# A Quadratic Problem (which looks very simple)

This arises as a part of my work. \begin{align} \min_{x^{H}x=1}~&x^{H}A_1x \\ subject~to~&x^{H}A_2x=0 \end{align} $A_1$ and $A_2$ are $N\times N$ hermitian matrices and $x$ is a unit norm complex vector to be found. Any recommendations on a iterative algorithm to solve this is also fine.

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Write $x=u+iv$ and $A_j=H_j+iK_j$ ($H_j$ is real symmetric and $K_j$ is skew-symmetric), then your minimisation problem can be rewritten as a (perhaps nonconvex) QCQP: \begin{align} \textrm{minimize} & (u^T,v^T)\begin{pmatrix}H_1&-K_1\\ K_1&H_1\end{pmatrix}\begin{pmatrix}u\\ v\end{pmatrix}\\ \textrm{subject to} & (u^T,v^T)\begin{pmatrix}H_2&-K_2\\ K_2&H_2\end{pmatrix}\begin{pmatrix}u\\ v\end{pmatrix}=0,\\ &\|(u^T,v^T)\|^2=1. \end{align} As I know very little optimisation theory, I am not sure how easy it is to solve the above problem.

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Thanks for the answer. But unfortunately this is a very well known technique of converting a complex quadratic problem into a real one. In fact, every complex quadratic problem can be converted in this manner. But it doesn't help more than that :(:( –  dineshdileep Nov 23 '12 at 17:00
@dineshdileep Nevermind, but what I really mean is this: since your problem is equivalent to a (perhaps nonconvex) QCQP, and nonconvex QCQP in general, if I understand correctly, are difficult problems, I think your seemingly simple problem is actually nothing simple at all. –  user1551 Nov 23 '12 at 17:12

After posting the same question in MO, I got an answer, and I am providing the link here.