MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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? – Federico Poloni Jan 15 '14 at 4:39
(EDIT: no, I am not sure anymore that it does). – Federico Poloni Jan 15 '14 at 4:40
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? – 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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