd(XTAX)dX=====EATA2X(A\+AT)X(33)(34)(35)(36)(37)  (33) d X T d X = d X d X T = E (34) d ( A X ) T d X = A T (35) d ( A X ) d X T = A (36) d ( X T X ) d X = 2 X (37) d ( X T A X ) d X = ( A + A T ) X

### Gradient: ###

假设有函数 f:Rm×n→R  f : R m × n → R  ,将 m×n  m × n  的矩阵映射到实数 R  R  空间去，则我们定义 f f 关于矩阵 A  A  的导数为：

∇Af(A)=⎡⎣⎢⎢⎢⎢∂f∂a11⋮∂f∂an1⋯⋱⋯∂f∂a1n⋮∂f∂ann⎤⎦⎥⎥⎥⎥ ∇Af ( A ) =\[∂ f∂a11⋯∂ f∂a1 n⋮⋱⋮∂ f∂an 1⋯∂ f∂an n\]

∇AtrAB∇A|A|∇ATf(A)∇AtrABATC====BT|A|(A−1)T(∇Af(A))TCAB\+CTABT(38)(39)(40)(41)  (38) ∇ A t r A B = B T (39) ∇ A | A | = | A | ( A − 1 ) T (40) ∇ A T f ( A ) = ( ∇ A f ( A ) ) T (41) ∇ A t r A B A T C = C A B + C T A B T

## 公式证明 ##

(1) trAB=trBA  t r A B = t r B A

假设

A=⎡⎣⎢⎢a11⋮an1…⋱…a1n⋮ann⎤⎦⎥⎥B=⎡⎣⎢⎢b11⋮bn1…⋱…b1n⋮bnn⎤⎦⎥⎥  A = \[ a 11 … a 1 n ⋮ ⋱ ⋮ a n 1 … a n n \] B = \[ b 11 … b 1 n ⋮ ⋱ ⋮ b n 1 … b n n \]

可得

trABtrBA==∑i=1n∑j=1naijbji∑j=1n∑i=1naijbji(96)(97)  (96) t r A B = ∑ i = 1 n ∑ j = 1 n a i j b j i (97) t r B A = ∑ j = 1 n ∑ i = 1 n a i j b j i

同理得式（6）、（7）

(2) ∇AtrAB=BT  ∇ A t r A B = B T

又因为trAB=∑i=1n∑j=1naijbji  t r A B = ∑ i = 1 n ∑ j = 1 n a i j b j i ，对于A 中的每一个aij  a i j  都有：

dtrABdaij=bji(98)  (98) d t r A B d a i j = b j i

又因为  B  B  由式（18）定义，所以  ∇AtrAB=BT ∇At r A B =B T

(3) ∇A|A|=|A|(A−1)T  ∇ A | A | = | A | ( A − 1 ) T

由行列式的性质得：|A|=∑jaijAij  | A | = ∑ j a i j A i j  ，其中 Aij  A i j  矩阵 (i,j)  ( i , j ) 处的代数余子式。所以：

∂|A|∂aij⟹∇A|A|==AijAij=(A∗)T=(|A|A−1)T=|A|(A−1)T(99)(100)  (99) ∂ | A | ∂ a i j = A i j (100) ⟹ ∇ A | A | = A i j = ( A ∗ ) T = ( | A | A − 1 ) T = | A | ( A − 1 ) T

(4)  ∇ATf(A)=(∇Af(A))T  ∇ A T f ( A ) = ( ∇ A f ( A ) ) T

等号左边对于A 中的每一个aij  a i j  都有：df(A)daji  d f ( A ) d a j i

等号右边对于A 中的每一个aij  a i j  可表示为：df(A)daij  d f ( A ) d a i j

可以发现他们正好是转置的关系。

(5) ∇AtrABATC=CAB\+CTABT  ∇ A t r A B A T C = C A B + C T A B T

∇AtrABATC====∇AtrA(BATC)\+∇Atr(CAB)AT(BATC)T\+∇Atr\[(CAB)AT\]TCTABT\+∇AtrA(CAB)CTABT\+CAB(101)(102)(103)(104)  (101) ∇ A t r A B A T C = ∇ A t r A ( B A T C ) + ∇ A t r ( C A B ) A T (102) = ( B A T C ) T + ∇ A t r \[ ( C A B ) A T \] T (103) = C T A B T + ∇ A t r A ( C A B ) (104) = C T A B T + C A B

[Derivative]: https://blog.csdn.net/lirika_777/article/details/79646453