matrix calculus - CCRMA

Appendix D

Matrix calculus From too much study, and from extreme passion, cometh madnesse. − Isaac Newton [168, §5]

D.1 D.1.1

Directional derivative, Taylor series Gradients

Gradient of a differentiable real function f (x) : RK → R with respect to its vector argument is defined uniquely in terms of partial derivatives  ∂f (x)  ∂x1

   ∇f (x) ,   

∂f (x) ∂x2

.. .

∂f (x) ∂xK

    ∈ RK  

(1860)

while the second-order gradient of the twice differentiable real function with respect to its vector argument is traditionally called the Hessian ;  ∂ 2f (x)  ∂ 2f (x) ∂ 2f (x) · · · ∂x ∂x1 ∂x2 ∂x21 1 ∂xK    ∂ 2f (x)  ∂ 2f (x) ∂ 2f (x)   · · · 2 K 2  ∂x ∂x ∂x ∂x ∂x 2 1 2 K  ∈ S ∇ f (x) ,  (1861)  .. .. 2 .. ..   . . . .  2  ∂ 2f (x) ∂ 2f (x) ∂ f (x) · · · 2 ∂xK ∂x1 ∂xK ∂x2 ∂x K

The gradient of vector-valued function v(x) : R → RN on real domain is a row-vector i h ∂v2 (x) ∂vN (x) 1 (x) ∈ RN (1862) ∇v(x) , ∂v∂x · · · ∂x ∂x Dattorro, Convex Optimization Euclidean Distance Geometry 2ε, Mεβoo, v2015.07.21.

577

578

APPENDIX D. MATRIX CALCULUS

while the second-order gradient is h 2 v1 (x) ∇ 2 v(x) , ∂ ∂x 2

∂ 2 v2 (x) ∂x2

···

i

∂ 2 vN (x) ∂x2

∈ RN

(1863)

Gradient of vector-valued function h(x) : RK → RN on vector domain is 

   ∇h(x) ,   

∂h1 (x) ∂x1

∂h2 (x) ∂x1

···

∂hN (x) ∂x1

∂h1 (x) ∂x2

∂h2 (x) ∂x2

···

∂hN (x) ∂x2

∂h1 (x) ∂xK

∂h2 (x) ∂xK

···

∂hN (x) ∂xK

.. .

.. .

.. .

      

(1864)

= [ ∇h1 (x) ∇h2 (x) · · · ∇hN (x) ] ∈ RK×N while the second-order gradient has a three-dimensional written representation dubbed cubix ;D.1  ∂h1 (x) ∂hN (x)  2 (x) ∇ ∂x1 ∇ ∂h∂x · · · ∇ ∂x 1 1    ∇ ∂h1 (x) ∇ ∂h2 (x) · · · ∇ ∂hN (x)    2 ∂x ∂x ∂x 2 2 2 ∇ h(x) ,   .. .. ..   (1865)   . . . ∂hN (x) ∂h2 (x) ∂h1 (x) ∇ ∂xK · · · ∇ ∂xK ∇ ∂xK =

£

¤ ∇ 2 h1 (x) ∇ 2 h2 (x) · · · ∇ 2 hN (x) ∈ RK×N ×K

where the gradient of each real entry is with respect to vector x as in (1860). The gradient of real function g(X) : RK×L→ R on matrix domain is  ∂g(X) ∂g(X)  · · · ∂g(X) ∂X11 ∂X12 ∂X1L    ∂g(X) ∂g(X) ∂g(X)  · · ·  ∂X21 ∂X22 ∂X2L  ∇g(X) ,   ∈ RK×L .. .. ..   . . .   ∂g(X) ∂XK1

=

£

∂g(X) ∂XK2

···

∂g(X) ∂XKL

(1866)

∇X(:,1) g(X) ∇X(:,2) g(X) .. .

∈ RK×1×L ∇X(:,L) g(X)

¤

where gradient ∇X(:, i) is with respect to the i th column of X . The strange appearance of (1866) in RK×1×L is meant to suggest a third dimension perpendicular to the page (not D.1 The word matrix comes from the Latin for womb ; related to the prefix matri- derived from mater meaning mother.

D.1. DIRECTIONAL DERIVATIVE, TAYLOR SERIES

579

a diagonal matrix). The second-order gradient has representation 

∇ ∂g(X) ∂X11

  ∂g(X)  ∇ ∂X21 ∇ g(X) ,  ..  .  ∂g(X) ∇ ∂XK1 2

=

£

∇ ∂g(X) ∂X12

···

∇ ∂g(X) ∂X1L

∇ ∂g(X) ∂X22 .. .

···

∇ ∂g(X) ∂X2L .. .

∇ ∂g(X) ∂XK2

···

∇ ∂g(X) ∂XKL



    ∈ RK×L×K×L  

(1867)

∇∇X(:,1) g(X) ∇∇X(:,2) g(X) .. .

∈ RK×1×L×K×L

∇∇X(:,L) g(X)

¤

where the gradient ∇ is with respect to matrix X . Gradient of vector-valued function g(X) : RK×L→ RN on matrix domain is a cubix

∇g(X) ,

£

∇X(:,1) g1 (X) ∇X(:,1) g2 (X) · · · ∇X(:,1) gN (X) ∇X(:,2) g1 (X) ∇X(:,2) g2 (X) · · · ∇X(:,2) gN (X) .. .. .. . . . ∇X(:,L) g1 (X) ∇X(:,L) g2 (X) · · · ∇X(:,L) gN (X)

¤

(1868)

= [ ∇g1 (X) ∇g2 (X) · · · ∇gN (X) ] ∈ RK×N ×L while the second-order gradient has a five-dimensional representation;

∇ 2 g(X) ,

£

∇∇X(:,1) g1 (X) ∇∇X(:,1) g2 (X) · · · ∇∇X(:,1) gN (X) ∇∇X(:,2) g1 (X) ∇∇X(:,2) g2 (X) · · · ∇∇X(:,2) gN (X) .. .. .. . . . ¤ ∇∇X(:,L) g1 (X) ∇∇X(:,L) g2 (X) · · · ∇∇X(:,L) gN (X)

(1869)

£ ¤ = ∇ 2 g1 (X) ∇ 2 g2 (X) · · · ∇ 2 gN (X) ∈ RK×N ×L×K×L

The gradient of matrix-valued function g(X) : RK×L→ RM ×N on matrix domain has a four-dimensional representation called quartix (fourth-order tensor ) ∇g11 (X)

∇g12 (X)

···

∇g1N (X)

  ∇g21 (X) ∇g(X) ,  ..  . ∇gM 1 (X)

∇g22 (X) .. .

···

∇gM 2 (X)

···

 ∇g2N (X)   ∈ RM ×N ×K×L ..  .



∇gMN (X)



(1870)

580


while the second-order gradient has a six-dimensional representation   ∇ 2 g11 (X) ∇ 2 g12 (X) · · · ∇ 2 g1N (X)    ∇ 2 g21 (X) ∇ 2 g22 (X) · · · ∇ 2 g2N (X)  ∇ 2 g(X) ,   ∈ RM ×N ×K×L×K×L (1871) .. .. ..   . . . 2 2 2 ∇ gM 1 (X) ∇ gM 2 (X) · · · ∇ gMN (X) and so on.

D.1.2

Product rules for matrix-functions

Given dimensionally compatible matrix-valued functions of matrix variable f (X) and g(X) ¡ ¢ ∇X f (X)T g(X) = ∇X (f ) g + ∇X (g) f (1872) while [54, §8.3] [333]

³ ¡ ¡ ¢ ¢ ¡ ¢´¯¯ ∇X tr f (X)T g(X) = ∇X tr f (X)T g(Z ) + tr g(X) f (Z )T ¯

Z←X

(1873)

These expressions implicitly apply as well to scalar-, vector-, or matrix-valued functions of scalar, vector, or matrix arguments. D.1.2.0.1 Example. Cubix. Suppose f (X) : R2×2→ R2 = X Ta and g(X) : R2×2→ R2 = Xb . We wish to find ¡ ¢ ∇X f (X)T g(X) = ∇X aTX 2 b (1874) using the product rule. Formula (1872) calls for

∇X aTX 2 b = ∇X (X Ta) Xb + ∇X (Xb) X Ta

(1875)

Consider the first of the two terms: ∇X (f ) g = £∇X (X Ta) Xb ¤ = ∇(X Ta)1 ∇(X Ta)2 Xb

(1876)

The gradient of X Ta forms a cubix in R2×2×2 ; a.k.a, third-order tensor.  

        ∇X (X Ta) Xb =        

∂(X Ta)1 ∂X11

JJ JJ JJ

∂(X Ta)2 ∂X11

∂(X Ta)1 ∂X12

∂(X Ta)1 ∂X21

JJ JJ JJ

JJ JJ JJ

∂(X Ta)2 ∂X12

∂(X Ta)2 ∂X21 ∂(X Ta)1 ∂X22

JJ JJ JJ

∂(X Ta)2 ∂X22

(1877)        (Xb)1     ∈ R2×1×2     (Xb)  2    


581

Because gradient of the product (1874) requires total change with respect to change in each entry of matrix X , the Xb vector must make an inner product with each vector in that second dimension of the cubix indicated by dotted line segments; 

a1

0 0

 ∇X (X Ta) Xb =   a 2 =

·



a1    0 a2

0

·

b1 X11 + b2 X12 b1 X21 + b2 X22

¸

∈ R2×1×2 (1878)

a1 (b1 X11 + b2 X12 ) a1 (b1 X21 + b2 X22 ) a2 (b1 X11 + b2 X12 ) a2 (b1 X21 + b2 X22 )

= abTX T

¸

∈ R2×2

where the cubix appears as a complete 2 × 2 × 2 matrix. In like manner for the second term ∇X (g) f 

b1

 ∇X (Xb) X Ta =   0

0 b2 b1 0

= X TabT ∈ R2×2



0   

b2

·

X11 a1 + X21 a2 X12 a1 + X22 a2

¸

∈ R2×1×2

(1879)

The solution ∇X aTX 2 b = abTX T + X TabT

(1880)

can be found from Table D.2.1 or verified using (1873). D.1.2.1

2

Kronecker product

A partial remedy for venturing into hyperdimensional matrix representations, such as the cubix or quartix, is to first vectorize matrices as in (37). This device gives rise to the Kronecker product of matrices ⊗ ; a.k.a, tensor product. Although it sees reversal in the literature, [344, §2.1] we adopt the definition: for A ∈ Rm×n and B ∈ Rp×q   B11 A B12 A · · · B1q A  B21 A B22 A · · · B2q A    (1881) B ⊗A ,  ∈ Rpm×q n .. .. ..   . . . Bp1 A Bp2 A · · ·

Bpq A

for which A ⊗ 1 = 1 ⊗ A = A (real unity acts like Identity). One advantage to vectorization is existence of the traditional two-dimensional matrix representation (second-order tensor ) for the second-order gradient of a real function with respect to a vectorized matrix. For example, from §A.1.1 no.33 (§D.2.1) for square A , B ∈ Rn×n [182, §5.2] [13, §3] 2

2 2 n T T T T T ∇vec X tr(AXBX ) = ∇vec X vec(X) (B ⊗A) vec X = B ⊗A + B ⊗A ∈ R

×n2

(1882)

582


To disadvantage is a large new but known set of algebraic rules (§A.1.1) and the fact that its mere use does not generally guarantee two-dimensional matrix representation of gradients. Another application of the Kronecker product is to reverse order of appearance in a matrix product: Suppose we wish to weight the columns of a matrix S ∈ RM ×N , for example, by respective entries wi from the main diagonal in   w1 0 ..  ∈ SN W , (1883) . T 0 wN

A conventional means for accomplishing column weighting is to multiply S by diagonal matrix W on the right-hand side:   w1 0 £ ¤ ..  = S(: , 1)w1 · · · S(: , N )wN ∈ RM ×N S W = S (1884) . 0T wN

To reverse product order such that diagonal matrix W instead appears to the left of S : for I ∈ SM (Law)   S(: , 1) 0 0 .   0 S(: , 2) . .   T (1885) S W = (δ(W ) ⊗ I )  ∈ RM ×N . . .. ..   0 0 0 S(: , N ) To instead weight the rows of S via  S(1 , :) 0  0 S(2 , :)  WS =  . .  . 0

diagonal matrix W ∈ SM , for I ∈ SN  0 ..  .  (δ(W ) ⊗ I ) ∈ RM ×N ..  . 0 0 S(M , :)

For any matrices of like size, S , Y ∈ RM ×N  S(: , 1) 0 £ ¤ 0 S(: , 2)  S ◦ Y = δ(Y (: , 1)) · · · δ(Y (: , N ))  ..  . 0

.. ..

0 .

. 0

0 S(: , N )

(1886)



   ∈ RM ×N (1887) 

which converts a Hadamard product into a standard matrix product. In the special case that S = s and Y = y are vectors in RM s ◦ y = δ(s)y

(1888)

sT ⊗ y = ysT s ⊗ y T = sy T

(1889)


D.1.3

583

Chain rules for composite matrix-functions

Given dimensionally compatible matrix-valued functions of matrix variable f (X) and g(X) [235, §15.7] ¡ ¢ ∇X g f (X)T = ∇X f T ∇f g (1890) ¡ ¢ ¡ ¢ ∇X2 g f (X)T = ∇X ∇X f T ∇f g = ∇X2 f ∇f g + ∇X f T ∇f2 g ∇X f (1891) D.1.3.1

D.1.3.1.1

Two arguments

¡ ¢ ∇X g f (X)T , h(X)T = ∇X f T ∇f g + ∇X hT ∇h g

Example.

Chain rule for two arguments.

¡ ¢ T g f (x)T , h(x)T = (f (x) + h(x)) A(f (x) + h(x)) · ¸ · ¸ x1 εx1 f (x) = , h(x) = εx2 x2 ¡ ¢ ∇x g f (x)T , h(x)T =

·

1 0

¡ ¢ ∇x g f (x)T , h(x)T =

·

0 ε

¸ · ε T (A + A )(f + h) + 0

1+ε 0

0 1+ε

0 1

¸ (A + AT )(f + h)

¸ µ· ¸ · ¸¶ x1 εx1 T (A + A ) + εx2 x2

¡ ¢ lim ∇x g f (x)T , h(x)T = (A + AT )x

ε→0

from Table D.2.1.

(1892) [43, §1.1] (1893) (1894)

(1895)

(1896)

(1897) 2

These foregoing formulae remain correct when gradient produces hyperdimensional representation:

D.1.4

First directional derivative

Assume that a differentiable function g(X) : RK×L→ RM ×N has continuous first- and second-order gradients ∇g and ∇ 2 g over dom g which is an open set. We seek simple expressions for the first and second directional derivatives in direction Y ∈ RK×L : →Y

→Y

respectively, dg ∈ RM ×N and dg 2 ∈ RM ×N .

Assuming that the limit exists, we may state the partial derivative of the mn th entry of g with respect to the kl th entry of X ; gmn (X + ∆t ek eT ∂gmn (X) l ) − gmn (X) = lim ∈R ∆t→0 ∂Xkl ∆t

(1898)

584


where ek is the k th standard basis vector in RK while el is the l th standard basis vector in RL . The total number of partial derivatives equals KLM N while the gradient is defined in their terms; the mn th entry of the gradient is   ∂gmn (X) ∂gmn (X) ∂gmn (X) · · · ∂X12 ∂X1L  ∂X11   ∂gmn (X) ∂gmn (X)  ∂g mn (X)   · · · ∂X ∂X ∂X   ∈ RK×L 21 22 2L ∇gmn (X) =  (1899)  .. .. ..   . . .   ∂gmn (X) ∂gmn (X) mn (X) · · · ∂g∂X ∂XK1 ∂XK2 KL while the gradient is a quartix 

∇g11 (X)

∇g12 (X)

···

 ∇g (X) ∇g (X) 21 22  ∇g(X) =  .. ..  . . ∇gM 1 (X) ∇gM 2 (X)

∇g1N (X)

··· ···

∇g2N (X) .. .

∇gMN (X)



   ∈ RM ×N ×K×L 

(1900)

By simply rotating our perspective of a four-dimensional representation of gradient matrix, we find one of three useful transpositions of this quartix (connoted T1 ):   ∂g(X) ∂g(X) · · · ∂g(X) ∂X11 ∂X12 ∂X1L    ∂g(X) ∂g(X) ∂g(X)  · · ·  ∂X22 ∂X2L  (1901) ∇g(X)T1 =  ∂X. 21  ∈ RK×L×M ×N . ..   .. .. .   ∂g(X) ∂g(X) · · · ∂g(X) ∂XK1 ∂XK2 ∂XKL

When the limit for ∆t ∈ R exists, it is easy to show by substitution of variables in (1898) gmn (X + ∆t Ykl ek eT ∂gmn (X) l ) − gmn (X) Ykl = lim ∈R (1902) ∆t→0 ∂Xkl ∆t which may be interpreted as the change in gmn at X when the change in Xkl is equal to Ykl , the kl th entry of any Y ∈ RK×L . Because the total change in gmn (X) due to Y is the sum of change with respect to each and every Xkl , the mn th entry of the directional derivative is the corresponding total differential [235, §15.8] dgmn (X)|dX→Y =

X ∂gmn (X) ∂Xkl

k,l

=

X k,l

¡ ¢ Ykl = tr ∇gmn (X)T Y

gmn (X + ∆t Ykl ek eT l ) − gmn (X) ∆t→0 ∆t lim

gmn (X + ∆t Y ) − gmn (X) lim ∆t ¯ ¯ d¯ = gmn (X + t Y ) dt ¯t=0 =

∆t→0

(1903) (1904) (1905) (1906)


585

where t ∈ R . Assuming finite Y , equation (1905) is called the Gˆ ateaux differential [42, App.A.5] [215, §D.2.1] [377, §5.28] whose existence is implied by existence of the Fréchet differential (the sum in (1903)). [266, §7.2] Each may be understood as the change in gmn at X when the change in X is equal in magnitude and direction to Y .D.2 Hence the directional derivative,  ¯ ¯ dg11 (X) dg12 (X) · · · dg1N (X) ¯ ¯   →Y  dg21 (X) dg22 (X) · · · dg2N (X)  ¯ dg (X) ,  ¯ ∈ RM ×N  .. .. ..  ¯ . . . ¯ dg (X) dg (X) · · · dg (X) ¯ M1



  =  

M2

MN

¡ ¢ tr ∇g11 (X)T Y ¡ ¢ tr ∇g21 (X)T Y .. . ¡ ¢ tr ∇gM 1 (X)T Y

 P ∂g11 (X) ∂Xkl

Ykl

 k,l  P ∂g (X) 21  Ykl  =  k,l ∂X. kl  ..   P ∂gM 1 (X) ∂Xkl Ykl k,l

from which it follows

¡ ¢ tr ∇g12 (X)T Y ¡ ¢ tr ∇g22 (X)T Y .. . ¡ ¢ tr ∇gM 2 (X)T Y

P ∂g12 (X) k,l

∂Xkl

P ∂g22 (X) k,l

∂Xkl

Ykl Ykl

.. .

P ∂gM 2 (X) k,l

→Y

∂Xkl

dg (X) =

Ykl

··· ···

···

dX→Y

··· ··· ···

P ∂g1N (X) k,l

∂Xkl

∂Xkl

P ∂g2N (X) k,l

∂Xkl

Ykl Ykl

.. .

P ∂gMN (X) k,l

X ∂g(X) k,l

¡ ¢ tr ∇g1N (X)T Y ¡ ¢ tr ∇g2N (X)T Y .. . ¡ ¢ tr ∇gMN (X)T Y

∂Xkl

Ykl

    

(1907)

        

Ykl

(1908)

Yet for all X ∈ dom g , any Y ∈ RK×L , and some open interval of t ∈ R →Y

g(X + t Y ) = g(X) + t dg (X) + o(t2 )

(1909)

which is the first-order Taylor series expansion about X . [235, §18.4] [166, §2.3.4] Differentiation with respect to t and subsequent t-zeroing isolates the second term of expansion. Thus differentiating and zeroing g(X + t Y ) in t is an operation equivalent to individually differentiating and zeroing every entry gmn (X + t Y ) as in (1906). So the directional derivative of g(X) : RK×L→ RM ×N in any direction Y ∈ RK×L evaluated at X ∈ dom g becomes ¯ →Y d ¯¯ g(X + t Y ) ∈ RM ×N (1910) dg (X) = dt ¯t=0 [294, §2.1, §5.4.5] [35, §6.3.1] which is simplest. In case of a real function g(X) : RK×L→ R →Y ¡ ¢ dg (X) = tr ∇g(X)T Y (1932) D.2 Although

Y is a matrix, we may regard it as a vector in RKL .

586

APPENDIX D. MATRIX CALCULUS ✡T ✡ ✡ ✡ ✡ (α , f (α))✡ ✡ ✡ ✡ ✡ υ

f (α + t y) 

 υ ,

∇x f (α) →∇x f (α) 1 2 df(α)

  

f (x)

∂H

Figure 174: Strictly convex quadratic bowl in R2 × R ; f (x) = xTx : R2 → R versus x on some open disc in R2 . Plane slice ∂H is perpendicular to function domain. Slice intersection with domain connotes bidirectional vector y . Slope of tangent line T at point (α , f (α)) is value of directional derivative ∇x f (α)Ty (1935) at α in slice direction y . Negative gradient −∇x f (x) ∈ R2 is direction of steepest descent. [414] [235, §15.6] [166] When·vector¸ υ ∈ R3 entry υ3 is half directional derivative in gradient direction at α and υ1 when = ∇x f (α) , then −υ points directly toward bowl bottom. υ2 In case g(X) : RK → R

→Y

dg (X) = ∇g(X)T Y

(1935)

Unlike gradient, directional derivative does not expand dimension; directional derivative (1910) retains the dimensions of g . The derivative with respect to t makes the directional derivative resemble ordinary calculus (§D.2); e.g, when g(X) is linear, →Y

dg (X) = g(Y ). [266, §7.2]

D.1.4.1

Interpretation of directional derivative

In the case of any differentiable real function g(X) : RK×L→ R , the directional derivative of g(X) at X in any direction Y yields the slope of g along the line {X + t Y | t ∈ R} through its domain evaluated at t = 0. For higher-dimensional functions, by (1907), this slope interpretation can be applied to each entry of the directional derivative. Figure 174, for example, shows a plane slice of a real convex bowl-shaped function f (x) along a line {α + t y | t ∈ R} through its domain. The slice reveals a one-dimensional real function of t ; f (α + t y). The directional derivative at x = α in direction y is the slope of f (α + t y) with respect to t at t = 0. In the case of a real function having vector argument h(X) : RK → R , its directional derivative in the normalized direction


587

of its gradient is the gradient magnitude. (1935) For a real function of real variable, the directional derivative evaluated at any point in the function domain is just the slope of that function there scaled by the real direction. (confer §3.6) Directional derivative generalizes our one-dimensional notion of derivative to a multidimensional domain. When direction Y coincides with a member of the standard Cartesian basis ek eT l (60), then a single partial derivative ∂g(X)/∂Xkl is obtained from directional derivative (1908); such is each entry of gradient ∇g(X) in equalities (1932) and (1935), for example. D.1.4.1.1 Theorem. Directional derivative optimality condition. [266, §7.4] Suppose f (X) : RK×L→ R is minimized on convex set C ⊆ RK×L by X ⋆ , and the directional derivative of f exists there. Then for all X ∈ C →X−X ⋆

df (X) ≥ 0

(1911) ⋄

D.1.4.1.2 Example. Simple bowl. Bowl function (Figure 174) f (x) : RK → R , (x − a)T (x − a) − b has function offset (1861) everywhere quadratic f (x) has anywhere in dom f

(1912)

−b ∈ R , axis of revolution at x = a , and positive definite Hessian in its domain (an open hyperdisc in RK ); id est, strictly convex unique global minimum equal to −b at x = a . A vector −υ based × R pointing toward the unique bowl-bottom is specified: υ ∝

·

x−a f (x) + b

¸

∈ RK × R

(1913)

Such a vector is 

since the gradient is

 υ=

∇x f (x) →∇x f (x) 1 2 df(x)

  

∇x f (x) = 2(x − a)

(1914)

(1915)

and the directional derivative in direction of the gradient is (1935) →∇x f (x)

df(x) = ∇x f (x)T ∇x f (x) = 4(x − a)T (x − a) = 4(f (x) + b)

(1916) 2

588


D.1.5

Second directional derivative

By similar argument, it so happens: the second directional derivative is equally simple. Given g(X) : RK×L→ RM ×N on open domain, 

   ∂∇gmn (X) ∂gmn (X) = = ∇  ∂Xkl ∂Xkl   

∂ 2gmn (X) ∂Xkl ∂X11




∂ 2gmn (X) ∂Xkl ∂XK1

∂ 2gmn (X) ∂Xkl ∂XK2

.. .

mn (X) ∇ ∂g∂X 11

  ∂gmn (X)  ∇ ∂X21 ∇ 2 gmn (X) =   ..  .  mn (X) ∇ ∂g∂X K1 

   =  

.. .

∂ 2gmn (X) ∂Xkl ∂XKL

···

···

mn (X) ∇ ∂g∂X 1L

mn (X) ∇ ∂g∂X 22

···

mn (X) ∇ ∂g∂X 2L

mn (X) ∇ ∂g∂X K2

···

mn (X) ∇ ∂g∂X KL

.. .

∂∇gmn (X) ∂X11

∂∇gmn (X) ∂X12

∂∇gmn (X) ∂X21

∂∇gmn (X) ∂X22

∂∇gmn (X) ∂XK1

∂∇gmn (X) ∂XK2

.. .

∂ 2gmn (X) ∂Xkl ∂X2L

···

.. .

mn (X) ∇ ∂g∂X 12

∂ 2gmn (X) ∂Xkl ∂X1L

···

.. .

···

.. .

∂∇gmn (X) ∂X1L

···

∂∇gmn (X) ∂X2L

···

∂∇gmn (X) ∂XKL

.. .



    ∈ RK×L   

(1917)



    ∈ RK×L×K×L   

(1918)

      

Rotating our perspective, we get several views of the second-order gradient: 

∇ 2 g11 (X)

 ∇ 2 g (X) 21  ∇ 2 g(X) =  ..  .

∇ 2 gM 1 (X)

2

T1

∇ g(X)

2

T2

∇ g(X)



∇ 2 g12 (X)

∇ 2 g22 (X) .. .

   =  

···

∇ 2 gM 2 (X) ∇ ∂g(X) ∂X11

  ∂g(X)  ∇ ∂X21 = ..  .  ∇ ∂g(X) ∂XK1 

···

···

∇ ∂g(X) ∂X12 ∇ ∂g(X) ∂X22 .. .

∇ 2 g1N (X)

∇ 2 g2N (X) .. .

∇ 2 gMN (X) ··· ···

∇ ∂g(X) ∂XK2

···

∂∇g(X) ∂X11

∂∇g(X) ∂X12

∂∇g(X) ∂X21

∂∇g(X) ∂X22

···

∂∇g(X) ∂XK1

∂∇g(X) ∂XK2

.. .

.. .

∇ ∂g(X) ∂X1L



   ∈ RM ×N ×K×L×K×L  

  ∇ ∂g(X) ∂X2L   ∈ RK×L×M ×N ×K×L ..  .  ∂g(X) ∇ ∂XKL ∂∇g(X) ∂X1L

···

∂∇g(X) ∂X2L

···

∂∇g(X) ∂XKL

.. .



    ∈ RK×L×K×L×M ×N  

(1919)

(1920)

(1921)


589

Assuming the limits exist, we may state the partial derivative of the mn th entry of g with respect to the kl th and ij th entries of X ; ∂ 2gmn (X) ∂Xkl ∂Xij

=

T T T gmn (X+∆t ek eT l +∆τ ei ej )−gmn (X+∆t ek el )−(gmn (X+∆τ ei ej )−gmn (X)) ∆τ ∆t ∆τ,∆t→0

lim

(1922)

Differentiating (1902) and then scaling by Yij ∂ 2gmn (X) ∂Xkl ∂Xij Ykl Yij

=

∂gmn (X+∆t Ykl ek eT l )−∂gmn (X) Yij ∂X ∆t ij ∆t→0

= lim

(1923)

T T T gmn (X+∆t Ykl ek eT l +∆τ Yij ei ej )−gmn (X+∆t Ykl ek el )−(gmn (X+∆τ Yij ei ej )−gmn (X)) ∆τ ∆t ∆τ,∆t→0

lim

which can be proved by substitution of variables in (1922). The mn th second-order total differential due to any Y ∈ RK×L is d 2gmn (X)|dX→Y =

³ X X ∂ 2gmn (X) ¡ ¢T ´ Ykl Yij = tr ∇X tr ∇gmn (X)T Y Y ∂Xkl ∂Xij i,j

(1924)

k,l

=

X i,j

= =

lim

∆t→0

lim

∂gmn (X + ∆t Y ) − ∂gmn (X) Yij ∂Xij ∆t

(1925)

gmn (X + 2∆t Y ) − 2gmn (X + ∆t Y ) + gmn (X) ∆t2

(1926)

∆t→0 ¯ 2 ¯

d ¯ gmn (X + t Y ) dt2 ¯t=0

Hence the second directional derivative,  2 d g11 (X) d 2g12 (X)  d 2g21 (X) d 2g22 (X) →Y  dg 2(X) ,  .. ..  . . d 2gM 1 (X)

d 2gM 2 (X)

(1927)

··· ··· ···

d 2g1N (X) d 2g2N (X) .. .

¯ ¯ ¯ ¯ ¯  ¯ ∈ RM ×N ¯ ¯ 2 d gMN (X) ¯dX→Y

³ ³ ³ ¡ ¢T ´  ¡ ¢T ´ ¡ ¢T ´ tr ∇tr ∇g12 (X)T Y Y · · · tr ∇tr ∇g1N (X)T Y Y tr ∇tr ∇g11 (X)T Y Y  ³ ³ ³ ¡ ¢T ´ ¡ ¢T ´ ¡ ¢T ´     tr ∇tr ∇g21 (X)T Y Y tr ∇tr ∇g22 (X)T Y Y · · · tr ∇tr ∇g2N (X)T Y Y   =   .. .. ..   . . .  ³ ³ ¡ ¢T ´ ³ ¡ ¢T ´ ¡ ¢T ´ T T T tr ∇tr ∇gM 1 (X) Y Y tr ∇tr ∇gM 2 (X) Y Y · · · tr ∇tr ∇gMN (X) Y Y 

 PP 2 ∂ g11 (X) Ykl Yij  i,j k,l ∂Xkl ∂Xij   P P ∂ 2g21 (X)  ∂Xkl ∂Xij Ykl Yij =  i,j k,l ..  .   P P ∂ 2gM 1 (X) ∂Xkl ∂Xij Ykl Yij i,j k,l

PP i,j k,l

PP i,j k,l

PP i,j k,l

∂ 2g12 (X) ∂Xkl ∂Xij Ykl Yij 2

∂ g22 (X) ∂Xkl ∂Xij Ykl Yij

.. .

∂ 2gM 2 (X) ∂Xkl ∂Xij Ykl Yij

··· ··· ···

PP i,j k,l

PP i,j k,l

∂ 2g1N (X) ∂Xkl ∂Xij Ykl Yij 2

∂ g2N (X) ∂Xkl ∂Xij Ykl Yij

.. .

P P ∂ 2gMN (X) i,j k,l

∂Xkl ∂Xij

Ykl Yij

         

(1928)

590


from which it follows →Y 2

dg (X) =

X X ∂ 2g(X) X ∂ →Y Ykl Yij = dg (X) Yij ∂Xkl ∂Xij ∂Xij i,j i,j

(1929)

k,l

Yet for all X ∈ dom g , any Y ∈ RK×L , and some open interval of t ∈ R →Y

g(X + t Y ) = g(X) + t dg (X) +

1 2 →Y2 t dg (X) + o(t3 ) 2!

(1930)

which is the second-order Taylor series expansion about X . [235, §18.4] [166, §2.3.4] Differentiating twice with respect to t and subsequent t-zeroing isolates the third term of the expansion. Thus differentiating and zeroing g(X + t Y ) in t is an operation equivalent to individually differentiating and zeroing every entry gmn (X + t Y ) as in (1927). So the second directional derivative of g(X) : RK×L → RM ×N becomes [294, §2.1, §5.4.5] [35, §6.3.1] ¯ →Y d 2 ¯¯ 2 (1931) dg (X) = 2 ¯ g(X + t Y ) ∈ RM ×N dt t=0

which is again simplest. (confer (1910)) Directional derivative retains the dimensions of g .

D.1.6

directional derivative expressions

In the case of a real function g(X) : RK×L→ R , all its directional derivatives are in R : →Y

¡ ¢ dg (X) = tr ∇g(X)T Y

→Y 2

¶ µ ³ →Y ¡ ¢T ´ dg (X) = tr ∇X tr ∇g(X)T Y Y = tr ∇X dg (X)T Y

Ã ! ¶ µ ³ →Y ¡ ¢T ´T 2 T T dg (X) = tr ∇X tr ∇X tr ∇g(X) Y Y Y = tr ∇X dg (X) Y

→Y 3

(1932)

(1933)

(1934)

In the case g(X) : RK → R has vector argument, they further simplify: →Y

dg (X) = ∇g(X)T Y

→Y 2

dg (X) = Y T ∇ 2 g(X)Y

(1936)

¡ ¢T dg (X) = ∇X Y T ∇ 2 g(X)Y Y

(1937)

→Y 3

and so on.

(1935)


D.1.7

591

Taylor series

Series expansions of the differentiable matrix-valued function g(X) , of matrix argument, were given earlier in (1909) and (1930). Assuming g(X) has continuous first-, second-, and third-order gradients over the open set dom g , then for X ∈ dom g and any Y ∈ RK×L the complete Taylor series is expressed on some open interval of µ ∈ R →Y

g(X + µY ) = g(X) + µ dg (X) +

1 3 →Y3 1 2 →Y2 µ dg (X) + µ dg (X) + o(µ4 ) 2! 3!

(1938)

or on some open interval of kY k2 →Y −X

g(Y ) = g(X) + dg(X) +

1 →Y2 −X 1 →Y3 −X dg (X) + dg (X) + o(kY k4 ) 2! 3!

(1939)

which are third-order expansions about X . The mean value theorem from calculus is what insures finite order of the series. [235] [43, §1.1] [42, App.A.5] [215, §0.4] These somewhat unbelievable formulae imply that a function can be determined over the whole of its domain by knowing its value and all its directional derivatives at a single point X . D.1.7.0.1 Example. Inverse-matrix function. Say g(Y ) = Y −1 . From the table on page 596, ¯ →Y d ¯¯ g(X + t Y ) = −X −1 Y X −1 dg (X) = dt ¯t=0 →Y 2

¯ d 2 ¯¯ dg (X) = 2 ¯ g(X + t Y ) = 2X −1 Y X −1 Y X −1 dt t=0 ¯ →Y d3 ¯ dg 3(X) = 3 ¯¯ g(X + t Y ) = −6X −1 Y X −1 Y X −1 Y X −1 dt t=0

(1940)

(1941) (1942)

Let’s find the Taylor series expansion of g about X = I : Since g(I ) = I , for kY k2 < 1 (µ = 1 in (1938)) g(I + Y ) = (I + Y )−1 = I − Y + Y 2 − Y 3 + . . .

(1943)

If Y is small, (I + Y )−1 ≈ I − Y .D.3 Now we find Taylor series expansion about X : g(X + Y ) = (X + Y )−1 = X −1 − X −1 Y X −1 + 2X −1 Y X −1 Y X −1 − . . . If Y is small, (X + Y )−1 ≈ X −1 − X −1 Y X −1 .

(1944) 2

D.1.7.0.2 Exercise. log det . (confer [63, p.644]) Find the first three terms of a Taylor series expansion for log det Y . Specify an open interval over which the expansion holds in vicinity of X . H D.3 Had

we instead set g(Y ) = (I + Y )−1 , then the equivalent expansion would have been about X = 0.

592

D.1.8


Correspondence of gradient to derivative

From the foregoing expressions for directional derivative, we derive a relationship between gradient with respect to matrix X and derivative with respect to real variable t : D.1.8.1

first-order

Removing evaluation at t = 0 from (1910),D.4 we find an expression for the directional derivative of g(X) in direction Y evaluated anywhere along a line {X + t Y | t ∈ R} intersecting dom g →Y d dg (X + t Y ) = g(X + t Y ) (1945) dt In the general case g(X) : RK×L→ RM ×N , from (1903) and (1906) we find ¡ ¢ d tr ∇X gmn (X + t Y )T Y = gmn (X + t Y ) dt

(1946)

which is valid at t = 0, of course, when X ∈ dom g . In the important case of a real function g(X) : RK×L→ R , from (1932) we have simply ¡ ¢ d tr ∇X g(X + t Y )T Y = g(X + t Y ) dt

(1947)

When, additionally, g(X) : RK → R has vector argument, ∇X g(X + t Y )T Y =

d g(X + t Y ) dt

(1948)

D.1.8.1.1 Example. Gradient. g(X) = wTX TXw , X ∈ RK×L , w ∈ RL . Using the tables in §D.2, ¡ ¢ ¡ ¢ tr ∇X g(X + t Y )T Y = tr 2wwT(X T + t Y T )Y T

T

T

= 2w (X Y + t Y Y )w

(1949) (1950)

Applying equivalence (1947), d T d g(X + t Y ) = w (X + t Y )T (X + t Y )w dt dt ¡ ¢ = wT X T Y + Y TX + 2t Y T Y w T

T

T

= 2w (X Y + t Y Y )w

which is the same as (1950). Hence, the equivalence is demonstrated. D.4 Justified

by replacing X with X + t Y in (1903)-(1905); beginning, dgmn (X + t Y )|dX→Y =

X ∂gmn (X + t Y ) Ykl ∂Xkl k, l

(1951) (1952) (1953)

D.1. DIRECTIONAL DERIVATIVE, TAYLOR SERIES It is easy to extract ∇g(X) from (1953) knowing only (1947): ¡ ¢ tr ∇X g(X + t Y )T Y = 2wT¡(X T Y + t Y T Y )w ¢ = 2 tr wwT(X T + t Y T )Y ¡ ¢ ¡ ¢ tr ∇X g(X)T Y = 2 tr wwTX T Y ⇔ ∇X g(X) = 2XwwT D.1.8.2

593

(1954)

2

second-order

Likewise removing the evaluation at t = 0 from (1931), →Y 2

dg (X + t Y ) =

d2 g(X + t Y ) dt2

(1955)

we can find a similar relationship between second-order gradient and second derivative: In the general case g(X) : RK×L→ RM ×N from (1924) and (1927), ³ ¡ ¢T ´ d2 tr ∇X tr ∇X gmn (X + t Y )T Y Y = 2 gmn (X + t Y ) dt

(1956)

In the case of a real function g(X) : RK×L→ R we have, of course, ³ ¡ ¢T ´ d2 tr ∇X tr ∇X g(X + t Y )T Y Y = 2 g(X + t Y ) dt

(1957)

From (1936), the simpler case, where real function g(X) : RK → R has vector argument, Y T ∇X2 g(X + t Y )Y =

d2 g(X + t Y ) dt2

(1958)

D.1.8.2.1 Example. Second-order gradient. We want to find ∇ 2 g(X) ∈ RK×K×K×K given real function g(X) = log det X having domain int SK + . From the tables in §D.2, h(X) , ∇g(X) = X −1 ∈ int SK + so ∇ 2 g(X) = ∇h(X). By (1946) and (1909), for Y ∈ SK ¯ ¡ ¢ d ¯¯ tr ∇hmn (X)T Y = hmn (X + t Y ) dt ¯t=0 µ ¯ ¶ d ¯¯ = h(X + t Y ) dt ¯ mn ¶ µ ¯t=0 d ¯¯ = (X + t Y )−1 dt ¯t=0 mn ¡ ¢ = − X −1 Y X −1 mn

(1959)

(1960) (1961) (1962) (1963)

594


K×K Setting Y to a member of {ek eT | k , l = 1 . . . K } , and employing a property (39) l ∈R of the trace function we find ¢ ¡ ¢ ¡ −1 ∇ 2 g(X)mnkl = tr ∇hmn (X)T ek eT = ∇hmn (X)kl = − X −1 ek eT (1964) l l X mn

¡ ¢ −1 ∇ 2 g(X)kl = ∇h(X)kl = − X −1 ek eT ∈ RK×K l X

(1965) 2

From all these first- and second-order expressions, we may generate new ones by evaluating both sides at arbitrary t (in some open interval) but only after the differentiation.

D.2

Tables of gradients and derivatives

Results may be numerically proven by Romberg extrapolation. [115] When proving results for symmetric matrices algebraically, it is critical to take gradients ignoring symmetry and to then substitute symmetric entries afterward. [182] [67] a , b ∈ Rn , x , y ∈ Rk , A , B ∈ Rm×n , X , Y ∈ RK×L , i, j , k , ℓ , K , L , m , n , M , N are integers, unless otherwise noted.

t , µ∈R ,

x µ means δ(δ(x)µ ) for µ ∈ R ; id est, entrywise vector exponentiation. δ is the main-diagonal linear operator (1504). x0 , 1, X 0 , I if square.  d  dx1

d dx

 ,

.. .

d dxk

 ,

→y

→y

dg(x) , dg 2(x) (directional derivatives §D.1), log x ,

ex,

|x| ,

√ sgn x , x/y (Hadamard quotient), ◦ x (entrywise square root), etcetera, are maps f : Rk → Rk that maintain dimension; e.g, (§A.1.1) d −1 x , ∇x 1T δ(x)−1 1 dx

(1966)

For A a scalar or square matrix, we have the Taylor series [80, §3.6] ∞ X 1 k e , A k! A

(1967)

k=0

Further, [348, §5.4] eA ≻ 0

∀ A ∈ Sm

(1968)

For all square A and integer k detk A = det Ak

(1969)

D.2. TABLES OF GRADIENTS AND DERIVATIVES

D.2.1

595

algebraic

∇x x = ∇x xT = I ∈ Rk×k

∇X X = ∇X X T , I ∈ RK×L×K×L

(Identity)

∇x (Ax − b) = AT ¡ ¢ ∇x xTA − bT = A

∇x (Ax − b)T(Ax − b) = 2AT(Ax − b) ∇x2 (Ax − b)T(Ax − b) = 2ATA ∇x kAx − bk = AT(Ax − b)/kAx − bk ∇x z T |Ax − b| = AT δ(z) sgn(Ax − b) , z¶ i 6= 0 ⇒ (Ax − b)i 6= 0 µ ¯ (y) ¯ sgn(Ax − b) ∇x 1T f (|Ax − b|) = AT δ dfdy ¯ y=|Ax−b|

¡ ¢ ∇x xTAx + 2xTB y + y TC y = A + AT x + 2B y T T ∇x (x ¡ + y) A(x + y) = (A +¢A )(x + y) ∇x2 xTAx + 2xTB y + y TC y = A + AT ¡

¢

∇X aTXb = ∇X bTX Ta = abT ∇X aTX 2 b = X T abT + abT X T ∇X aTX −1 b = −X −T abT X −T

∇x aTxTxb = 2xaT b

confer ∂X −1 −1 (1901) = −X −1 ek eT X , l ∂Xkl (1965) ∇X aTX TXb = X(abT + baT )

∇x aTxxT b = (abT + baT )x

∇X aTXX T b = (abT + baT )X

∇x aTxTxa = 2xaTa

∇X aTX TXa = 2XaaT

∇x aTxxTa = 2aaTx

∇X aTXX T a = 2aaT X

∇x aTyxT b = b aTy

∇X aT Y X T b = baT Y

∇x aTy Tx b = y bTa

∇X aT Y TXb = Y abT

∇x aTxy T b = a bTy

∇X aTXY T b = abT Y

∇x aTxTy b = y aT b

∇X aTX T Y b = Y baT

∇X (X −1 )kl =

596


algebraic continued

d dt (X +

tY ) = Y

d T dt B (X + d T dt B (X + d T dt B (X +

t Y )−1 A = −B T (X + t Y )−1 Y (X + t Y )−1 A t Y )−TA = −B T (X + t Y )−T Y T (X + t Y )−TA t Y )µ A = ... , −1 ≤ µ ≤ 1, X , Y ∈ SM +

d2 B T (X + dt2 3 d B T (X + dt3

t Y )−1 A =

2B T (X + t Y )−1 Y (X + t Y )−1 Y (X + t Y )−1 A

t Y )−1 A = −6B T (X + t Y )−1 Y (X + t Y )−1 Y (X + t Y )−1 Y (X + t Y )−1 A

d (X + t Y )TA(X + t Y ) = Y TAX + dt ¡ ¢ 2 d (X + t Y )TA(X + t Y ) = 2 Y TAY dt2¡ ¢−1 d T dt (X¡+ t Y ) A(X + t Y ) ¢−1 T T

¡

¢

X TAY + 2 t Y TAY

¡ ¢−1 = − (X + t Y ) A(X + t Y ) (Y AX + X TAY + 2 t Y TAY ) (X + t Y )TA(X + t Y ) d dt ((X + t Y )A(X + t Y )) = YAX + XAY + 2 t YAY d2 ((X + t Y )A(X + t Y )) = 2 YAY dt2

D.2.1.0.1 Exercise. Expand these tables. Provide unfinished table entries indicated by . . . throughout §D.2.

H

D.2.1.0.2 Exercise. log . (§D.1.7) Find the first four terms of the Taylor series expansion for log x about x = 1. Prove that log x ≤ ¸x −·1 ; alternatively, plot the supporting hyperplane to the hypograph of log x at · ¸ x 1 = . H log x 0

D.2.2

trace Kronecker

∇vec X tr(A XBX T ) = ∇vec X vec(X)T (B T ⊗ A) vec X = (B ⊗ AT + B T ⊗ A) vec X 2 2 T T T T T ∇vec X tr(A XBX ) = ∇vec X vec(X) (B ⊗ A) vec X = B ⊗ A + B ⊗ A


D.2.3

597

trace

∇x µ x = µI

∇X tr µX = ∇X µ tr X = µI

d −1 ∇x 1T δ(x)−1 1 = dx x = −x−2 T −1 ∇x 1 δ(x) y = −δ(x)−2 y

∇X tr X −1 = −X −2T ∇X tr(X −1 Y ) = ∇X tr(Y X −1 ) = −X −T Y TX −T

d µ dx x

∇X tr X µ = µX µ−1 ,

= µx µ−1

X ∈ SM

∇X tr X j = jX (j−1)T ∇x (b − aTx)−1 = (b − aTx)−2 a ∇x (b − aTx)µ = −µ(b − aTx)µ−1 a ∇x xTy = ∇x y Tx = y

¡ ¢ ¡ ¢T ∇X tr (B − AX)−1 = (B − AX)−2 A ∇X tr(X T Y ) = ∇X tr(Y X T ) = ∇X tr(Y TX) = ∇X tr(XY T ) = Y ∇X tr(AXBX T ) = ∇X tr(XBX TA) = ATXB T + AXB ∇X tr(AXBX) = ∇X tr(XBXA) = ATX TB T + B TX TAT ∇X tr(AXAXAX) = ∇X tr(XAXAXA) = 3(AXAXA )T ∇X tr(Y X k ) = ∇X tr(X k Y ) =

k−1 P¡

X i Y X k−1−i

i=0

¢T

∇X tr(Y TXX T Y ) = ∇X tr(X T Y Y TX) = 2 Y Y TX ∇X tr(Y TX TXY ) = ∇X tr(XY Y TX T ) = 2XY Y T

¡ ¢ ∇X tr (X + Y )T (X + Y ) = 2(X + Y ) = ∇X kX + Y k2F ∇X tr((X + Y )(X + Y )) = 2(X + Y )T ∇X tr(ATXB) = ∇X tr(X TAB T ) = AB T T −1 −T T −T ∇X tr(A X B) = ∇X tr(X AB ) = −X AB T X −T ∇X aTXb = ∇X tr(baTX) = ∇X tr(XbaT ) = abT ∇X bTX Ta = ∇X tr(X TabT ) = ∇X tr(abTX T ) = abT ∇X aTX −1 b = ∇X tr(X −T abT ) = −X −T abT X −T µ ∇X aTX b = ...

598


trace continued d dt

d tr g(X + t Y ) = tr dt g(X + t Y )

d dt

tr(X + t Y ) = tr Y

d dt

tr j(X + t Y ) = j tr j−1(X + t Y ) tr Y

d dt

¡ ¢ tr(X + t Y )j = j tr (X + t Y )j−1 Y

d dt

[219, p.491]

(∀ j)

tr((X + t Y )Y ) = tr Y 2

d dt

¡ ¢ tr (X + t Y )k Y =

d dt

¡ ¢ tr(Y (X + t Y )k ) = k tr (X + t Y )k−1 Y 2 ,

d dt

¡ ¢ tr (X + t Y )k Y =

d dt

tr(Y (X + t Y )k ) = tr

d dt d dt d dt d dt d dt

k−1 P

k ∈ {0, 1, 2}

(X + t Y )i Y (X + t Y )k−1−i Y

i=0

¡ ¢ ¡ ¢ tr¡(X + t Y )−1 Y ¢ = − tr¡ (X + t Y )−1 Y (X + t Y )−1 Y ¢ tr¡B T (X + t Y )−1 A¢ = − tr¡B T (X + t Y )−1 Y (X + t Y )−1 A ¢ tr¡B T (X + t Y )−TA ¢ = − tr B T (X + t Y )−T Y T (X + t Y )−TA tr B T (X + t Y )−k A = ... , k > 0 ¡ ¢ tr B T (X + t Y )µ A = ... , −1 ≤ µ ≤ 1, X , Y ∈ SM +

d2 dt2

¡ ¢ ¡ ¢ tr B T (X + t Y )−1 A = 2 tr B T (X + t Y )−1 Y (X + t Y )−1 Y (X + t Y )−1 A

d (X + t Y )TA(X + t Y ) = tr Y TAX + X TAY dt tr ¡ ¢ ¡ ¢ d2 tr (X + t Y )TA(X + t Y ) = 2 tr Y TAY dt2 ³ ¡ ¢−1 ´ d + t Y )TA(X + t Y ) dt tr (X ³¡ ¢−1 T T T

¡

¢

= − tr (X + t Y ) A(X + t Y )

¡

+ 2 t Y TAY

¢

¡ ¢−1 ´ (Y AX + X AY + 2 t Y TAY ) (X + t Y )TA(X + t Y )

d dt tr((X + t Y )A(X + t Y )) = tr(YAX + XAY d2 tr((X + t Y )A(X + t Y )) = 2 tr(YAY ) dt2

+ 2 t YAY )


D.2.4

599

logarithmic determinant

x ≻ 0, det X > 0 on some neighborhood of X , and det(X + t Y ) > 0 on some open interval of t ; otherwise, log( ) would be discontinuous. [86, p.75] d dx

log x = x−1

∇X log det X = X −T ∇X2 log det(X)kl =

¡ ¢ ∂X −T −1 T = − X −1 ek eT , confer (1918)(1965) l X ∂Xkl

d dx

log x−1 = −x−1

∇X log det X −1 = −X −T

d dx

log x µ = µx−1

∇X log detµ X = µX −T µ

∇X log det X = µX −T ∇X log det X k = ∇X log detk X = kX −T ∇X log detµ (X + t Y ) = µ(X + t Y )−T 1 ∇x log(aTx + b) = a aTx+b

∇X log det(AX + B) = AT(AX + B)−T ∇X log det(I ± ATXA) = ±A(I ± ATXA)−TAT ∇X log det(X + t Y )k = ∇X log detk (X + t Y ) = k(X + t Y )−T d dt

log det(X + t Y ) = tr ((X + t Y )−1 Y )

d2 dt2 d dt

log det(X + t Y )−1 = − tr ((X + t Y )−1 Y )

d2 dt2 d dt

log det(X + t Y ) = − tr ((X + t Y )−1 Y (X + t Y )−1 Y )

log det(X + t Y )−1 = tr ((X + t Y )−1 Y (X + t Y )−1 Y )

log det(δ(A(x + t y) + a)2 + µI) ³

= tr (δ(A(x + t y) + a)2 + µI)

−1

´ 2δ(A(x + t y) + a)δ(Ay)

600


D.2.5

determinant

∇X det X = ∇X det X T = det(X)X −T ∇X det X −1 = − det(X −1 )X −T = − det(X)−1 X −T ∇X detµ X = µ detµ (X)X −T µ

µ

∇X det X = µ det(X )X −T ¡ ¢ ∇X det X k = k detk−1(X) tr(X)I − X T ,

X ∈ R2×2

∇X det X k = ∇X detk X = k det(X k )X −T = k detk (X)X −T ∇X detµ (X + t Y ) = µ detµ (X + t Y )(X + t Y )−T ∇X det(X + t Y )k = ∇X detk (X + t Y ) = k detk (X + t Y )(X + t Y )−T d dt

det(X + t Y ) = det(X + t Y ) tr((X + t Y )−1 Y )

d2 dt2 d dt

det(X + t Y )−1 = − det(X + t Y )−1 tr((X + t Y )−1 Y )

d2 dt2 d dt

¡ ¢ det(X + t Y ) = det(X + t Y )(tr 2 (X + t Y )−1 Y − tr((X + t Y )−1 Y (X + t Y )−1 Y )) det(X + t Y )−1 = det(X + t Y )−1 (tr 2 ((X + t Y )−1 Y ) + tr((X + t Y )−1 Y (X + t Y )−1 Y ))

detµ (X + t Y ) = µ detµ (X + t Y ) tr((X + t Y )−1 Y )

D.2.6

logarithmic

Matrix logarithm. d dt log(X + d dt log(I −

t Y )µ = µY (X + t Y )−1 = µ(X + t Y )−1 Y ,

XY = YX

t Y )µ = −µY (I − t Y )−1 = −µ(I − t Y )−1 Y

[219, p.493]


D.2.7

601

exponential

Matrix exponential. [80, §3.6, §4.5] [348, §5.4] ∇X etr(Y

T

X)

= ∇X det eY

∇X tr eY X = eY

T

XT

T

X

= etr(Y

Y T = Y T eX

T

T

X)

Y

(∀ X , Y )

YT

∇x 1T eAx = ATeAx ∇x 1T e|Ax| = AT δ(sgn(Ax))e|Ax|

(Ax)i 6= 0

1 ex 1T e x µ ¶ 1 1 x x xT 2 T x ∇x log(1 e ) = T x δ(e ) − T x e e 1 e 1 e ∇x log(1T e x ) =

∇x ∇x2

k Q

i=1 k Q

µ

xik = −

1 k

1

i=1

d tY dt e

1 k

1

xik =

k Q

i=1

µ

¶ 1 xik 1/x

k Q

i=1

1

xik

¶ ¶µ 1 δ(x)−2 − (1/x)(1/x)T k

= etY Y = Y etY

d X+ t Y dt e d 2 X+ t Y e dt2

= eX+ t Y Y = Y eX+ t Y , = eX+ t Y Y 2 = Y eX+ t Y Y = Y 2 eX+ t Y ,

d j tr(X+ t Y ) e dt j

= etr(X+ t Y ) tr j(Y )

XY = YX XY = YX