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I have a Hermitian matrix $H=H^\dagger$, and some complex matrix $B$ (not Hermitian).
I want to do the following: $$minimize\ \ \ \ \xi^\dagger H\ \xi$$ $$s.t.\ \ \ \ \ \xi^\dagger B\xi=0,\ \xi^\dagger\xi=1$$ In order to solve I want to use Lagrange multipliers, and vary with respect to $\xi^\dagger$: $$(H-\mu B)\xi=\lambda\xi$$ Where $\mu,\ \lambda$ are Lagrange multipliers.
The problem is I am not sure about if I can do that because $\xi^\dagger B\xi$ is not a real function outside of the restricted plane.

So, The question is: Can I use Lagrange multipliers like this?
If not, would the answer change if $B$ were a real matrix?

Any help will be greatly appreciated.

btw - I know I can just replace the condition to $\xi^\dagger\ (B+B^\dagger)\xi=0$, but the note in the top answer to this question: Lagrangian multipliers in complex optimization made me think it is not neccesary. If not I wonder whats working there that is not working here

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Considering $\xi = \xi_r+j\xi_i$ and $B = B_r+jB_i$

we have

$$ (\xi_r+j\xi_i)^{\dagger}(B_r+jB_i)(\xi_r+j\xi_i)= \xi_r^{\top} B_r \xi_r+\xi_i^{\top} B_r\xi_i + j(\xi_r^{\top} B_i\xi_r+\xi_r^{\top} B_i\xi_r) = 0 $$

which is equivalent to

$$ \xi_r^{\top} B_r \xi_r+\xi_i^{\top} B_r\xi_i = 0\\ \xi_r^{\top} B_i\xi_r+\xi_r^{\top} B_i\xi_r = 0 $$

The stationary conditions are

$$ \xi^{\dagger}H+\lambda_1 \xi^{\dagger}B+\lambda_2\xi^{\dagger} = 0 $$

or

$$ \xi^{\dagger}(H+\lambda_1 B+\lambda_2 I)=0 $$

Those conditions can be established over $\xi_r,\xi_i$ instead. The multipliers $\lambda_k = \lambda_{kr}+j\lambda_{ki}$

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  • $\begingroup$ First you got some $i$'s instead of $r's$ in several places. Second I can not see why your second line is correct - you suppose to have mix terms like $\xi_r^\dagger \ B_r \xi_i$ where did they go? last thing I did not understand how is this answer my question. you care to clarify? thanks for trying to help $\endgroup$ – Reguar May 26 '18 at 17:46
  • $\begingroup$ First, thanks for checking the indexes. Now they are fixed. Second a complex restriction can be developed into two real restrictions. This is the best way to handle those problems when $B$ is not Hermitian. $\endgroup$ – Cesareo May 26 '18 at 19:26

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