Derivatives of matrix

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matrix V scalar
scalar V matrix
vector V vector
Examples
- 1 Derivative of inverse matrix
- 2 Derivative of trace
- 3 Applications
- 4 Derivative of determinant
Reference

1. matrix V scalar

\partial A ( x ) \partial x = [\partial A ( x ) i j \partial x] i, j

f ( A ) \partial A = [\partial f ( x ) \partial A i j] i, j

g \to ( V \to ) \partial V \to = ⎡ ⎣ ⎢ g \to ( V \to ) i \partial V \to j ⎤ ⎦ ⎥ i, j

\Rightarrow \Rightarrow \Rightarrow A - 1 A = I \partial A - 1 A \partial x = 0 \partial A - 1 \partial x A + A - 1 \partial A \partial x = 0 \partial A - 1 \partial x = - A - 1 \partial A \partial x A - 1

For x=Aij with A being nonsymmetric, we have

\partial A \partial A i j = I i j

and

\partial A - 1 \partial A i j = - A - 1 I i j A - 1

and

\partial 2 A - 1 \partial A i j \partial A k r = \partial \partial A - 1 \partial A k r \partial A i j = - \partial A - 1 I k r A - 1 \partial A i j = - \partial A - 1 \partial A i j I k r A - 1 - A - 1 I k r \partial A - 1 I k r A - 1 \partial A i j = - (- A - 1 I i j A - 1) I k r A - 1 - A - 1 I k r (- A - 1 I i j A - 1) I k r = A - 1 I i j A - 1 I k r A - 1 + A - 1 I k r A - 1 I i j A - 1

For x=Aij with A being symmetric, we have

\partial A \partial A i j = I i j + I j i - δ j i I j i = d I * i j

Then all can be replaced by I∗ij

If A is not symmetric:

\partial tr ( A ) \partial A = [\partial tr ( A ) \partial A i j] i j = [tr (\partial A \partial A i j)] i j = tr [I i j] i j = I n

If A is symmetric:

\partial tr ( A ) \partial A = [\partial tr ( A ) \partial A i j] i j = [tr (\partial A \partial A i j)] i j = tr [I i j + I j i + δ j i I j i] i j = I n

If S is not symmetric:

\Rightarrow \Rightarrow f = tr (S - 1 B) \partial f \partial S i j = \partial tr ( S - 1 B ) \partial S i j = tr (\partial S - 1 B S i j) = - tr (S - 1 I i j S - 1 B) = tr (I i j S - 1 B S - 1) =    Θ = S - 1 B S - 1 tr (I i j Θ) = Θ j i \partial tr ( S - 1 B ) \partial S = - (S - 1 B S - 1) T

If S is symmetric:

\Rightarrow \Rightarrow f = tr (S - 1 B) \partial f \partial S i j = \partial tr ( S - 1 B ) \partial S i j = tr (\partial S - 1 B S i j) = - tr (I * i j Θ) = - [Θ j i + Θ i j - δ i j Θ i j] \partial tr ( S - 1 B ) \partial S = - [S - 1 (B + B T) S - 1 - diag (S - 1 B S - 1)]

\partial tr ( A B ) \partial A = B T

\partial tr ( A T B ) \partial A = B

\partial tr ( A T S A ) \partial A = [\partial tr ( A T S A ) \partial A i j] i j = [tr (I j i S A + A T S I i j)] i j = S A + S A T A = (S + S T) A

S is symmetric

\partial tr ( ( A T S A ) - 1 R ) \partial A = ⎡ ⎣ ⎢ ⎢ ⎢ \partial tr ( ( A T S A ) - 1 R ) \partial A i j ⎤ ⎦ ⎥ ⎥ ⎥ i j = ⎡ ⎣ ⎢ ⎢ tr ⎛ ⎝ ⎜ ⎜ \partial ( A T S A ) - 1 R \partial A i j ⎞ ⎠ ⎟ ⎟ ⎤ ⎦ ⎥ ⎥ i j = [tr (- (A T S A) - 1 \partial A T S A \partial A i j (A T S A) - 1 R)] i j = [tr (- (A T S A) - 1 (I j i S A + A T S I i j) (A T S A) - 1 R)] i j = [- tr (I j i S A (A T S A) - 1 R (A T S A) - 1)] i j + [- tr ((A T S A) - 1 R (A T S A) - 1 A T S I i j)] i j = - S A (A T S A) - 1 R (A T S A) - 1 - S A (A T S A) - 1 R T (A T S A) - 1 = - S A (A T S A) - 1 (R + R T) (A T S A) - 1

\partial tr ( ( A T S 2 A ) - 1 ( A T S 1 A ) ) \partial A = \partial tr ( ( A T 1 S 2 A 1 ) - 1 ( A T S 1 A ) ) \partial A ∣ ∣ ∣ ∣ ∣ ∣ A 1 = A + \partial tr ( ( A T S 2 A ) - 1 ( A T 2 S 1 A 2 ) ) \partial A ∣ ∣ ∣ ∣ ∣ ∣ A 2 = A = - 2 S 2 A (A T S 2 A) - 1 (A T S 1 A) (A T S 2 A) - 1 + 2 S 1 A (A T S 2 A) - 1

\partial | R | \partial R i j = C i j

which is its cofactor(from the cofactor expansion). Then

\partial | R | \partial R = C

which is the adjoint matrix. Then from the fact that

| R | I = R C T

we have

\partial | R | \partial R = | R | R - 1 T

Furthermore

\partial ln | R | \partial R = 1 | R | \partial | R | \partial R = R - 1 T

reference of book “Introduction to Statistical Pattern Recognition, second edition.”

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