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Differentiation by a complex vector

differentiate by complex vector complex vector complex operator

Complex numbers frequently arise in physical problems, particularly those with Fourier series. Let us extend the multivariable least-squares theory to the use of complex-valued unknowns $\bold x$.First recall how complex numbers were handled with single-variable least squares; i.e., as in the discussion leading up to equation (5). Use a prime, such as $\bold x'$, to denote the complex conjugate of the transposed vector $\bold x$.Now write the positive quadratic form as  
 \begin{displaymath}
Q(\bold x', \bold x) \eq
(\bold F\bold x - \bold d)'
(\bold ...
 ...)
\eq
(\bold x' \bold F' - \bold d')
(\bold F\bold x - \bold d)\end{displaymath} (34)
After equation (4), we minimized a quadratic form $Q(\bar X,X)$by setting to zero both $\partial Q/\partial \bar X$ and $\partial Q/\partial X$.We noted that only one of $\partial Q/\partial \bar X$ and $\partial Q/\partial X$is necessarily zero because they are conjugates of each other. Now take the derivative of Q with respect to the (possibly complex, row) vector $\bold x'$.Notice that $\partial Q/\partial \bold x'$ is the complex conjugate transpose of $\partial Q/\partial \bold x$.Thus, setting one to zero sets the other also to zero. Setting $\partial Q/\partial \bold x' =\bold 0$ gives the normal equations:  
 \begin{displaymath}
\bold 0 \eq {\partial Q \over \partial \bold x'} \eq
\bold F' (\bold F\bold x - \bold d)\end{displaymath} (35)
The result is merely the complex form of our earlier result (32). Therefore, differentiating by a complex vector is an abstract concept, but it gives the same set of equations as differentiating by each scalar component, and it saves much clutter.


next up previous print clean
Next: From the frequency domain Up: MULTIVARIATE LEAST SQUARES Previous: Normal equations
Stanford Exploration Project
4/27/2004