$$ S \left( 1-Q\left(\frac{\log \det \left(\mathbf{I}+\left( \mathbf{I} + \mathbf{U} \mathbf{X} \mathbf{X}^H \mathbf {U}^{H} \right)^{-1} \left(\mathbf {U} \mathbf {Y} \mathbf {Y}^H \mathbf {U}^{H}\right)\right)-c}{\sqrt{\left(1-\left( \det \left(\mathbf{I}+\left( \mathbf{I} + \mathbf{U} \mathbf{X} \mathbf{X}^H \mathbf {U}^{H} \right)^{-1} \left(\mathbf {U} \mathbf {Y} \mathbf {Y}^H \mathbf {U}^{H}\right)\right) \right)^{-2}\right) (\log_2 e)^2 /n}} \right) \right) \geq \zeta $$

I am trying to calculate gradient of the function as mentioned above with respect to the matrix $\mathbf{X}$.

Here the $S$, $c$, $n$, $\zeta$ and $e$ are constants. $Q$ refers to the Q-function. The matrices $\mathbf{U} \in \mathbb{C}^{R \times T}$, $\mathbf{X} \in \mathbb{C}^{T \times K}$ and $\mathbf{Y} \in \mathbb{C}^{T \times K}$, $\mathbf{I}$ refers to the identity matrix. I referred to matrix cookbook for any leads, firstly I tried applying chain rule for the same but the expression becomes too complex to handle. I tried applying Woodbury identity to the term $(\mathbf{I} + \mathbf{U} \mathbf{X} \mathbf{X}^H \mathbf {U}^{H} )^{-1}$ in hopes of simplifying the expression but it leads to a much more complicated expression. I have been stuck with this for a while any help would be much appreciated.

  • $\begingroup$ Which function? All I see is an inequality. $\endgroup$ Feb 18 at 19:29

1 Answer 1


Given the expression $$ \small{ S \left[ 1-Q\left(\frac{\log \det \left(\mathbf{I}+\left( \mathbf{I} + \mathbf{U} \mathbf{X} \mathbf{X}^H \mathbf {U}^{H} \right)^{-1} \left(\mathbf {U} \mathbf {Y} \mathbf {Y}^H \mathbf {U}^{H}\right)\right)-c}{\sqrt{\left(1-\left( \det \left(\mathbf{I}+\left( \mathbf{I} + \mathbf{U} \mathbf{X} \mathbf{X}^H \mathbf {U}^{H} \right)^{-1} \left(\mathbf {U} \mathbf {Y} \mathbf {Y}^H \mathbf {U}^{H}\right)\right) \right)^{-2}\right) (\log_2 e)^2 /n}} \right) \right] } $$ First, build the common matrix expression $$\eqalign{ \def\LR#1{\left(#1\right)} \def\qiq{\quad\implies\quad} \def\A{A^{-1}} A &= {I+UXX^HU^H} &\qiq dA=U\,dX\,X^HU^H \\ B &= {UYY^HU^H} &\qiq dB=0\quad\{{\rm constant}\} \\ Z &= {I+\A B} &\qiq dZ=-\A dA\,\A \\ }$$ Then build the common scalar expression $$\eqalign{ \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\dgrad#1#2{\frac{d #1}{d #2}} \def\Z{Z^{-1}} \def\ZZ{Z^{-T}} \def\AA{A^{-T}} \def\a{\alpha} \def\b{\beta} \def\g{\gamma} \def\l{\lambda} \def\e{e^{-2\a}} \def\ee{\LR{1-\e}} \a &= \log\det(Z) &\qiq d\a= \ZZ:dZ \\ \b &= {\sqrt n}\log(2)&\qiq d\b=0\qquad\{\b\:{\rm is\:constant}\} \\ \g &= \frac {\b\LR{\a-c}}{\ee^{1/2}} &\qiq \dgrad{\g}{\a} = \frac{\b\ee+\a\b\LR{c-\a}\e} {\ee^{3/2}} \\ }$$ Using these variables, the objective function simplifies to $$\l = S\Big[1-Q\LR{\g}\Big]$$ I have no idea what your Q-function is, but I'll assume that you know how to calculate its derivative $$P(\g) = \dgrad{Q(\g)}{\g}\qiq dQ = P\,d\g $$ Now calculate the differential and gradient of $\l$ $$\eqalign{ d\l &= -S\,dQ \\ &= -SP\:d\g \\ &= -SP\LR{\dgrad{\g}{\a}\:d\a} \\ &= -\dgrad{\g}{\a}\:SP\:{\ZZ:dZ} \\ }$$ Let's collect everything on the LHS of $dZ$ into a new matrix variable before continuing $$\eqalign{ d\l &= -M:dZ \\ &= M:\LR{\A\,dA\,\A} \\ &= {\AA M\AA}:dA \\ &= \LR{\AA M\AA}:\LR{U\,dX\,X^HU^H} \\ &= \LR{U^T\AA M\AA U^*X^*}:dX \\ \grad{\l}{X} &= {U^T\AA M\AA U^*X^*} \;=\; G \\ }$$ So $G$ is the gradient wrt $X.\:$ The gradients wrt the conjugate variables are $$\eqalign{ G^* = \grad{\l}{X^*} \qquad G^H = \grad{\l}{X^H} \\ \\ }$$

The above derivation utilizes this well-known gradient and calculates derivatives in the Wirtinger sense.

It also uses a colon to denote the Frobenius product $$\eqalign{ A:B &= \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} \;=\; {\rm Tr}\!\LR{A^TB} \\ A:A &= \|A\|^2_F \qquad \{{\rm Frobenius\:norm}\} \\ }$$ The properties of the underlying trace function allow the terms in such a product to be rearranged in many fruitful ways, e.g. $$\eqalign{ A:B &= B:A \\ A:B &= A^T:B^T \\ C:\LR{AB} &= \LR{CB^T}:A &= \LR{A^TC}:B \\ }$$


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