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Consider the "Complex" Quadratic minimization problem \begin{align} \min_{\mathbb{x}\in \mathbb{C}^{N \times 1}}~\mathbf{{x}}^H\mathbf{Q}\mathbf{x}-2~\Re{(\mathbf{x}^H\mathbf{b})}+1 \end{align}

$\Re(.)$ denotes the real part. Here $\mathbf{Q}$ is a $N \times N$ positive definite matrix. $\mathbf{b}$ is a $N \times 1$ vector. I am familiar with the technique of putting this problem in the real domain (where it becomes in $2N \times 1$ dimension) and then using the lagrangian technique to solve the resulting problem. I was looking for some analytic technique which would solve it in the complex domain itself. Applying lagrangian technique for complex vectors is also fine.

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2 Answers

up vote 2 down vote accepted

For problems of this kind it is often convenient to treat the $\mathbf{x}$ and $\mathbf{x}^H$ as independent variables (in the end they are linearly related to real and imaginary part of $x$). So we want to minimize $$E= \mathbf{x}^H Q \mathbf{x} - (\mathbf{x}^H \mathbf{b} + \mathbf{b}^H \mathbf{x}) +1.$$

In this case as $E$ is a real function (as it should otherwise minimization does not make too much sense), the equations $$ \nabla_{\mathbf{x}} E =0 $$ and $$ \nabla_{\mathbf{x}^H} E=0$$ are complex conjugates of each other. So it is enough to solve one of these. We choose $$ \nabla_{\mathbf{x}^H} E = Q \mathbf{x} - \mathbf{b} =0$$ with the solution $$\mathbf{x} = Q^{-1} \mathbf{b}.$$

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Let $Q^{1/2}$ be a Hermitian square root of $Q$. Then by completing square, we get \begin{align*} x^HQx-2~\Re{(x^Hb)}+1 &=x^HQx-x^Hb-b^Hx+1\\ &=\|Q^{1/2}x - Q^{-1/2}b\|^2 + (1 - b^HQ^{-1}b). \end{align*} Hence the minimum occurs at $x = Q^{-1}b$ and the minimum value is $1 - b^HQ^{-1}b$.

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(+1) elegant, but i forgot to include that I was familiar with this solution also. I was looking for something in the lines of "Fabian"'s answer. –  dineshdileep Jan 23 '13 at 8:48
I don't know if you are from a signal processing background, but it is really common place to use this technique you showed in minimum mean square error (MMSE) estimation theory. I feel happy to see that answer yet again. –  dineshdileep Jan 23 '13 at 8:52
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