| 2.1 | d(X-1) =
-X-1dX X-1 
0 = dI = d (XX-1) =
dX X-1 + X d
(X-1)d(X-1) = -X-1dX
X-1 | 
| 2.2 | dX/dxij =
eiejT 
where ei is the ith column of
I. Note that
eiejT
is a matrix containing a 1 in position i, j and zeros
elsewhere. | 
| 2.3 | d/dxij
(X-1) =
-X-1eiejT
X-1 =
-X-1ei(X-Tej)T This follows straightforwardly from 2.1 and
2.2. | 
| 2.4 | d/dX (tr(AXB)) =
ATBT 
d/dxij (tr(AXB)) =
tr(AeiejT
B) = tr(ejT
BAei) =
ejT
BAei = (BA)ji =
(ATBT)ij[2.3] | 
| 2.5 | d/dX
(tr(AX-1B)) =
-X-TATBTX-T 
d/dxij (tr(AX-1B)) =
-tr(AX-1eiejT
X-1B) =
-tr(ejT
X-1BAX-1ei) =
-ejT
X-1BAX-1ei =
-(X-1BAX-1)ji =
-(X-TATBTX-T)ij
[2.3] | 
| 2.6 | [D=DH] d{tr((AXB+C)D(AXB+C)H)}
=
{(2AH(AXB+C)DBH):H
dX:}R 
d{tr((AXB+C)D(AXB+C)H)}
=
tr{(A(dX)B)D(AXB+C)H}
+
tr{(AXB+C)D(A(dX)B)H}=
tr{A(dX)BD(AXB+C)H}
+
tr{((AXB+C)DBH(dX)HAH}=
tr{BD(AXB+C)HA(dX)}
+
tr{AH(AXB+C)DBH(dX)H} 
 [1.17]=
(BD(AXB+C)HA)H:HdX:
+
((AH(AXB+C)DBH):HdX:)C
  [1.18]=
((AH(AXB+C)DBH):HdX:)
+ 
((AH(AXB+C)DBH):HdX:)C=
{(2AH(AXB+C)DBH):H
dX:}R | 
| 2.7 | [D=DH] argminX{tr((AXB+C)D(AXB+C)H}
=
-(AHA)-1AHCDBH(BDBH)-1 
d{tr((AXB+C)D(AXB+C)H)}
= 0⇒
{(2AH(AXB+C)DBH):H
dX:}R = 0  [2.6]⇒
(2AH(AXB+C)DBH):
= 0  since it must be true for any dX⇒
AH(AXB+C)DBH
= 0  removing the vectorization⇒ AHAXBDBH
=  -AHCDBH⇒  X =
-(AHA)-1AHCDBH(BDBH)-1 | 
| 2.8 | [D=DH] argminX{tr((AXB+C)D(AXB+C)H
| EXF-G=0} =
(AHA)-1(EH{E(AHA)-1EH}-1{E(AHA)-1AHCDBH(BDBH)-1F+G}{FH(BDBH)-1F}-1FH
-
AHCDBH)(BDBH)-1 
∂{tr((AXB+C)D(AXB+C)H)+tr(KH(EXF-G))+tr((EXF-G)HK)}/∂X
= 0⇒
AH(AXB+C)DBH)
+EHKFH
= 0    [2.4]⇒ AHAXBDBH =
-(AHCDBH
+EHKFH
)⇒ X =
-(AHA)-1(AHCDBH
+EHKFH
)(BDBH)-1Substituting this into the constraint, EXF-G=0, gives⇒
-E(AHA)-1(AHCDBH
+EHKFH
)(BDBH)-1F =
G⇒
 E(AHA)-1EHKFH
(BDBH)-1F = -(G
+
E(AHA)-1AHCDBH(BDBH)-1F)⇒ K =
-(E(AHA)-1EH)-1(
E(AHA)-1AHCDBH(BDBH)-1F+G)(FH
(BDBH)-1F)-1Finally, substituting this back into the previous expression for
X givesX = (AHA)-1(
EH{E(AHA)-1EH}-1{
E(AHA)-1AHCDBH(BDBH)-1F+G}{FH
(BDBH)-1F}-1FH
-
AHCDBH)(BDBH)-1 | 
| 2.9 | d/dX
(aTX-1b) =
-X-TabTX-T 
d/dxij
(aTX-1b) =
-aTX-1eiejTX-1b
=
-aTX-1ei
*
ejTX-1b
=
-eiTX-Ta
*
bTX-Tej
= 
-eiTX-TabTX-TejHence d/dX
(aTX-1b) =
-X-TabTX-T
[2.3]
 | 
| 2.10 | d(det(X)) =
ADJ(X)T:T
dX:  = [X:
nonsingular] det(X)
(X-T):T
dX: 
det(X) I = X ADJ(X) so it follows that det(X)
=  sumj(xij
ADJ(X)ji) for each id/dxij det(X) =
ADJ(X)ji =
(ADJ(X)T)ij since
ADJ(X)ji does not depend on
xijd(det(X)) =
ADJ(X)T:T
dX: | 
| 2.11 | d(det(ATXB)) =
d(det(BTXTA)) =
(A ADJ(ATXB)TBT):T 
dX: =  [A,B:
nonsingular] det(ATXB) ×
(X-T):T
dX: 
d(det(ATXB)) =
ADJ(ATXB)T:T
d(ATXB): [2.10]
=
ADJ(ATXB)T:T
(B ⊗ A)T d(X): =
((B ⊗ A)
ADJ(ATXB)T:)T
d(X): = (A
ADJ(ATXB)TBT):T 
dX:(A
ADJ(ATXB)TBT):T 
dX: = [A,B,X:
nonsingular] det(ATXB) ×
(A
(ATXB)-TBT):T 
dX: =  det(ATXB)
× (A
A-1X-TB-TBT):T 
dX: =  det(ATXB)
× (X-T):T 
dX: | 
| 2.12 | d(ln(det(ATXB))) =
[A,B: nonsingular] (X-T):T
dX: 
d(ln(det(ATXB))) =
det(ATXB)-1 ×
d(det(ATXB)) = [A,B,X:
nonsingular] det(ATXB)-1
× det(ATXB) ×
(X-T):T 
dX: [2.11] =
(X-T):T
dX: | 
| 2.13 | d(det(X)k) = k ×
det(Xk) ×
(X-T):T
dX: 
d(det(X)k) =  k ×
det(X)k-1 × d(det(X))
=  k × det(X)k-1
× det(X) ×
(X-T):T dX:
[2.10] | 
| 2.14 | d(det(XTCX)) =
[C=CT]
2det(XTCX)×(CX(XTCX)-1):T
dX: 
d(det(XTCX)) =
det(XTCX)×(XTCX)-T:T
(d(XTCX)): [2.10] =
det(XTCX)×(XTCX)-T:T
( (I ⊗ XTC) dX:
+ (XTCT
⊗ I) dXT: ) = 
det(XTCX)×(
(CTX
(XTCX)-T):T 
dX: + ((XTCX)-T
XTCT):T
dXT: )  = 
det(XTCX)×(
(CTX
(XTCTX)-1) 
 + (CX(XTCX)-1)
):T  dX:  =  [C=CT] 
2det(XTCX) ×
(CX(XTCX)-1):T
dX: | 
| 2.15 | d(det(XHCX))  = 
det(XHCX)×
((CTXC
(XTCTXC)-1):TdX: 
 +
(CX(XHCX)-1):T
dXC:) 
d(det(XHCX)) =
det(XHCX)×(XHCX)-T:T
(d(XHCX)): [2.10] =
det(XHCX)×(XHCX)-T:T
( (I ⊗ XHC) dX:
+ (XTCT
⊗ I) dXH: ) = 
det(XHCX)×(
(CTXC
(XHCX)-T):T 
dX: + ((XHCX)-T
XTCT):T
dXH: )  = 
det(XHCX)×
(CTXC
(XTCTXC)-1):TdX: 
 +
(CX(XHCX)-1):T
dXC:) | 
| 2.16 | d(ln(det(XHCX))) =
(CTXC
(XTCTXC)-1):TdX: 
 +
(CX(XHCX)-1):T
dXC: 
d(ln(det(XHCX))) =
(det(XHCX))-1×
d(det(XHCX)) =
(CTXC
(XTCTXC)-1):TdX: 
 +
(CX(XHCX)-1):T
dXC: [2.15] | 
| 2.17 | If C=CH,
then Hx
(Ax+b)HC(Ax+b) =
(AHCA)T 
(d/dx (d/dx
(Ax+b)HC(Ax+b))H)T
= (d/dx
((Ax+b)HCA)H)T
= (d/dx
(AHC(Ax+b)))T =
(AHCA)T |