I have a 4x1 vector with complex elements $\mathbf{c}=\left[c_{1}, \dots, c_{N}\right]^{ T }$ and the following $\ell2$-norm function $f(\mathbf{c}) = \|\textbf{A}.\textbf{c}\|_{2}^2 = \sum_{m=1}^{M}\left|\mathbf{a}_{m}^{H}\mathbf{c}\right|^{2}=\sum_{m=1}^{M}\mathbf{c}^{H}\mathbf{a}_{m}\mathbf{a}_{m}^{H}\mathbf{c}$ , where the elements of c are complex and have the form $c_i = r_i\cdot exp(j\Phi_i)$ I am trying to find the derivatives the aforementioned function given $r_i$ and $\Phi_i$

I tried this by decomposing my complex numbers and sums but it takes too long and I cannot shorten my formulas at the end. I think there is an alternative way as this post states but I am unable to find it. Any help or hints are much appreciated. Thank you in advance.


1 Answer 1


Given the real vectors $(r,\phi)$, define the complex vectors $(p,c,b)$ as $$\eqalign{ p &= \exp(j\phi) &\implies dp = j\,p\odot d\phi \cr c &= r\odot p &\implies dc = r\odot dp + p\odot dr \cr b &= A^HAc \cr }$$ where $\odot$ denotes the elementwise/Hadamard product.
Further, a colon will denote the trace/Frobenius product, i.e. $\,\,A:B = {\rm Tr}(A^TB)$
(NB: The exp() function is applied elementwise)

Calculate the differential of the real function $(f)$ in terms of the real variables $(r,\phi)$. $$\eqalign{ f &= (Ac)^* : Ac \cr df &= (Ac)^* : (A \, dc) + (Ac) : (A \, dc)^* \cr &= b^*:dc + b:dc^* \cr &= b^*:(r\odot dp + p\odot dr) + b:(r\odot dp + p\odot dr)^* \cr &= (r\odot b^*):dp + (r^*\odot b):dp^* + (p\odot b^*):dr + (p^*\odot b):dr^* \cr &= (r\odot b^*):(j\,p\odot d\phi) + (r^*\odot b):(j\,p\odot d\phi)^* + (p\odot b^*):dr + (p^*\odot b):dr^* \cr &= (j\,p)\odot(r\odot b^*):d\phi + (j\,p)^*\odot(r^*\odot b):d\phi^* + (p\odot b^*):dr + (p^*\odot b):dr^* \cr &= 2\,{\mathcal Re}(j\,p\odot r\odot b^*):d\phi + 2\,{\mathcal Re}(p\odot b^*):dr \cr }$$ The fact that $(dr^*=dr,\,\,d\phi^*=d\phi)$ and ${\mathcal Re(z)=\tfrac{1}{2}(z+z^*)}$ allows terms to be combined in that last line.

In this form, the gradients can be identified as
$$\eqalign{ \frac{\partial f}{\partial\phi} &= 2\,{\mathcal Re}(j\,p\odot r\odot b^*) \cr &= 2r\odot {\mathcal Im}(p^*\odot b) \cr \frac{\partial f}{\partial r} &= 2\,{\mathcal Re}(p^*\odot b) \cr }$$

  • $\begingroup$ Thank you for your answer but what do you mean exactly by "the real and imaginary components of a variable are treated independently" and the c you are using here is the vector, I assume? or is it an element of the vector? And how is the $\mathbf{c}*$ relevant here? Anyway my variables are $r_i$ & $\Phi_i$ so my derivatives are $\frac{\partial f}{\partial \Phi_i}$ & $\frac{\partial f}{\partial r_i}$ and not $\frac{\partial f}{\partial \mathbf{c}}$ $\endgroup$ Mar 27, 2019 at 15:29
  • $\begingroup$ Sorry. the answer has been re-worked to find the gradients with respect to $(r,\phi)$ instead of $(c,c^*)$. $\endgroup$
    – greg
    Mar 27, 2019 at 16:35
  • $\begingroup$ Thank you so much @greg . This is exactly what I needed. Just one question: Is the $A^{*}$ the Conjugate transpose or just the Conjugate? $\endgroup$ Apr 1, 2019 at 13:45
  • $\begingroup$ In my post, the symbols $(A^T,A^*,A^H)$ denote the $($transpose, complex conjugate, hermitian conjugate$)$ of $A$, respectively. $\endgroup$
    – greg
    Apr 3, 2019 at 21:58
  • $\begingroup$ I figured so, but I wanted to be sure. Again thank you so much, you are a hero sir :) $\endgroup$ Apr 3, 2019 at 22:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.