Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $A$ and $B$ be two given hermitian positive semi-definite matrices, then what is the solution for

$$\max_{x\neq 0}\frac{x^HAx}{x^HBx+1}.$$

I am looking for closed form solutions. If the denominator didn't have that $1$, this is standard generalized rayleigh quotient and would be unbounded.

I know how to solve it numerically. The trick is to re-write it as

$$\max_{x,t}~t\\ \text{s.t.}~~x^H(A-tB)x \ge t$$

Then find the largest $t$ such that there exists a $x$ which satisfies $x^H(A-tB)x\gt0$. A Bi-section search on $t$ will do the job.

I have already asked this question before in math stack exchange. Since I didn't get any answers, I thought I will post it up here.

share|improve this question

1 Answer 1

up vote 7 down vote accepted

Replacing any nonzero $x$ by $tx$ with real $t$, $ \dfrac{(tx)^H A (tx)}{(tx)^H B (tx) + 1}$ increases to $\dfrac{x^H A x}{x^H B x}$ as $t \to \infty $. Thus the supremum is your generalized Rayleigh quotient. If $B$ is positive definite, the supremum (not maximum, as it is not attained) is the largest eigenvalue of $B^{-1/2} A B^{-1/2}$. If $\text{ker}(B)$ is not contained in $\text{ker}(A)$, it is unbounded.

share|improve this answer

Your Answer

 
discard

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.