Let $\mathbf{A}$ and $\mathbf{B}$ be two complex rank-one $N\times N$ positive semi-definite matrices. Let the matrix $\mathbf{C}$ be defined as \begin{align} \mathbf{C}=\left(\mathbf{I}*\frac{1}{\alpha}+\mathbf{B}\right)^{-\frac{1}{2}}\mathbf{A}\left(\mathbf{I}*\frac{1}{\alpha}+\mathbf{B}\right)^{-\frac{1}{2}} \end{align} where $\mathbf{I}$ is the $N\times N$ identity matrix, $\alpha$ is a positive variable. The question is to find the $\alpha$ which maximizes the largest eigenvalue of $\mathbf{C}$.

Background: This comes from a real world wireless problem where the largest eigenvalue represents the best power allocation at the transmitter. $\mathbf{A}$ and $\mathbf{B}$ represent the channel matrices.

  • $\begingroup$ If $A=vv^*$, then $C$ has $N-1$ zero eigenvalues and the last one equal to $v^*(I/\alpha+B)^{-1}v$ --- does that help? $\endgroup$ – Federico Poloni Jan 15 '14 at 4:39
  • $\begingroup$ (EDIT: no, I am not sure anymore that it does). $\endgroup$ – Federico Poloni Jan 15 '14 at 4:40
  • $\begingroup$ In any case, can you please see if by using the above trick and the Sherman-Morrison formula you can manage to reduce it to a scalar problem? $\endgroup$ – Federico Poloni Jan 15 '14 at 5:03

Frederico Poloni's approach does work indeed, and it shows that there is no positive $\alpha$ for which the eigenvalue is maximal.

If one denotes $A=vv^*$ and $B=ww^*$ then the Sherman-Morrison formula applied to $(I/\alpha+B)^{-1}$ leads to the following formula for the eigenvalue $\lambda$: $$ \lambda(\alpha)= \alpha v^*\left(I-\frac{\alpha}{1+\alpha \lVert w\rVert^2}B\right) v $$ For maximizing this expression you look at the zeros of the derivative $\lambda'(\alpha)$. If I did the calculations correct you obtain for $\alpha$ the quadratic equation $$\lVert w\rVert^2 s\alpha^2+ 2s\alpha + \lVert v\rVert^2=0.$$ where $s=\lVert w\rVert^2\lVert v\rVert^2-v^*Bv=\lVert w\rVert^2\lVert v\rVert^2-\lvert \langle v,w\rangle\rvert^2$. The value $s$ is non-negative by the Cauchy-Schwarz-inequality and even positive if $v$ and $w$ are linearly independent. The (complex) solutions of the quadratic equation, however, are $$-s \pm \lvert \langle v,w\rangle\rvert \sqrt{-s}.$$ This means one only gets a real solution, if $s=0$ (in which case $\alpha=0$ which is forbidden), or if $\langle v,w\rangle=0$, in which case the solution is $-s$ which is negative.

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