Matrix differentiation - Signal Processing and Optimization for Transceiver Systems - page 667

It can similarly be shown that

T )

∂ Tr( XAX

T .

= XA + XA

∂ X

By a slight modification of the steps in the preceding example we can show that

T

∂ Tr( X AXB )

∂ X

T T + AXB .

= A XB

Example 20.7 .

Let Y = AX − 1 B . To find ∂ (Tr Y ) /∂ X observe that

= r ∂ ( AX − 1 B )
∂x rs

∂ Tr( AX − 1 B )
∂x rs

= r

∂ ( X − 1 )
∂x rs B

A

− Tr AX − 1 I rs X − 1 B

= (from Eq. (20.5))

Tr I rs X − 1 BAX − 1 (from Eq. (20.13))

= −

X − 1 BAX − 1

= − (from Eq. (20.11)) ,

sr

which shows that

X − 1 BAX − 1 T

∂ Tr( AX − 1 B )
∂ X

− X −T

T T

X −T .

= − = A B (20 . 14)

Example 20.8.

T ) − 1 . To compute ∂ (Tr Y ) /∂ X notice first that

Let Y =( XAX

=Tr ∂ ( XAX

.

T ) − 1 ) T ) − 1

∂x rs

∂ Tr(( XAX

(20 . 15)

∂x rs

Using the product rule we write

= ∂ X
∂x rs

∂ X =

∂ ( XAX

T ) T

∂x rs

T + XA T + XA I T

AX I rs AX

rs .

∂x rs

T ) we therefore obtain

Using Eq. (20.6) with Y =( XAX

T ) − 1
∂x rs

T )

∂ ( XAX ∂ ( XAX

T ) − 1 T ) − 1

= − ( XAX ( XAX

∂x rs

T ) − 1 rs ( XAX T + XA I T ) − 1 .

= − ( XAX I rs AX