关于复矩阵W求迹函数trace(CWW^H)导数的技术问询

Hey there! Let's work through this derivative step by step—this is a common problem in matrix calculus for complex variables, so I'll break it down clearly, focusing on conventions used in most engineering and optimization contexts.

First, a quick recap: when dealing with derivatives of functions involving complex matrices, we typically use Wirtinger calculus, which distinguishes between derivatives with respect to the matrix (W) itself and its conjugate (\overline{W}) (or conjugate transpose (W^H)). This simplifies handling the real and imaginary components of complex variables.

Step 1: Rewrite the trace using matrix properties #

We can use the cyclic property of the trace (({\rm trace}(AB) = {\rm trace}(BA)) for any compatible matrices (A,B)) to rewrite our function for easier differentiation:
[
c = {\rm trace}(CWW^H) = {\rm trace}(W^HCW)
]
This form doesn't change the value of (c), but it's often more straightforward to apply differentiation rules to.

Step 2: Compute the differential of (c) #

Using the basic rule for differentiating traces ((d({\rm trace}(AB)) = {\rm trace}(dA \cdot B) + {\rm trace}(A \cdot dB))), we calculate the differential (dc):
[
dc = {\rm trace}(C \cdot dW \cdot W^H) + {\rm trace}(C W \cdot d(W^H))
]
Note that the differential of the conjugate transpose is the conjugate transpose of the differential: (d(W^H) = (dW)^H). Substituting this in, we get:
[
dc = {\rm trace}(C dW W^H) + {\rm trace}(C W (dW)^H)
]

Step 3: Map the differential to Wirtinger derivatives #

Wirtinger derivatives split the differential into terms involving (dW) (for (\frac{\partial c}{\partial W})) and terms involving ((dW)^H) (for (\frac{\partial c}{\partial \overline{W}})).

Derivative with respect to (W) ((\frac{\partial c}{\partial W}))

For the term involving (dW), we use the cyclic trace property again to rearrange:
[
{\rm trace}(C dW W^H) = {\rm trace}(W^H C dW)
]
After expanding and simplifying, we find the ((i,j))-th element of (\frac{\partial c}{\partial W}) is the conjugate of the ((i,j))-th element of (C^H W). In matrix form:
[
\frac{\partial c}{\partial W} = \overline{C^H W} = C^T \overline{W}
]
(Where (C^T) is the transpose of (C), and (\overline{W}) is the element-wise conjugate of (W).)

Derivative with respect to (\overline{W}) ((\frac{\partial c}{\partial \overline{W}}))

For the term involving ((dW)^H), we rearrange it similarly:
[
{\rm trace}(C W (dW)^H) = {\rm trace}((dW)^H C W)
]
The ((i,j))-th element of (\frac{\partial c}{\partial \overline{W}}) is exactly the ((i,j))-th element of (C W), so:
[
\frac{\partial c}{\partial \overline{W}} = C W
]

Key Note for Optimization #

If you're working on an optimization problem where (c) is real-valued (this happens when (C) is Hermitian, i.e., (C^H = C)), the relevant gradient for updating (W) is usually (\frac{\partial c}{\partial \overline{W}}), as it aligns with the direction of steepest descent in complex space.

备注：内容来源于stack exchange，提问作者Nnn A