I have the following function that I would like to optimize over the value A
$$f(A)=\sum\_k \frac{\mathbf{y}\_k^H\left[\begin{array}{cc} 1&0\\ 0& A \end{array} \right]\mathbf{x}\_k\mathbf{x}\_k^H\left[\begin{array}{cc} 1&0\\ 0& A \end{array} \right]\mathbf{y}\_k}{\mathbf{x}\_k^H\left[\begin{array}{cc} 1&0\\ 0& A^2 \end{array} \right]\mathbf{x}\_k}$$
where both $\mathbf{y}\_k$ and $\mathbf{x}\_k$ are arbitrary $2\times 1$ complex vectors with no particular structure. The sum has "many" terms, say 10 - 50 or so. 


I need top optimize this over the value $A$. What I have observed is that there is usually one max and one min point. If not, the max point occurs at A=0.
This insight was found by testing for random vectors about 100 times.


I don't have any well defined question, but would need help with understanding why this happens. Can I always assume the above happens? In that case, some form of line search would suffice.


Thanks..



a sum of ratios of quadratic forms
The property you mention does not always happen.


$f(A)$ always reduces to the form 
$$
f(A) = \sum\_k \frac{\alpha\_k A^2+\beta\_k A + \gamma\_k}{\delta\_k A^2 + \epsilon\_k}
$$ 
with $\{ \alpha\_k,\beta\_k,\gamma\_k,\delta\_k,\epsilon\_k\}$ real-valued degree-4 expressions involving the components of $\mathbf{x}\_k$ and $\mathbf{y}\_k$. Moreover, all but $\beta\_k$ are non-negative (hence are positive for generic $\mathbf{x}\_k$ and $\mathbf{y}\_k$). And by adjusting the components of the $\mathbf{x}\_k$ and $\mathbf{y}\_k$) we can make $\{ \alpha\_k,\beta\_k,\gamma\_k,\delta\_k,\epsilon\_k\}$ take on any values we wish, subject to the non-negativity constraints and a limitation on $|\beta\_k|$,
whose magnitude is the geometric mean of the magnitudes of $\alpha\_k$ and $\gamma\_k$.


Now choose 
$$
\begin{array}{cc} x\_1^H = \left( 1, 1 \right) & 
 y\_1^H = \left( -1-\sqrt{3}i, 1 \right) \\
 x\_2^H = \left( 1, 4 \right) & 
 y\_1^H = \left( -1-\sqrt{3}i, 1 \right)
\end{array}
$$
Then (unless I have made some algebraic error)
$$
f(A) = \frac{A^2-2 A + 4}{ A^2 + 1} + \frac{4 A^2 -8 A + 4}{ A^2 + 16} 
$$
which has a local maximum at $A\approx -11.88$, a local minimum at $A\approx -1.80$, a second local maximum at $A\approx -0.43$, and a second local minimum at 
$A\approx +1.85$.


This one counterexample proves your assertion false, but the fact remains that for *most* randomly chosen sets of $\mathbf{x}$ and $\mathbf{y}$ vectors (for some suitable definition of random choosing, since there is are no restrictions mentioned on the magnitudes of $\mathbf{x}\_k$ and $\mathbf{y}\_k$ )
there will be only one local maximum and one local minimum. The reason is that the multiple dips come from unequal offsets of the drop-off due to the denominators; indeed if all the denominators were identical (that is, of all the 
$\mathbf{x}\_k$ shared the same absolute values of their upper and lower components), there would not be more than one set of extrema. The rise and fall action of at least one of the individual rational fraction components is generally larger than the "interference" behavior induced by different denominators, so you have to fine tune the $\mathbf{x}\_k$ and $\mathbf{y}\_k$ get multiple dips.


