# Derivative of the l2-norm of a multivariate complex matrix

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.

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 }

• 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 $\frac{\partial f}{\partial \mathbf{c}}$ Mar 27, 2019 at 15:29
• Sorry. the answer has been re-worked to find the gradients with respect to $(r,\phi)$ instead of $(c,c^*)$.
– greg
Mar 27, 2019 at 16:35
• Thank you so much @greg . This is exactly what I needed. Just one question: Is the $A^{*}$ the Conjugate transpose or just the Conjugate? Apr 1, 2019 at 13:45
• In my post, the symbols $(A^T,A^*,A^H)$ denote the $($transpose, complex conjugate, hermitian conjugate$)$ of $A$, respectively.
– greg
Apr 3, 2019 at 21:58
• I figured so, but I wanted to be sure. Again thank you so much, you are a hero sir :) Apr 3, 2019 at 22:00